WBBSE Solutions For Class 10 Maths Solid Geometry Chapter 5 Similarity

Similar Objects

We see infinite number of objects in our daily life. There exists many similarities among these objects.

Again, sometimes there exists no similarity between them. The size and shape of some objects are likely to be the same. These objects are said similar objects.

Definition of similar objects

If two objects be of same shape though their sizes may or may not be the same, then the objects are called similar objects.

 WBBSE Solutions for Class 10 Maths

For example, all squares (whatever their sizes may be ) are similar, all circles are similar, all equilateral triangles are similar.

The squares ABCD and PQRS are similar, The circles with centres A and B are similar and the equilateral triangles ΔABC and ΔPQR are also similar.

Thus, if the shapes of two or more than two objects, although their sizes may or may not be the same, then they can be said to be similar objects.

Equiangular triangles 

Three angles of two or more than two triangles may or may not be equal.

If they are equal, then we call them equiangular triangles. For example, the three angles of the triangles ABC and PQR are equal, i,e., ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R.

So the triangles are equiangular.

Definition

If the three angles of a triangle are equal to the three angles of another triangle, then the two triangles are called equiangular triangles.

Obviously in other words we can say that if any two amgles of a triangles be equal to any two angles A of another triangle, then the two triangles are called equiangular triangles.

Since in this case, the rest third angle of the two triangles are always equal.

The sizes of equiangular triangles may be different even though the shapes of them are similar.

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So, one of the two equiangular triangles may be called increased or decreased form of the another triangle, i.e, if a triangle is increased or decreased by its size, then it will be always equiangular to the previous triangle.

For example, ΔABC is the increased form, of ΔPQR and conversely ΔPQR is the decreased form of ΔABC.

In both the cases, the triangles are equiangular.

Solid Geometry Chapter 5 Similarity

Similar Triangles

Definition

If two or more than two triangles be equiangular and the ratios of their corresponding sides be equal, then the triangles are called similar triangles.

i,e., if ΔABC and ΔDEF be similar triangles, then ∠A = ∠D, ∠B = ∠E and ∠C = ∠F and \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{CA}}{\mathrm{DF}}\)

Axioms of similarity of triangles

Axiom 1: If the ratios of the three pair of corresponding sides are equal, then the two triangles are similar to each other.

It is called S-S-S similarities of two triangles.

Axiom 2: If two angles of a triangle are equal to two angles of another triangle, then the triangles are similar to each other.

It is called angle-angle (A – A) similarity.

Axiom 3: If one pair of angles of two triangles are equal and their adjacent sides arc proportionate, then the triangles are similar to each other.

It is called S-A-S similarity.

The shape of two similar triangles are same, even though their sizes may or may not be the same.

Congruent triangles

If two or more than two triangles are equal in all respects, then the triangles are called congruent triangles.

For example, both the shapes and sizes of two triangles ΔABC and ΔPQR are equal.

So, they are congruent triangles.

Definition

If the shapes and sizes of two or more than two triangles be the same, then they are called congruent triangles.

For example, the shapes and sizes of ΔABC and ΔPQR are equal.

Hence they are congruent triangles.

Sign of similarity and congruency

The sign “∼” is known as the sign of similarity.

Such as, if the triangles ΔABC and ΔPQR are similar, then we write ΔABC ∼ ΔPQR which means that ΔABC and ΔPQR are similar.

ΔABC ∼ ΔPQR, i.e., ΔABC and ΔPQR are similar to each other.

Because, ∠A = ∠P, ∠B = ∠Q, and ∠C = ∠R, But AB ≠ PQ, BC ≠ QR, and AC ≠ PR, even though \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AC}}{\mathrm{PR}}\).

In ΔABC = ΔPQR i.e., ΔABC and ΔPQR are congruent to each other; since, ∠A = ∠P, ∠B = ∠Q, and ∠C = ∠R and the corresponding sides AB = PQ, BC = QR, and AC = PR.

[Note: The areas of two similar triangles may or may not be equal. But the areas of two bcongruent triangles are always equal.

Solid Geometry Chapter 5 Similarity

Conditions Of Congruency

Condition 1. If three sides of any triangle be equal to the corresponding three sides of another triangle, then the triangles are congruent. It is called the S-S-S condition of congruency.

Condition 2. If any two sides of a triangle and their included angle be equal to two sides of another, triangle, then the two triangles are congruent to each other.

Condition 3. If two angles and one Side of a triangle be equal to the two angles and the corresponding side of another triangle, then the triangles are congruent to each other. It is called the A-A-S condition of congruency.

Condition 4. If the hypotenuse and any other side of a right-angled traiangle be equal to the hypotenuse and the corresponding side of another right-angled triangle, then the triangles are congruent to each other.

It is called the R-H-S condition of congruency.

If two triangles satisfies and obey any one ot the above four conditions, then the triangles are called congruent triangles.

Solid Geometry Chapter 5 Similarity

Thales Theorems

Thales was a famous mathematician and philosopher of ancient Greece.

He used an important theorem regarding equiangular mangles’. The theorem was :“The corresponding sides of two equiangular triangles are proportionate.”

We shall now state and prove Thale’s theorem.

Thales’ Theorem:

Statement: A straight line parallel to any side of any triangle divides other two sides (or the extended two sides) proportionally.

Proof: Let in ΔABC, the straight line DE parallel to BC intersects the sides AB and AC at the points D and E respectively and extended AB and AC in the points D and E respectively.

To prove: AD: BD = AE; CE

Construction: Let us join B, E and C, D.

Proof: ΔBED and ΔCED lie on the same base DE and between same parallels DE and BC.

∴ areas of the triangles of ΔBED and ΔCED are equal, i.e, area of ΔBED = area of ΔCED.

or, \(\frac{1}{\text { area of } \Delta \mathrm{BED}}=\frac{1}{\text { area of } \Delta \mathrm{CED}}\)

or, \(\frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of } \Delta \mathrm{BED}}=\frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of } \Delta \mathrm{CED}}\)

But AD and BD are the bases of ΔADE and ΔBED and lie on the same straight line and their vertices are the same point E.

So, the heights of the triangles are equal.

We know that the ratio of the areas of triangles having same heights is equal to the ratio of their bases.

∴ \(\frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of } \triangle \mathrm{BED}}=\frac{\mathrm{AD}}{\mathrm{BD}}\)…..(1)

Similarly, it can be proved that

\(\frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of } \triangle \mathrm{BED}}=\frac{\mathrm{AE}}{\mathrm{CE}}\)….(2)

So, from (1) and (2) we get, \(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{CE}}\)

or, AD : BD = AE : CE

Hence AD : BD = AE : CE (Proved)

[Note: This theorem is also known as theorem of basic proportionality.]

We shall now prove logically the converse theorem of Thales’ theorem by the method of geometry.

Converse Of Thales’ Theorem

Statement: If a straight line divides any two sides (or their extended sides) in the same ratio, it will be parallel to the third side.

 

Given: Let the straight line DE divides two sides AB and AC of ΔABC (or the extended AB and AC) in equal proportion, i.e.,

\(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{CE}}\)

To prove DE || BC.

Construction: Let us join B, E and C, D.

Proof: The bases AD and BD of ΔADE and ΔBDE lie on the same straight line and their common vertex is E.

∴ the heights of the triangles are equal.

∴ the ratio of the areas of the triangles is equal to the ratio of their bases.

\(\quad \frac{\text { area of } \triangle \mathrm{ADE}}{\text { area of } \triangle \mathrm{BED}}=\frac{\mathrm{AD}}{\mathrm{BD}}\)…..(1)

Similarly, it can be proved that

\(\frac{\text { area of } \triangle \mathrm{ADE}}{\text { area of } \triangle \mathrm{CED}}=\frac{\mathrm{AE}}{\mathrm{CE}}\)….(2)

But given that \(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{CE}}\)

∴ \(\quad \frac{\text { area of } \triangle \mathrm{ADE}}{\text { area of } \triangle \mathrm{BED}}=\frac{\text { area of } \triangle \mathrm{ADE}}{\text { area of } \triangle \mathrm{CED}}\)

or, area of ΔBED = area of ΔCED.

But the two ΔBED and ΔCED of equal areas lie on the same base DE and on the same side of DE.

So, the triangles lie between the same pair of parallel straight lines DE and BC.

Hence DE || BC. (Proved) .

“WBBSE Class 10 similarity solved examples”

Corollary 1. The straight line parallel to side BC of ΔABC intersects the sides AB and AC at points D and E respectively. Prove that 

1. \(\frac{A B}{B D}=\frac{A C}{C E}\)

Proof: From Thale’s theorem we get, \(\frac{\mathrm{A D}}{\mathbf{B D}}=\frac{\mathrm{A E}}{\mathrm{C E}}\)

or, \(\frac{\mathrm{AD}}{\mathrm{BD}}+1=\frac{\mathrm{AE}}{\mathrm{CE}}+1\)

or, \(\frac{\mathrm{AD}+\mathrm{BD}}{\mathrm{BD}}=\frac{\mathrm{AE}+\mathrm{CE}}{\mathrm{CE}}\)

or, \(\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{AC}}{\mathrm{CE}}\)

∴ \(\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{AC}}{\mathrm{CE}}\) (proved)

2. \(\frac{\mathrm{A D}}{\mathrm{A B}}=\frac{\mathrm{A E}}{\mathrm{A C}}\)

Proof: From Thale’s theorem we get, \(\frac{\mathrm{A D}}{\mathbf{B D}}=\frac{\mathrm{A E}}{\mathrm{C E}}\)

or, \(\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{CE}}{\mathrm{AE}}\)

or, \(\frac{\mathrm{BD}}{\mathrm{AD}}+1=\frac{\mathrm{CE}}{\mathrm{AE}}+1\)

or, \(\frac{\mathrm{BD}+\mathrm{AD}}{\mathrm{AD}}=\frac{\mathrm{CE}+\mathrm{AE}}{\mathrm{AE}}\)

or, \(\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AE}}\)

∴ \(\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AE}}\)

or, \(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}\)(Proved)

Corollary 2. The line segment joining the mid-points of any two sides of a triangle is parallel to the third side of the triangle.

Solution: Let D and E be the mid-points of the sides AB and AC of a ΔABC.

 

To prove DE || BC.

Construction: Let us join B, E and C, D.

Proof: ΔADE = \(\frac{1}{2}\) x AD x h…….(1), where h = distance of AD from E = height of ΔADE.

Again, ΔBED = \(\frac{1}{2}\) x BD x h= \(\frac{1}{2}\) x AD x h…….(2), [AD = BD]

[Here the value of h in both cases will be the same.]

∴ from (1) and (2) we get, ΔADE = ΔBED ……..(3)

Similarly, it can be proved that ΔADE = ΔCED ……… (4)

From (3) and (4) we get, ΔBED = ΔCED.

But these are on the same base DE and on the same side of DE.

So, the triangles will lie between the same parallels.

∴ DE || BC. (Proved)

“Similarity theorems for Class 10 Maths”

Corollary 3. Prove that the internal or external bisector of any angle of a triangle divides its opposite side in the ratio of the lengths of the adjacent sides of the angle internally or externally.

Solution: AD is the internal bisector of ∠BAC or external bisector which intersects BC or extended BC of ΔABC at the point D.

To prove: BD: DC = AB: AC

Construction: Let us draw a straight line through C parallel to DA which intersects BA and extended BA at a point E.

Proof: DA || CE [by construction], ∴ ∠DAC = alternate ∠ACE.

Again, DA || CE, ∴ ∠BAD (or, ∠FAD) = similar ∠AEC.

But  ∠BAD (or ∠FAD) = ∠DAC.

∴ ∠ACE = ∠AEC. ∴ AC = AE

Now, in ΔBEC or in ΔBDA

DA || CE; So, \(\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{BA}}{\mathrm{AE}}\) [by Thales’ theorem]

i.e., BD : DC = AB : AE

or, BD : DC = AB : AC [AE = AC]

∴ BD: DC = AB: AC (Proved)

In the following examples how the above theorems are applied to solve the real problems is discussed thoroughly.

Solid Geometry Chapter 5 Similarity Multiple Choice Questions

“Chapter 5 similarity exercises WBBSE solutions”

Example 1. A line parallel to the side BC of ΔABC intersects the sides AB and AC at the points X and Y respectively. If AX = 2.4 cm, AY = 3.2 cm, and YC = 4.8 cm, then the length of AB is 

1. 3.6 cm
2. 6 cm
3. 6.4 cm
4. 7.2 cm

 

 

Solution:

Given

A line parallel to the side BC of ΔABC intersects the sides AB and AC at the points X and Y respectively. If AX = 2.4 cm, AY = 3.2 cm, and YC = 4.8 cm

The side BC of ΔABC is parallel to XY and XY intersects AB and AC at X and Y respectively.

∴ by Thales’ theorem,

\(\frac{A X}{B X}=\frac{A Y}{C Y}\)

or, \(\frac{B X}{A X}=\frac{C Y}{A Y}\)

or, \(\frac{B X+A X}{A X}=\frac{C Y+A Y}{A Y}\)

or, \(\frac{A B}{2 \cdot 4}=\frac{4 \cdot 8+3 \cdot 2}{3 \cdot 2}\)

[AX = 2.4 cm, AY = 3.2 cm and CY = 4.8 cm]

or, \(\frac{\mathrm{AB}}{2 \cdot 4}=\frac{8}{3 \cdot 2}\)

or, AB=\(\frac{8 \times 2 \cdot 4}{3 \cdot 2}=6\)

∴ AB = 6 cm

∴ 2. 6 cm is correct

Example 2. The point D and E are situated on the sides AB and AC of ΔABC in such a way that DE || BC and AD : DB = 3 : 1; If EA = 3.3 cm, then the length of AC is 

1. 1.1 cm
2. 4 cm
3. 4.4 cm
4. 5.5 cm

Solution:

Given

The point D and E are situated on the sides AB and AC of ΔABC in such a way that DE || BC and AD : DB = 3 : 1; If EA = 3.3 cm

In ΔABC, DE || BC,

∴ by Thales’ theorem, we get, \(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{CE}}\)

or, \(\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{CE}}{\mathrm{AE}}\)

or, \(\frac{\mathrm{BD}+\mathrm{AD}}{\mathrm{AD}}=\frac{\mathrm{CE}+\mathrm{AE}}{\mathrm{AE}}\)

or, \(\frac{4}{3}=\frac{\mathrm{AC}}{3 \cdot 3}\)

[AD: DB=3 :1 or, \(\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{1}{3}\)

or, 3AC=4 x 3.3

or, \(\frac{\mathrm{AD}+\mathrm{BD}}{\mathrm{AD}}=\frac{3+1}{3}=\frac{4}{3}\)]

or, AC =\(\frac{4 \times 3 \cdot 3}{3}\)

or, AC = 4. 4

∴ AC = 4.4 cm

∴ 3. 4.4 cm is correct

“Class 10 Maths properties of similar triangles”

Example 3. If DE || BC, then the value of x is

1. 4
2. 1
3. 3
4. 2

Solution: In ΔABC, DE || BC

∴ by Thales’ theorem, we get, \(\frac{\mathrm{AD}+\mathrm{BD}}\) = \(\frac{\mathrm{AE}+\mathrm{CE}}\)

or, \(\frac{x+3}{3 x+19}\) = \(\frac{x}{3 x+4}\)

or, 3x² + + 4x + 9x+ 12 = 3x² + 19x

or, 3x² + 13x – 3x² -19x + 12 = 0

or, -6x + 12 = 0 or, 6x = 12 or, x = \(\frac{12}{6}\)

Hence the required value of x  is 2

∴ 4. 2 is correct

Example 4. In the trapezium ABCD, AB || DC and the two points P and Q are situated on the sides AD and BC in such a way that PQ || DC; if PD = 18 cm, BQ = 35 cm, QC = 15 cm, then the length of AD is 

1. 60 cm
2. 30 cm
3. 12 cm
4. 15 cm

 

Solution:

Given

In the trapezium ABCD, AB || DC and the two points P and Q are situated on the sides AD and BC in such a way that PQ || DC; if PD = 18 cm, BQ = 35 cm, QC = 15 cm,

In trapezium ABCD, AB || DC.

Construction: The sides DA and CB of the trapezium are extended.

Let extended DA and extended CB intersect at the point x.

Now, in ΔXPQ, AB || PQ [PQ || DC and AB || DC

∴ by Thales Theorem, \(\frac{X A}{A P}=\frac{X B}{B Q}\)

or, \(\frac{X A}{X B}=\frac{A P}{B Q}\)……(1)

Again, in ΔXCD, AB || DC

∴ by Thales’ theorem, \(\frac{X A}{A D}=\frac{X B}{B C}\)

or, \(\frac{X A}{X B}=\frac{A D}{B C}\) ……(2)

From (1) and (2) we get, \(\frac{A P}{B Q}=\frac{A D}{B C}\)

or, \(\frac{A D-P D}{B Q}=\frac{A D}{B Q+C Q} \text { or. } \frac{A D-18}{35}=\frac{A D}{35+15}\)

[PD = 18 cm, BQ = 35 cm and CQ = 15 cm]

or, \(\frac{A D-18}{35}=\frac{A D}{50}\)

or, 50 AD – 900 = 35 AD

or, 15 AD = 900 or, AD = \(\frac{900}{15}\) = 60

∴ AD = 60

∴ 1. 60 cm  is correct

Example 5. If DP = 5 cm, DE = 15 cm, DQ = 6 cm and QF = 18 cm, then

1. PQ = EF
2. PQ || EF
3. PQ ≠ EF
4. PQ ∦ EF.

Solution:

Given

If DP = 5 cm, DE = 15 cm, DQ = 6 cm and QF = 18 cm

We know, 15 cm = \(\frac{5 \mathrm{~cm}}{15 \mathrm{~cm}}\) = \(\frac{1}{3}\)

and \(\frac{6 \mathrm{~cm}}{18 \mathrm{~cm}}\) = \(\frac{1}{3}\)

∴ \(\frac{5 \mathrm{~cm}}{15 \mathrm{~cm}}=\frac{6 \mathrm{~cm}}{18 \mathrm{~cm}}\)

or, \(\frac{\mathrm{DP}}{\mathrm{DE}}=\frac{\mathrm{DQ}}{\mathrm{QF}}\),

i.e.. in ΔDEF, \(\frac{\mathrm{DP}}{\mathrm{DE}}=\frac{\mathrm{DQ}}{\mathrm{QF}}\)

⇒ \(\frac{\mathrm{DP}}{\mathrm{PE}} \neq \frac{\mathrm{DQ}}{\mathrm{QF}}\)

∴ by the converse of Thales’ theorem,

∴ PQ ∦EF

∴ 4. PQ ∦EF is correct.

Solid Geometry Chapter 5 Similarity True Or False

Example 1. Two similar triangles are always congruent.

Solution: False, since the angles of two similar triangles are equal, even though their sides are not equal, but proportionate.

Example 2. In the adjoining If  DE || BC, then \(\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{AC}}{\mathrm{CE}}\)

Solution: True, since DE || BC

∴ by Thales theorem \(\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{CE}}\)

or, \(\frac{A D+B D}{B D}=\frac{A E+C E}{C E}\) or, \(\frac{A B}{B D}=\frac{A C}{C E}\)

Solid Geometry Chapter 5 Similarity Fill In The Blanks

Example 1. The line segment parallel to any side of a triangle divides other two sides or the extended two sides ______

Solution: Proportional

Example 2. If the bases of two triangles are situated on a same line and other vertex of the two triangles are common, then the ratio of the areas of two triangles are ______ to the ratio of their baes.

Solution: Equal

Example 3. The straight line parallel to the parallel sides of a trapezium divides ______ others two sides.

Solution: Proportional

Solid Geometry Chapter 5 Similarity Short Answer Type Questions

“Understanding similarity in geometry for Class 10”

Example 1. If in ΔABC, \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\) and if ∠ADE = ∠ACB, then write what type of triangle according to side ΔABC is? 

Solution:

Given:

If in ΔABC, \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\) and if ∠ADE = ∠ACB

In ΔABC, \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)

∴ by the converse of Thales’ theorem DE || DC.

∴ ∠ADE = similar ∠ABC

or, ∠ACB = ∠ABC [Given that ∠ADE = ∠ACB]

∴ AB = AC, ∴ ΔABC is an isosceles triangle.

Example 2. In DE || and if AD : BD = 3: 5, then find  (area of ΔADE) : (area of ΔCDE).

Solution:

Given:

In DE || and if AD : BD = 3: 5

In ΔABC, DE || BC,

∴ by Thales’ theorem,

\(\frac{A D}{B D}=\frac{A E}{C E}\) …….(1)

Also given that AD : BD = 3 : 5 or, \(\frac{A D}{B D}=\frac{A E}{C E}\) = \(\frac{3}{5}\)

∴ \(\frac{\mathrm{AE}}{\mathrm{CE}}=\frac{3}{5}\)

Now, AE and CE lie on the same straight line and D is the common vertex of ΔADE and ΔCDE,

∴ the heights of both triangles are the same. Let the equal heights be h.

∴ \(\frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of } \Delta \mathrm{CDE}}=\frac{\frac{1}{2} \times \mathrm{AE} \times h}{\frac{1}{2} \times \mathrm{CE} \times h}=\frac{\mathrm{AE}}{\mathrm{CE}}=\frac{3}{5}\) [from{2}]

Hence the required ratio is 3: 5.

Example 3. The LM || AB and if AL = (x – 3) unit, AC = 2x unit, BM = (x-2) unit and BC = (2x + 3) unit, then find the value of x.

Solution:

Given:

The LM || AB and if AL = (x – 3) unit, AC = 2x unit, BM = (x-2) unit and BC = (2x + 3) unit

In ΔABC, LM || AB.

∴ by Thales theorem, \(\frac{C L}{A L}=\frac{C M}{B M}\)

or, \(\frac{\mathrm{CL}+\mathrm{AL}}{\mathrm{AL}}=\frac{\mathrm{CM}+\mathrm{BM}}{\mathrm{BM}}\)

or, \(\frac{\mathrm{AC}}{\mathrm{AL}}=\frac{\mathrm{BC}}{\mathrm{BM}}\)

or, 2x² – 4x = 2x² – 6x + 3x – 9 or, – 4x = – 3x – 9

or, – x = – 9 or, x = 9

Hence the required value of x = 9

Example 4. If in ΔABC, DE || PQ || BC and AD = 3 cm, DP = x cm, PB = 4 cm, AE = 4 cm, EQ = 5 cm, QC = y cm, then determine the value of x and y

Solution:

Given:

If in ΔABC, DE || PQ || BC and AD = 3 cm, DP = x cm, PB = 4 cm, AE = 4 cm, EQ = 5 cm, QC = y cm

In ΔAPQ, DE || PQ,

∴ by Thales’ theorem, \(\frac{\mathrm{AD}}{\mathrm{DP}}=\frac{\mathrm{AE}}{\mathrm{EQ}}\)

or, \(\frac{3 \mathrm{~cm}}{x \mathrm{~cm}}=\frac{4 \mathrm{~cm}}{5 \mathrm{~cm}}\)

[AD =3 cm, DP = x cm, AE = 4 cm and EQ = 5 cm]

or, \(\frac{3}{x}=\frac{4}{5}\) or, 4 x=15 or, \(x=\frac{15}{4}\)

Again, in ΔABC, PQ || BC, ∴ by Thales’ theorem

\(\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AQ}}{\mathrm{QC}}\)

or, \(\frac{\mathrm{AD}+\mathrm{DP}}{\mathrm{PB}}=\frac{\mathrm{AE}+\mathrm{EQ}}{\mathrm{QC}}\)

or, \(\frac{3 \mathrm{~cm}+x \mathrm{~cm}}{4 \mathrm{~cm}}=\frac{4 \mathrm{~cm}+5 \mathrm{~cm}}{y \mathrm{~cm}}\)

or, \(\frac{3+x}{4}=\frac{9}{y}\)

or, \(\frac{3+\frac{15}{4}}{4}=\frac{9}{y}\)

[because x = \(\frac{15}{4}\)]

or, \(\frac{27}{16}=\frac{9}{y}\) or, 3 y=16

or, y = \(\frac{16}{3}\)

Hence the required value of x = \(\frac{15}{4}\) and y= \(\frac{16}{3}\)

Example 5. In the adjoining, if DE || BC, BE || XC and \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{2}{1}\) then value of \(\frac{\mathrm{AD}}{\mathrm{DB}}\) 

Solution:

Given:

In the adjoining, if DE || BC, BE || XC and \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{2}{1}\)

In ΔABC, DE || BC,

∴ by Thales’ theorem,

\(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)

or, \(\frac{\mathrm{AE}}{\mathrm{EC}}=\frac{2}{1}\) [because \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{2}{1}\)]

or, AE = 2EC …..(1)

Again, in ΔAXC, BE || XC,

∴ by Thales’ Theorem,

\(\frac{\mathrm{AB}}{\mathrm{BX}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)

or, \(\frac{\mathrm{AB}}{\mathrm{XB}}=\frac{2 \mathrm{EC}}{\mathrm{EC}}\)[from (1)]

or, \(\frac{\mathrm{AB}}{\mathrm{XB}}=2\)

or, \(\frac{\mathrm{AB}}{\mathrm{XB}}=\frac{2}{1}\)

or, \(\frac{\mathrm{AB}+\mathrm{XB}}{\mathrm{XB}}=\frac{2+1}{1}\)

or, \(\frac{\mathrm{AX}}{\mathrm{XB}}=\frac{3}{1}\),

∴ the value of \(\frac{\mathrm{AX}}{\mathrm{XB}}\) is \(\frac{3}{1}\).

Example 6. Write the correct answer 

1. All squares are (congruent/similar)

Solution: Similar

2. All squares are (congruent/similar)

Solution: Similar

3. All (equilateral/isosceles) triangles are always similar.

Solution: Equilateral

4. Two quadrilaterals will be similar if their similar angles are (equal/proportional) and similar sides are (unequal/proportional)

Solution: Equal, Proportional

Example 7. Write whether the following statements are true or false 

1. Any two congruent are similar.

Solution: True

2. Any two similar are always congruent.

Solution: False

3. The corresponding angles of any two polygonal are equal.

Solution: True

4. The corresponding sides of any two polygonal are proportional.

Solution: True

5. The square and rhombus are always similar.

Solution: False

Example 8. Write an example of a pair of similar images.

Solution: A pair of similar figures is two circles of unequal diameter as shown.

 

Example 9. Construct a pair of dissimilar images.

Solution: The Images are not similar.

 

Solid Geometry Chapter 5 Similarity Long Answer Type Questions

Example 1. A line parallel to the side BC of AABC intersects the sides AB and AC at points P and Q respectively.

1. If PB = AQ, AP = 9 units, QC = 4 units, then calculate the length of PB.

2. The length of PB is twice of AP and the length of QC is 3 units more than the length of AQ, then calculate the length of AC. 

3. If AP = QC, the length of AB is 12 units and the length of AQ is 2 units, then calculate the length of CQ. 

Solution:

1. In ΔABC, PQ || BC.

∴ by Thales’ theorem \(\frac{A P}{B P}=\frac{A Q}{C Q}\) …….(1)

or, \(\frac{9}{\mathrm{AQ}}=\frac{\mathrm{AQ}}{4}\)

[ because AP = 9 units, QC = 4 units]

or, AQ² = 36 or, AQ = √36 or, AQ = 6

∴ from (1) we get, \(\frac{9}{P B}=\frac{6}{4}\)

∴ the length of PB = 6 units.

2. From (1) of ΔABC, PQ || BC we get, \(\frac{A P}{B P}=\frac{A Q}{C Q}\)

or, \(\frac{A P}{2 A P}=\frac{A Q}{A Q+3}\) [as per question]

or, \(\frac{1}{2}\) = \(\frac{A P}{2 A P}=\frac{A Q}{A Q+3}\)

or, 2AQ = AQ + 3

or, AQ = 3

∴ QC = AQ + 3 = (3 + 3) units = 6 units.
∴ AC = AQ + QC = (3 + 6) units = 9 units.

∴ the length of AC = 9 units.

3. From (1) of ΔABC, PQ || BC we get, \(\frac{A P}{B P}=\frac{A Q}{C Q}\)

or, \(\frac{\mathrm{AP}+\mathrm{BP}}{\mathrm{BP}}=\frac{\mathrm{AQ}+\mathrm{CQ}}{\mathrm{CQ}}\)

or, \(\frac{\mathrm{AB}}{\mathrm{PB}}=\frac{2+\mathrm{CQ}}{\mathrm{CQ}}\)

or, \(\frac{12}{\mathrm{~PB}}=\frac{2+\mathrm{CQ}}{\mathrm{CQ}}\)………(2)

Again, \(\frac{\mathrm{AP}}{\mathrm{BP}}=\frac{\mathrm{AQ}}{\mathrm{CQ}}\)

or, \(\frac{\mathrm{CQ}}{\mathrm{BP}}=\frac{2}{\mathrm{CQ}}\)

or, 2 PB= CQ²

or, PB= \(\frac{\mathrm{CQ}^2}{2}\)

From (2) we get, \(\frac{12}{\frac{\mathrm{CQ}^2}{2}}=\frac{2+\mathrm{CQ}}{\mathrm{CQ}}\)

or, \(\frac{24}{\mathrm{CQ}^2}=\frac{2+\mathrm{CQ}}{\mathrm{CQ}}\)

or, \(\frac{24}{\mathrm{CQ}}=2+\mathrm{CQ}\) [because \(\mathrm{CQ} \neq 0\)]

or, 2CQ + CQ² = 24 or, CQ² + 2CQ – 24 = 0 or, CQ² + 6CQ – 4CQ – 24 = 0

or, CQ (CQ + 6) – 4 (CQ + 6) = 0 or, (CQ + 6) (CQ – 4) = 0

either CQ + 6 = 0 or, CQ – 4 = 0

⇒ CQ = – 6 ⇒ CQ = 4

But the value of CQ cannot be negative.

∴ CQ = 4 units.

∴ the required length of CQ = 4 units

Example 2. X and Y are two points on the sides PQ and PR respectively of the ΔPQR.

1. If PX = 2 units, XQ = 3.5 units, YR = 7 units, and PY = 4.25 units, then find whether XY and QR are parallel or not. 

2. If PQ = 8 units, YR =12 units, PY = 4 units and the length of PY is 2 units less than that of XQ, then find whether XY and QR are parallel or not.

Solution:

 

1. In ΔPQR, XY and QR will be parallel if \(\frac{\mathrm{PX}}{\mathrm{QX}}=\frac{\mathrm{PY}}{\mathrm{RY}}\)

Now, \(\frac{P X}{Q X}=\frac{2 \text { units }}{3: 5 \text { units }}=\frac{2}{\frac{35}{10}}=\frac{4}{7}\)

Again, \(\frac{P Y}{R Y}=\frac{4 \cdot 25 \text { units }}{7 \text { units }}=\frac{17}{28}\)

∴ \(\frac{\mathrm{PX}}{\mathrm{QX}} \neq \frac{\mathrm{PY}}{\mathrm{RY}}\)

∴ \(\mathrm{XY}||\mathrm{QR}\)

 

2. In ΔPQR, XY and QR will be parallel if \(\frac{\mathrm{PX}}{\mathrm{QX}}=\frac{\mathrm{PY}}{\mathrm{RY}}\) …….(1)

or, \(\frac{\mathrm{PX}+\mathrm{QX}}{\mathrm{QX}}=\frac{\mathrm{PY}+\mathrm{RY}}{\mathrm{RY}}\)

or, \(\frac{\mathrm{PQ}}{\mathrm{QX}}=\frac{4+12}{12}\)

or, \(\frac{8}{6}=\frac{16}{12}\) [because QX = PY + 2 = 4 + 2 = 6]

or, \(\frac{4}{3}=\frac{4}{3}\)

∴ \( \frac{\mathrm{PX}}{\mathrm{QX}}=\frac{\mathrm{PY}}{\mathrm{RY}}\)

∴ XY || QR.

Hence according to the given conditions XY and QR are parallel.

Example 3. With the help of Thales’ theorem prove that the line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

Solution:

 

Let the mid-point of AB of the ΔABC be E and a straight line EF passing through E and parallel to BC which intersects AC at point F.

To prove: F is the mid-point of AC, i.e., AF = CF.

Proof: In ΔABC, EF || BC. (Given).

∴ by Thales’ theorem we get, \(\frac{\mathrm{AE}}{\mathrm{BE}}=\frac{\mathrm{AF}}{\mathrm{CF}}\)

or, \(\frac{\mathrm{AE}}{\mathrm{AE}}=\frac{\mathrm{AF}}{\mathrm{CF}}\)

or, \(1=\frac{\mathrm{AF}}{\mathrm{CF}}\)

or, AF = CF

Hence F is the mid-point of AC, i.e., EF has bisected AC (Proved)

Example 4. In ΔABC, P is a point on the median AD. Extended BP and CP intersect AC and AB at Q and R respectively. Prove that RQ || BC.

Solution:

Given:

In ΔABC, P is a point on the median AD. Extended BP and CP intersect AC and AB at Q and R respectively.

Let P be a point on the median AD of the ΔABC.

Extended BP and CP intersect AC and AB at the points Q and R respectively.

To prove RQ || BC.

Construction: Let us extend PD to T in such a way that PD = DT.

Let us join B, T, and C, T.

Proof: In the quadrilateral BPCT, BC and PT are two diagonals. Also, BD = CD.

[because AD is the median] and PD = DT [by construction]

i..e, the diagonals of the quadrilateral BPCT bisect each other.

∴ BPCT is a parallelogram.

∴ BP || TC and TB || CP.

Now, in ΔABT, RP || BT  [because TB || CP]

∴ by Thales’ theorem, \(\frac{\mathrm{AR}}{\mathrm{BR}}=\frac{\mathrm{AP}}{\mathrm{TP}}\) ……(1)

Again, in ΔACT, QP || CT  [because BP || TC]

∴ by Thales’theorem, \(\frac{\mathrm{AQ}}{\mathrm{CQ}}=\frac{\mathrm{AP}}{\mathrm{TP}}\)…….(2)

Then from (1) and (2) we get, \(\frac{\mathrm{AR}}{\mathrm{BR}}=\frac{\mathrm{AQ}}{\mathrm{CQ}}\)

∴ in ΔABC, \(\frac{\mathrm{AR}}{\mathrm{BR}}=\frac{\mathrm{AQ}}{\mathrm{CQ}}\)

∴ by the converse of Thales’ theorem, RQ || BC.

Hence RQ || BC (proved)

“Step-by-step solutions for similarity problems Class 10”

Example 5. The two medians BE and CF of AABC intersect each other at the point G and if the line segment FE intersects the line segment AG at the point O, then prove that AO = 30G. 

Solution:

Given:

The two medians BE and CF of AABC intersect each other at the point G and if the line segment FE intersects the line segment AG at the point O

In ΔABC, BE and CF medians of ΔABC intersect each other at G.

FE intersects AG at point O.

To prove: AO = 3OG.

Construction: Let us extend AG.

Also, let extended AG intersects BC at D.

Then G is centroid of ΔABC and AD is its median [because the medians of any triangle are concurrent.]

Proof: G is the centroid of ΔABC and AD is its median.

AG: GD = 2:1 [since the centroid of the triangle divides the medians in the ratio 2:1]

or, \(\frac{\mathrm{AG}}{\mathrm{GD}}=\frac{2}{1}\)

or, \(\frac{\mathrm{AO}+\mathrm{OG}}{\mathrm{OD}-\mathrm{OG}}=\frac{2}{1}\)

or, 2OD – 2OG = AO + OG

or, 2OD – AO = OG + 2OG

or, 2 AO – AO = 3OG

Hence AO = 3OG (proved)

[because in ΔABD, FO || BD, line segment joining the midpoints of any two sides of a traiangle is parallel to its third side.

∴ \(\frac{\mathrm{AF}}{\mathrm{BF}}=\frac{\mathrm{AO}}{\mathrm{OD}}\) or, \(\frac{\mathrm{AF}}{\mathrm{AF}}\)=\(\frac{\mathrm{AO}}{\mathrm{OD}}\) (because AF = BF)

or, 1 = \(\frac{\mathrm{AO}}{\mathrm{OD}}\) or, OD = AO.]

AO = 30G

Example 6. Prove that the line segment joining the mid-point of two transverse sides of a trapezium is parallel to its parallel sides.

Solution:

Let ABCD be a trapezium of which AD || BC.

AB and DC are two of its transverse sides, the midpoints of which are E and F respectively.

To prove EF || AD and EF || BC.

Construction: Let us extend BA and CD.

Also, let extended BA and CD intersect each other at point P.

Proof: In ΔPCB, AD || BC,

∴ by Thales’ theorem.

\(\frac{P A}{A B}=\frac{P D}{D C} \text { or, } \frac{P A}{2 A E}=\frac{P D}{2 D F}\)

[because E and F are the midpoints of AB and DC respectively.]

or, \(\frac{P A}{A E}=\frac{P D}{D F}\)

∴ in ΔPEF, \(\frac{P A}{A E}=\frac{P D}{D F}\)

∴ by the converse of Thales’ theorem, AD || EF.

Again, ∵ AD || BC, ∴AD || EF || BC. (Proved)

Example 7. D is any point on the side BC of ΔABC. P and Q are centroids of ΔABD and ΔADC respectively. Prove that PQ || BC. 

Solution:

Given:

D is any point on the side BC of ΔABC. P and Q are centroids of ΔABD and ΔADC respectively.

Let D is any point on the side BC of ΔABC.

The medians BE and DF of ΔABD intersect each other at P and two medians CE and DS of ΔACD intersect each other at Q.

∴ P and Q are two centroids of ΔABD and ΔADC respectively.

Let us join P and Q.

To prove: PQ || BC

Proof: P is the centroid of ΔABD.

∴ \(\frac{\mathrm{BP}}{\mathrm{PE}}=\frac{2}{1}\) [∵ centroid divides internally each median in the ratio 2:1]

or, BP = 2PE………..(1)

Similarly, CQ = 2QE……….(2)

Now, in ΔEBC, \(\frac{\mathrm{EP}}{\mathrm{BP}}=\frac{\mathrm{EP}}{2 \mathrm{EP}}=\frac{1}{2}\) [from(1)]…….(3)

Again, in ΔEBC, \(\frac{\mathrm{EQ}}{\mathrm{CQ}}=\frac{\mathrm{EQ}}{2 \mathrm{EQ}}=\frac{1}{2}\) [from(2)]…….(4)

∴ from (3) and (4) we get, \(\frac{E P}{B P}=\frac{E Q}{C Q}\)

i.e., in ΔEBC the straight line segment PQ has divided both BE and CE in such a way that \(\frac{E P}{B P}=\frac{E Q}{C Q}\)

∴ by the converse of Thales’ theorem we get, PQ || BC.

Hence PQ || BC. (Proved)

“WBBSE Mensuration Chapter 5 practice questions on similarity”

Example 8. Two triangles ΔPQR and ΔSQR are drawn on the same base QR and on the same side of QR and their areas are equal. If F and G are two centroids of two triangles, then prove that FG || QR.

Solution:

Given:

Two triangles ΔPQR and ΔSQR are drawn on the same base QR and on the same side of QR and their areas are equal. If F and G are two centroids of two triangles,

 

Let ΔPQR and ΔSQR have the same base QR and lie on the same side of QR so that the areas are equal.

F and G are the centroids of ΔPQR and ΔSQR respectively.

Let us join F, G.

To prove FG || QR.

Construction: Let T is the midpoint of QR.

∴ F must lie on the median PT of ΔPQR. Let us join P, T.

Similarly, G must lie on the median ST of ΔSQR. Let us join S, T.

Proof: F is the centroid of ΔPQR and PT is one of its medians,

∴ PF : FT = 2: 1 [∵ the centroid ofa triangle divides its median at the ratio 2:1]

or, \(\frac{\mathrm{PF}}{\mathrm{FT}}=\frac{2}{1}\) = or, PF = 2 FT …….(1)

Similarly, SG = 2GT…….(2)

Now, in ΔPST, the line segment FG intersects PT at F and intersect ST at G,

when \(\frac{P F}{F T}=\frac{2 F T}{F T}=2 \text { and } \frac{S G}{G T}=\frac{2 G T}{G T}=2 \text {, i.e., } \frac{P F}{F T}=\frac{S G}{G T}\)

∴ FG || PS……(3)

Again, ΔPQR and ΔSQR lie on the same base QR and they lie on the same side of QR, the areas of the triangles being equal.

∴ they must lie between same parallel couples.

∴ PS || QR ………(4).

Then from (3) and (4) we get, FG || PS || QR.

Hence FG || QR (Proved)

Example 9. Prove that two adjacent angles of any parallel side of an isosceles trapezium are equal.

Solution:

 

Given: Let ABCD be an isosceles trapezium of which AD = BC, AB || DC, and ∠ADC and ∠BCD are two adjacent angles of its parallel side DC.

To prove ∠ADC = ∠BCD.

Construction: Let us extend BA and CB and let extend DA and extended CB intersect each other at O.

Proof: In AODC, AB || DC (given),

∴ by Thales’ theorem, \(\frac{\mathrm{OA}}{\mathrm{AD}}=\frac{\mathrm{OB}}{\mathrm{BC}}\)

or, \(\frac{\mathrm{OA}}{\mathrm{BC}}=\frac{\mathrm{OB}}{\mathrm{BC}}\) [because AD = BC] or, OA = OB

Now, OD = OA + AD = OB + BC [because OA = OB and AD = BC]

∴ OD = OC, ∴ ∠ODC = ∠OCD or, ∠ADC = ∠BCD.

Hence ∠ADC =∠BCD. (Proved)

Example 10. ΔABC and ΔDBC are situated on the same base BC and on the same side of BC. E is any point on the side of BC. Two line through the point E and parallel to AB and BD intersects the sides AC and DC at the points F and G respectively. Prove that AD || FG.

Solution:

Given:

ΔABC and ΔDBC are situated on the same base BC and on the same side of BC. E is any point on the side of BC. Two line through the point E and parallel to AB and BD intersects the sides AC and DC at the points F and G respectively.

 

Given: Let ΔABC and ΔDBC have the same base BC and lie on the same side of BC.

E is any point on the side BC.

Two straight line passing through E and parallel to AB and BD intersect AC and DC at the points F and G respectively.

To prove AD || FG.

Proof: In ΔABC, FE || AB.

∴ by Thales’ theorem,

\(\frac{\mathrm{CF}}{\mathrm{FA}}=\frac{\mathrm{CE}}{\mathrm{EB}}\)……. (1)

Similarly in ΔBCD, GE || DB,

∴ By Thales’ theorem, \(\frac{\mathrm{CG}}{\mathrm{GD}}=\frac{\mathrm{CE}}{\mathrm{EB}}\)……(2)

From (1) and (2) we get, \(\frac{\mathrm{CF}}{\mathrm{FA}}=\frac{\mathrm{CG}}{\mathrm{GD}}\).

∴ in ΔACD, FG divides AC and DC in such a way that \(\frac{\mathrm{CF}}{\mathrm{FA}}=\frac{\mathrm{CG}}{\mathrm{GD}}\).

∴ By the converse of Thales’ theorem, FG || AD.

Hence AD || FG. (Proved)

Example 11. In a right-angled triangle ∠A is a right-angle and AO is perpendicular to BC at the point O. Prove that AO² = BO x CO.

Solution:

Given:

In a right-angled triangle ∠A is a right-angle and AO is perpendicular to BC at the point O.

Let in ΔABC, ∠A is a right angle, AO is perpendicular to BC at O.

∴ ∠AOC = 1 right angle.

Now, in ΔAOB and ΔAOC, ΔAOB = ΔAOC [∵ right angle]

∵ ∠BAO = ∠ACO [∵ ∠BAO = ∠BAC – ∠CAO = 90° – ∠CAO = ∠ACO ∵∠AOC = 90°]

∴ ΔAOB and ΔAOC are equiangular.

∴ by Thales’ theorem, \(\frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{BO}}{\mathrm{AO}}\)

or, AO² = BO x CO

Hence AO² = BO x CO. (proved)

“Examples of similar figures for WBBSE Class 10 Maths”

Example 12. A straight line drawn through D of the parallelogram ABCD intersects AB and the extended part of CB at points E and F respectively. Prove that AD: AE = CF: CD.

Solution:

Given:

A straight line drawn through D of the parallelogram ABCD intersects AB and the extended part of CB at points E and F respectively.

Let the straight line DF drawn through D of the parallelogram ABCD intersect AB and the extended CB at points E and F respectively.

To prove: AD: AE = CF: CD

Proof: In ΔADE and ΔDCF,

∠DAE = ∠DCF [opposite angles of a parallelogram are equal.]

∠ADE = ∠CFD [∵ AD || FC and DF is their transversal, ∠ADE = alternate ∠CFD.]

∴ ΔADE and ΔDCF are equiangular.

∴ by Thales’ theorem,

\(\frac{\mathrm{AD}}{\mathrm{CF}}=\frac{\mathrm{AE}}{\mathrm{CD}}\)

or, \(\frac{\mathrm{AD}}{\mathrm{AE}}=\frac{\mathrm{CF}}{\mathrm{CD}}\)

Hence AD: AE = CF: CD (Proved)

Example 13. Two chords AB and CD of a circle intersect at a internal point P of the circle. Prove that AP x BP = CP x DP.

Solution:

Given:

Two chords AB and CD of a circle intersect at a internal point P of the circle.

Let AB and CD be two chords of a circle with centre at O and they intersect each other at an internal point P of the circle.

To prove: AP x BP = CP x DP.

Construction: Let us join A, D and B, C.

Proof: In ΔAPD and ΔBPC, ∠APD = ∠BPC, [∵ opposite angles]

∠PAD = ∠PCB [∵ two angles in circle produced by the arc BD.]

∴ ΔAPD and ΔBPC are equiangular.

∴ by Thales’ theorem we get, \(\frac{\mathrm{AP}}{\mathrm{CP}}=\frac{\mathrm{DP}}{\mathrm{BP}}\)

∴ AP x BP = CP x DP. (Proved)

Example 14. AB is a diameter of a circle. BP is a tangent to the circle at B. A straight line passing through A intersects BP at C and the circle at D. Prove that BC² = AC x CD.

Solution:

Given:

AB is a diameter of a circle. BP is a tangent to the circle at B. A straight line passing through A intersects BP at C and the circle at D.

Let AB is a diameter of the circle with centre at O.

BP is tangent to the circle at B.

The straight line AC passing through A intersects BP at C and the circle at D.

To prove BC² = AC x CD.

Construction: Let us join B and D.

Proof: In ΔABD, ∠BAD = 90° – ∠ABD [∵ ∠ADB = semicircular angle = 90°]

= ∠ABC – ∠ABD [∵ BP is a tangent and OB is a radius through point of contact, ∴ ∠ABC = 90°] = ∠DBC

Then in ΔABC and ΔBCD,

∠ABC = ∠BDC [∵ each is a right angle]

∠BAC = ∠BAD = ∠DBC [∵ Proved]

∴ ΔABC and ΔBCD are equiangular

∴ by Thales’theorem we get, \(\frac{\mathrm{BC}}{\mathrm{CD}}=\frac{\mathrm{AC}}{\mathrm{BC}}\)

or, BC² = AC x CD .

Hence BC² = AC x CD (Proved)

Example 15. In the cyclic quadrilateral ABCD, the diagonal BD bisects the diagonal AC. prove that AB x AD = CB x CD

Solution:

Given:

In the cyclic quadrilateral ABCD, the diagonal BD bisects the diagonal AC.

Let ABCD be a cyclic quadrilateral in the circle with centre at O.

The diagonal BD of ABCD intersects the diagonal AC of ABCD at P in such a way that AP = CP.

To prove AB x AD = CB x CD.

Proof: Since diagonal BD bisects the diagonal AC at P, ∴ AP = CP.

Now, in ΔAPB and ΔCPD, ∠APB = ∠CPD [∵ opposite angles]

and ∠BAP = ∠CDP [∵ these are two angles in the circle produced by the chord BC.]

∴ ΔAPB and ΔCPD are equiangular.

∴ by Thales’ theorem we get, \(\frac{\mathrm{AB}}{\mathrm{AP}}=\frac{\mathrm{CD}}{\mathrm{DP}}\)…..(1)

Again, ΔAOD and ΔBOC are equiangular

[∵ ∠APD = opposite ∠BPC and ∠PAD = ∠PBC (same angles in circle)]

∴ by Thales’ theorem, \(\frac{\mathrm{AD}}{\mathrm{DP}}=\frac{\mathrm{BC}}{\mathrm{CP}}\)

Now, multiplying (1) by (2) we get,

\(\frac{\mathrm{AB}}{\mathrm{AP}} \times \frac{\mathrm{AD}}{\mathrm{DP}}=\frac{\mathrm{CD}}{\mathrm{DP}} \times \frac{\mathrm{BC}}{\mathrm{CP}}\)

or, \(\frac{\mathrm{AB} \times \mathrm{AD}}{\mathrm{CD} \times \mathrm{BC}}=\frac{\mathrm{AP} \times \mathrm{DP}}{\mathrm{DP} \times \mathrm{CP}}=1\)

[∵ AP = CP] or, AB x AD = CD x BC

Hence AB x AD = CB x CD (proved)

Example 16. AD is a diameter of the circumcircle of ΔABC. AE is perpendicular to BC. Prove that AB x AC – AD x AE.

Solution:

Given:

AD is a diameter of the circumcircle of ΔABC. AE is perpendicular to BC.

Let the circle with centre O be the circumcircle of ΔABC.

AD is its diameter. AE ⊥ BC.

To prove AB x AC = AD x AE.

Construction: Let us join C and D.

Proof: in ΔABE and ΔACD,

∠AEB = ∠ACD [∵ AE ⊥ BC and ∠ACD = semicircular angle = 1 right angle]

∠ABE = ∠ADC [∵ ∠ABE and ∠ADC are two angles in circle produced by the arc \(\overparen{A C}\)]

∴ ΔABE and ΔACD are equiangular.

∴ by Thales’ theorem, \(\frac{\mathrm{AB}}{\mathrm{AE}}=\frac{\mathrm{AD}}{\mathrm{AC}}\)

∴ AB x AC = AD x AE (proved)

Example 17. XY is a straight line parallel to the side MT of the parallelogram MNOT, which intersects the side MN at X and the side TO at Y. E and F are two points on XY. If ME and TF are extended, they intersect at P and the extended NE and OF intersect each other at Q. Prove that PQ || MN || TO. 

Solution:

Given:

XY is a straight line parallel to the side MT of the parallelogram MNOT, which intersects the side MN at X and the side TO at Y. E and F are two points on XY. If ME and TF are extended, they intersect at P and the extended NE and OF intersect each other at Q.

Given: XY is a straight line parallel to the side MT of the parallelogram MNOT, which intersects the side MN at X and the side TO at Y.

E and F are two points on XY.

If ME and TF are extended, they intersect at P and the extended NE and OF intersect each other at Q.

To prove PQ || MN || TO.

Proof: In ΔPMT, EF || MT [∵ XY || MT]

by Thales’ theorem, \(\frac{\mathrm{PE}}{\mathrm{ME}}=\frac{\mathrm{PF}}{\mathrm{TF}}=\frac{\mathrm{EF}}{\mathrm{MT}}\)…….(1)

Again, in ΔQON, EF || NO [∵ XY || MT || NO]

∴ by Thales’ theorem, \(\frac{\mathrm{QE}}{\mathrm{EN}}=\frac{\mathrm{QF}}{F O}=\frac{\mathrm{EF}}{\mathrm{NO}}\)

From (1) and (2) we get,

\(\frac{\mathrm{PE}}{\mathrm{ME}}=\frac{\mathrm{EF}}{\mathrm{MT}}\) and \(\frac{\mathrm{QE}}{\mathrm{EN}}=\frac{\mathrm{EF}}{\mathrm{NO}}\)……..(3)

But ∵ MNOT is a parallelogram, ∴ MT = NO

[∵ Opposite sides of any parallelogram are equal]

From (3) we get, \(\frac{P E}{M E}=\frac{\mathrm{QE}}{\mathrm{EN}}\)

∴ ΔMEN – ΔPEQ, i.e., ∠MNE = ∠PQE,

∠NME = ∠QPM and ∠MNQ = ∠PQN.

But these are alternate angles when PQ and MN are intersected by the transversal NQ.

∴ PQ || MN ………(4)

Again, since MNOT is a parallelogram, ∴MN || TO……(5)

Then from (4) and (5) we get, PQ || MN || TO.

Hence PQ || MN || TO. (Proved)

Example 18. In ΔPQR, ∠Q = 2∠R; The bisector of ∠PQR intersects PR at the point D. Prove that PQ.QR = QD.PR.

Solution:

Given:

In ΔPQR, ∠Q = 2∠R; The bisector of ∠PQR intersects PR at the point D.

In ΔPQR, ∠Q = 2∠R; Bisector QD of ∠PQR intersects PR at D.

To prove: PQ.QR = QD.PR ∠Q = 2∠R (given)

Proof: Since

Again, ∠PQD = ∠DQR [∵ QD is the bisector of ∠PQR]

∴ ∠PQD = ∠DRQ.

Then in ΔPQR and ΔPQD, ∠PRQ = ∠PQD, ∠QPR =∠QPD

∴ ΔPQR and ΔPQD are equiangular.

∴ by Thales’ theorem, \(\frac{\mathrm{PQ}}{\mathrm{QD}}=\frac{\mathrm{PR}}{\mathrm{QR}}\)

∴ PQ.QR = QD.PR. (Proved)

“WBBSE Class 10 Maths solved problems on similarity”

Example 19. PQ is a diameter of a circle and AB is such a chord of it that it is perpendicular to PQ. If C be the point of intersection of PQ and AB, then prove that PC.QC = AC.BC.

Solution:

Given:

PQ is a diameter of a circle and AB is such a chord of it that it is perpendicular to PQ. If C be the point of intersection of PQ and AB

Let PQ is a diameter of the circle with centre at O.

The chord AB is perpendicular to PQ and has intersected PQ at C.

To prove: PC.QC = AC.BC.

Construction: Let us join B, Q and A, P.

In ΔACP and ΔBCQ, ∠ACP = ∠BCQ [∵ AB ⊥ PQ, each is right-angled.]

and ∠APC = ∠QBC [∵ these are two angles in circle produced by the arc \(\overparen{A Q}\)]

∴ ΔACP ~ ΔBCQ.

∴ by Thales’ theorem, \(\frac{\mathrm{PC}}{\mathrm{BC}}=\frac{\mathrm{AC}}{\mathrm{QC}}\)

or, PC.QC = AC.BC

Hence PC. QC = AC. BC (Proved)

Example 20. PQRS is a cyclic quadrilateral. Extended PQ and SR intersect each other at A. Prove that AP.AQ = AR.AS.

Solution:

Given:

PQRS is a cyclic quadrilateral. Extended PQ and SR intersect each other at A.

Let PQRS is a cyclic quadrilateral in the circle with centre at O.

Extended PQ and RS intersect each other at A.

To prove: AP.AQ = AR.AS.

Proof: In ΔAPS and ΔAQR, ∠APS = ∠ARQ and

[∵ ∠QPS + ∠QRS = 180° or, ∠QPS + 180° – ∠QRA = 180° or, ∠QPS = ∠QRA]

∠ASP – ∠AQR [for similar reason]

∴ ΔAPS and ΔAQR are equiangular.

∴ by Thales’ theorem, \(\frac{\mathrm{AP}}{\mathrm{AR}}=\frac{\mathrm{AS}}{\mathrm{AQ}}\)

or, AP.AQ = AR.AS.

Hence AP AQ = AR.AS (Proved)

Solid Geometry Chapter 5 Similarity

Relation Between The Sides Of Two Similar Triangles

In the previous part of this chapter, you have studied about similar triangles, congruent triangles and about their properties and conditions.

In the present part we shall discuss about the relations between the sides of two similar triangles and the theorems related to them.

If we measure the lengths of the sides of two similar triangles and determine their ratios, then we shall see that the corresponding sides of two similar triangles are proportional, i.e., the corresponding sides of two similar triangles are proportional.

We shall now prove this theorem logically with the help of the geometric method.

Solid Geometry Chapter 5 Similarity

Relation Between The Sides Of Two Similar Triangles Theorems

Theorem 1. If two triangles are similar then prove that their corresponding sides are in same ratio, i.e., their corresponding sides are proportional.

Given: Let ΔABC and ΔDEF are two similar triangles, i.e., ∠A = ∠D, ∠B = ∠E and ∠C = ∠F.

To prove \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{DF}}\)

Construction: Let us cut off AP and AQ from AB and AC of the ΔABC equal to DE and DF respectively. Let us join P and Q.

Proof: In ΔAPQ and ΔDEF, AP = DE, AQ = DF [by construction] and ∠PAQ = ∠EDF.

∴ ΔAPQ ≅ ΔDEF [by the S-A-S condition of congruency]

∴ ∠APQ = ∠DEF [∵ corresponding angles of congruent triangles.]

or, ∠APQ = ∠ABC [∵ ∠DEF = ∠ABC (given)] ,

But these are similar angles.

∴ PQ || BC.

∴ \(\quad \frac{\mathrm{AP}}{\mathrm{BP}}=\frac{\mathrm{AQ}}{\mathrm{CQ}}\) [by Thales’ theorem]

or, \(\frac{\mathrm{BP}}{\mathrm{AP}}=\frac{\mathrm{CQ}}{\mathrm{AQ}}\)

or, \(\frac{\mathrm{BP}+\mathrm{AP}}{\mathrm{AP}}=\frac{\mathrm{CQ}+\mathrm{AQ}}{\mathrm{AQ}}\)

or, \(\frac{\mathrm{AB}}{\mathrm{AP}}=\frac{\mathrm{AC}}{\mathrm{AQ}}\)

or, \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AC}}{\mathrm{DF}}\)……(1)

[∵ AP = DE and AQ = DF.]

Similarly, cutting parts from BA and BC respectively equal to DE and EF, it can be prove that

\(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}\)……(2)

∴ from (1) and (2) we get, \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}\)= \(\frac{\mathrm{AC}}{\mathrm{DF}}\)

Hence \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}\)= \(\frac{\mathrm{AC}}{\mathrm{DF}}\), i.e.,

The corresponding sides of two similar triangles are proportional. (Proved)

Theorem 2. If the sides of two triangles are in same ratio, then their corresponding angles are equal, i.e., two triangles are similar.

 

Given: Let the sides of the ΔABC and ΔDEF are in same ratio,

i.e., \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}\)= \(\frac{\mathrm{CA}}{\mathrm{DF}}\),

To prove ΔABC ~ ΔDEF, i.e., ∠A =∠D, ∠B – ∠E and ∠C = ∠F.

Construction: Let the parts AP and AQ are cut offfrom AB and AC in such a way that the two parts are equal to DE and DF respectively.

Let us join P and Q.

Proof: Given that \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AC}}{\mathrm{DF}}\)

or, \(\frac{\mathrm{AB}}{\mathrm{AP}}=\frac{\mathrm{AC}}{\mathrm{AQ}}\)[∵ DE = AP and DF = AQ by construction]

∴ by the converse of Thales’ theorem, PQ || BC.

∴ ∠B = ∠APQ [∵ similar angles] and ∠C = ∠AQP [∵ similar angles]

∴ ΔABC and ΔAPQ are similar.

∴ \(\frac{\mathrm{AB}}{\mathrm{AP}}=\frac{\mathrm{BC}}{\mathrm{PQ}}\)…….. (1),

But \(\frac{\mathrm{AB}}{\mathrm{AP}}=\frac{\mathrm{AB}}{\mathrm{DE}}\) [∵ AP = DE] = \(\frac{\mathrm{BC}}{\mathrm{EF}}\)…….(2)

From (1) and (2) we get, \(\frac{\mathrm{BC}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{EF}}\), ∴ PQ = EF.

Then in ΔAPQ and ΔDEF, AP = DE, AQ = DF and PQ = EF.

∴ ΔAPQ = ΔDEF [by the S-S-S condition of congruency]

∴ ∠APQ = ∠DEF = ∠E corresponding angles of similar triangles]

But ∠APQ = ∠B, ∴ ∠B = ∠E

Again, ∠AQP = ∠DFE = ∠F [for similar reason]

But ∠AQP =∠C, ∴ ∠C = ∠F

It is then obvious that ∠A = ∠D.

i.e, in ΔABC and ΔDEF, ∠A – ∠D, ∠B = ∠E and ∠C = ∠F.

∴ ΔABC ~ ΔDEF. (Proved)

From above discussion we can say that two triangles will be similar if

1. Their sides are proportional; or
2. Their corresponding angles are equal.

Similarly, if the sides of two polygons be proportional and the corresponding angles are equal, then the polygons are similar.

Again, if any one angle of a triangle be equal to one angle of another triangle and their sides adjacent to that angle be proportional, then the triangles will be similar.

We shall now prove this theorem logically by the method of geometry.

Theorem 3. If in two triangles, an angle of one triangle is equal to the angle of another triangle and the adjacent sides of the angle are proportional, then two triangles are similar.

 

 

Given: Let in ΔABC and ΔDEF, ∠A = ∠D and \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AC}}{\mathrm{DF}}\)

To prove ΔABC ~ ΔDEF.

Construction: Two parts DP and DQ equal to AB and AC are cut off respectively from DE and DF.

Let us join P and Q.

Proof: Given that \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AC}}{\mathrm{DF}}\)

or, \(\frac{\mathrm{DP}}{\mathrm{DE}}=\frac{\mathrm{DQ}}{\mathrm{DF}}\) [∵ AB = DP, AC = DQ]

∴ PQ || EF [by the converse of Thales’ theorem]

∴ ∠DPQ = ∠DEF …..(1) [∵ similar angles]

∴ ∠DQP = ∠DFE ……(2) [ similar angles]

Again, ΔDPQ = ΔABC [∴ ∠D = ∠A, DP = AB and DQ = AC]

∴ ∠DPQ = similar ∠B, from (1) we get ∠B = ∠DEF, i.e. ∠B = ∠E.

Again, ∠DQP = similar ∠C, from (2) we get, ∠C = ∠DQP, i.e., ∠C = ∠F.

in ΔABC and ΔDEF, ∠A = ∠D, ∠B = ∠E and ∠C = ∠F.

∴ ΔABC ~ ΔDEF. (Proved)

We know that if a perpendicular is drawn from the right angular point of any right-angled triangle to its hypotenuse, then the triangle is divided into two right-angled triangles.

We shall now prove logically that the two right-angled triangle thus produced are similar to each other and each of them is also similar to the original right-angled triangle.

Theorem 4. If any right-angled triangle if a perpendicular is drawn from right angular point on the hypotenuse then prove that the two triangles on both sides of this perpendicular are similar and each of them is similar to original triangle.

Given: Let ΔBC be a right-angled triangle of which A = 90°.

So, the hypotenuse is BC. AD is perpendicular to the hypotenuse BC from the right-angular point A.

Then two right-angled triangles ΔABD and ΔACD have been produced.

To prove

1. ΔABD ~ ΔABC
2. ΔACD ~ ΔABC; and
3. ΔABD ~ ΔACD

Construction: Let us draw a perpendicular AD from the right angular point A to the hypotenuse BC which intersects BC at the point D.

Proof:

1. In ΔABD and ΔABC, ∠ADB = ∠BAC each is right-angle] ∠ABD is common to both the triangles.

Obviously, ∠BAD = ∠ACB [∵ sum of three angles of any triangle is 2 right angles or 180°]

So, in ΔABD and ΔABC, ∠ADB = ∠A, ∠ABD = ∠B and ∠BAD = ∠C.

Hence ΔABD ~ ΔABC [(1)Proved]

2. In ΔACD and ΔABC, ∠ADC = ∠BAC, [∵ each is a right angle.]∠ACD is common to both the triangles.

Obviously, ∠CAD = ∠ABC [∵ sum of three angles of any triangle is 2 right angles or 180°]

So in ΔACD and ΔABC, ∠ADC = ∠A, ∠ACD = ∠C and ∠CAD = ∠B.

Hence ΔACD ~ ΔABC. [(2) Proved]

3. Now, ΔABD and ΔACD, ∠ADB = ∠ADC [∵ each is a right angle]

∴ ∠ABD + ∠BAD = 90° and ∠CAD + ∠ACD = 90°

∴ ∠ABD = 90° – ∠BADor, ∠ABD = ∠CAD [∵ ∠A = 90°]

i.e., in ΔABD and ΔACD, ∠ADB – ∠ADC, ∠ABD = ∠CAD and ∠BAD = ∠ACD, i.e., three angles of each of the triangles are equal.

Hence ΔABD ~ ΔACD [(3) Proved]

Corollary 1. According to theorem 4, ΔABC ~ ΔABD.

∴ \(\frac{B C}{B A}=\frac{B A}{B D}\) = or, BA² = BC. BD.

∴ BA is the mean-proportional of BC and BD.

Corollary 2. According to the theorem 4, ΔABD ~ ΔACD,

∴ \(\frac{D B}{D A}=\frac{D A}{D C}\) or, DA² = DB.DC

∴ DA is the mean proportional of DB and DC.

Corollary 3. According to the theorem 4, ΔABC ~ ΔACD,

∴ \(\frac{C B}{C A}=\frac{C A}{C D}\) or.CA² = CB.CD.

∴ CA is the mean-proportional of CB and CD

Solid Geometry Chapter 5 Similarity

Some Essential Theorems

Theorem 1. (1) Prove that the ratio of the areas of two similar triangles is equal to the ratio of the square of the corresponding sides.

Given: Let ΔABC and ΔPQR are two similar triangles.

To prove 

\(\frac{\text { Area of } \triangle \mathrm{ABC}}{\text { Area of } \Delta \mathrm{PQR}}=\frac{\mathrm{AB}^2}{\mathrm{PQ}^2}=\frac{\mathrm{BC}^2}{\mathrm{QR}^2}=\frac{\mathrm{AC}^2}{\mathrm{PR}^2}\)

Construction: Let us draw a perpendicular AD from A to BC and a perpendicular PM from P to QR.

Proof: ∵ ΔABC ~ ΔPQR,

∴ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}\)……(1)

Again, ΔABD ~ΔPQM [∵ ∠ADB = ∠PMQ and ∠B = ∠Q]

∴ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)…….(2)

Now, from (1) and (2) we get, \(\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)…….(3)

Then, \(\frac{\text { Area of } \Delta \mathrm{ABC}}{\text { Area of } \Delta \mathrm{PQR}}=\frac{\frac{1}{2} \times \mathrm{BC} \times \mathrm{AD}}{\frac{1}{2} \times \mathrm{QR} \times \mathrm{PM}}\)

[∵ Base = BC and height = AD]

[∵ Base = QR and height = PM]

= \(\frac{\mathrm{BC} \times \mathrm{AD}}{\mathrm{QR} \times \mathrm{PM}}=\frac{\mathrm{BC}}{\mathrm{QR}} \times \frac{\mathrm{AD}}{\mathrm{PM}}\)

= \(\frac{\mathrm{BC}}{\mathrm{QR}} \times \frac{\mathrm{BC}}{\mathrm{QR}}\)[from (3)]

= \(\frac{\mathrm{BC}^2}{\mathrm{QR}^2}\)…..(4)

Again, since ΔABC and ΔPQR are similar to each other,

∴ \(\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}\)

or,\(\frac{\mathrm{BC}^2}{\mathrm{QR}^2}=\frac{\mathrm{AB}^2}{\mathrm{PQ}^2}=\frac{\mathrm{AC}^2}{\mathrm{PR}^2}\)

Hence from (4) we get,

\(\frac{\text { Area of } \triangle \mathrm{ABC}}{\text { Area of } \triangle \mathrm{PQR}}=\frac{\mathrm{AB}^2}{\mathrm{PQ}^2}=\frac{\mathrm{BC}^2}{\mathrm{QR}^2}=\frac{\mathrm{AC}^2}{\mathrm{PR}^2}\) (Proved)

2. The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding heights.

Given: Let ΔABC and ΔDEF are two similar triangles, the heights of them are AM and DN respectively.

To prove

\(\frac{\text { Area of } \Delta \mathrm{ABC}}{\text { Area of } \Delta \mathrm{DEF}}=\frac{\mathrm{AM}^2}{\mathrm{DN}^2}\)

Proof: ΔABC ~ ΔDEF, by theorem 1(1) we get,

\(\frac{\text { Area of } \Delta \mathrm{ABC}}{\text { Area of } \Delta \mathrm{DEF}}=\frac{\mathrm{AB}^2}{\mathrm{DE}^2}\)……(1)

Again, in ΔABM and ΔDEN, ∠AMB = ∠DNE [∵ each is a right angle] and ∠ABM = ∠DEN (given)

∴ ΔABM ~ ΔDEN

∴ \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AM}}{\mathrm{DN}}\)

or, \(\frac{\mathrm{AB}^2}{\mathrm{DE}^2}=\frac{\mathrm{AM}^2}{\mathrm{DN}^2}\)

From (1) we get,

\(\frac{\text { Area of } \triangle \mathrm{ABC}}{\text { Area of } \Delta \mathrm{DEF}}=\frac{\mathrm{AM}^2}{\mathrm{DN}^2}\) (Proved)

3. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding medians.

 

Given: Let ΔABC and ΔPQR be two similar triangles of which AD and PM are two corresponding medians.

To prove 

\(\frac{\text { Area of } \Delta \mathrm{ABC}}{\text { Area of } \Delta \mathrm{PQR}}=\frac{\mathrm{AD}^2}{\mathrm{PM}^2}\)

Proof: ΔABC ~ ΔPQR,

∴ by theorem 4 (1), \(\frac{\text { Area of } \Delta \mathrm{ABC}}{\text { Area of } \Delta \mathrm{PQR}}=\frac{\mathrm{AB}^2}{\mathrm{PQ}^2}\)….(1)

Again, ∵ ΔABC ∼ ΔPQR,

∴ \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}\)….(2)

From (2) we get, \(\frac{A B}{P Q}=\frac{B C}{Q R}\)

or, \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{2 \mathrm{BD}}{2 \mathrm{QM}}=\frac{\mathrm{BD}}{\mathrm{QM}}\)…….(3)

Again in ΔABD and ΔPQM, ∠ABD = ∠PQM [∵ ΔABC ~ ΔPQR.]

and \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{2 \mathrm{BD}}{2 \mathrm{QM}}\) [from (3)]

∴ ΔABD ∼ ΔPQM

∴ \(\frac{A B}{P Q}=\frac{B D}{Q M}=\frac{A D}{P M}\)  [by Thales’ theorem]

∴ from (1) we get,

\(\frac{\text { Area of } \Delta \mathrm{ABC}}{\text { Area of } \Delta \mathrm{PQR}}=\frac{\mathrm{AD}^2}{\mathrm{PM}^2}\)(proved)

4. The ratio of the areas of two triangles having same heights will be equal to the ratio of their corresponding bases.

Given: Let the heights of ΔABC and ΔDEF be each h units.

BC and EF are the two corresponding bases of them.

To prove 

\(\frac{\text { Area of } \Delta \mathrm{ABC}}{\text { Area of } \Delta \mathrm{DEF}}=\frac{\mathrm{BC}}{\mathrm{EF}}\)

Proof: We know, area of a triangle = \(\frac{1}{2}\) x Base x Height

∴ area of ΔABC = \(\frac{1}{2}\) x BC x h…..(1)

area of ΔDEF =\(\frac{1}{2}\) x  EF x h……. (2)

Now, dividing (1) by (2) we get

\(\frac{\text { Area of } \Delta \mathrm{ABC}}{\text { Area of } \Delta \mathrm{DEF}}=\frac{\frac{1}{2} \times \mathrm{BC} \times h}{\frac{1}{2} \times \mathrm{EF} \times h}=\frac{\mathrm{BC}}{\mathrm{EF}}\)

Hence \(\frac{\text { Area of } \triangle \mathrm{ABC}}{\text { Area of } \triangle \mathrm{DEF}}=\frac{\mathrm{BC}}{\mathrm{EF}}\) (Proved)

In the following examples how the above theorems are applied in real problems have been discussed thoroughly.

Solid Geometry Chapter 5 Similarity

Relation Between The Sides Of Two Similar Triangles Multiple Choice Questions

Example 1. In ΔABC and ΔDEF if \(\frac{A B}{D E}=\frac{B C}{D F}=\frac{A C}{E F}\) then

1. ∠B = ∠E
2. ∠A = ∠D
3. ∠B = ∠D
4. ∠A = ∠F

Solution: ∵ in ΔABC and ΔDEF, \(\frac{A B}{D E}=\frac{B C}{D F}=\frac{A C}{E F}\)

ΔABC ~ ΔDEF and the corresponding sides of the sides AB, BC and AG are respectively DE, FD and EF.

∴ ∠B = ∠D

∴ 3. ∠B = ∠D is correct.

Example 2. If in ΔDEF and ΔPQR, ∠D = ∠Q and ∠E = ∠R , then which one of the following is not correct?

1. \(\frac{E F}{P R}=\frac{D F}{P Q}\)
2. \(\frac{Q R}{P Q}=\frac{E F}{D F}\)
3. \(\frac{D E}{Q R}=\frac{D F}{P Q}\)
4. \(\frac{E F}{R P}=\frac{D E}{Q R}\)

Solution: ∵ in ΔDEF and ΔPQR, ∠D = ∠Q and ∠E = ∠R, ∴ ΔDEF ~ ΔPQR.

∴ by Thales’ theorem, the ratios of the corresponding sides of ΔDEF and ΔPQR are equal.

∴ \(\frac{E F}{P R}=\frac{D F}{P Q}=\frac{D E}{Q R}\)

∴ 2. \(\frac{Q R}{P Q}=\frac{E F}{D F}\) is not correct

Example 3. In ΔABC and ΔDEF, ∠A = ∠E = 40°, AB: ED = AC: EF and∠F = 65°, then the value of ∠B is 

1. 35°
2. 65°
3. 75°
4. 85°

Solution: In ΔABC and ΔDEF, ∠A = ∠E, AB : ED = AC : EF

∴ ΔABC ~ ΔDEF

∴ ∠A = ∠E, ∠B = ∠D, ∠C = ∠F

∴ in ΔABC, ∠A = 40° [∵ ∠A = ∠E = 40°] and ∠C = 65° [∵ ∠C = ∠F = 65°]

∴ ∠B =180° – (∠A + ∠C) = 180° – (40° + 65°) = 180° – 105° = 75°

∴ 3. 75° is correct.

Example 4.  In ΔABC and ΔPQR, if \(\frac{A B}{Q R}=\frac{B C}{P R}=\frac{C A}{P Q}\) then

1. ∠A = ∠Q
2. ∠A = ∠P
3. ∠A = ∠R
4. ∠B = ∠Q

 

Solution: ∵ in ΔABC and ΔPQR,

\(\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{PR}}=\frac{\mathrm{CA}}{\mathrm{PQ}}\),

∴ ΔABC ∼ ΔPQR

∴ ∠A = ∠Q; ∠B = ∠R, ∠C = ∠P

∴ 1. ∠A = ∠Q is correct.

Example 5. In ΔABC, AB = 9 cm, BC = 6 cm and CA = 7.5 cm. In ΔDEF, the corresponding side of BC is EF; EF = 8 cm and if ΔDEF ~ ΔABC, then the perimeter of ΔDEF will be 

1. 22.5 cm
2. 25 cm
3. 27 cm
4. 30 cm

Solution:

Given

In ΔABC, AB = 9 cm, BC = 6 cm and CA = 7.5 cm. In ΔDEF, the corresponding side of BC is EF; EF = 8 cm and if ΔDEF ~ ΔAB

∵ ΔADEF ~ ΔABC and the corresponding side of BC is EF,

∴ \(\frac{\mathrm{AB}}{\mathrm{DE}}\) = \(\frac{\mathrm{BC}}{\mathrm{EF}}\) = \(\frac{\mathrm{AC}}{\mathrm{DF}}\)

Now, \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}\)

or, \(\frac{9 \mathrm{~cm}}{\mathrm{DE}}\) = \(\frac{6 \mathrm{~cm}}{8 \mathrm{~cm}}\)

[∵ AB = 9 cm, BC = 6 cm and EF = 8 cm]

or, DE = 12 cm

Again, \(\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{DF}}\)

or, \(\frac{6 \mathrm{~cm}}{8 \mathrm{~cm}}=\frac{7 \cdot 5 \mathrm{~cm}}{\mathrm{DF}}\)

[∵ AC = 7.5 cm] or, DF = 10 cm

∴ Perimeter of ΔDEF = DE + EF + FD = (12 + 8 + 10) cm = 30 cm

Hence the required perimeter of ΔDEF = 30 cm

∴ 4. 30 cm is correct.

Solid Geometry Chapter 5 Similarity

Relation Between The Sides Of Two Similar Triangles True Or False

Example 1. If the corresponding angles of two quadrilaterals are equal, then they are similar.

Solution: False

Since here though the angles are equal, their shapes may not be equal, such as four angles of a rectangle and of a square are equal (each is a right angle), but their shapes are not the same.

Example 2. In the adjoining, if ∠ADE = ∠ACB, then ΔADE ~ ΔACB.

Solution: True

Since in ΔADE and ΔABC, ∠ADE = ∠ACB (given) and ∠DAE = ∠BAC, i.e. ∠A is common to both.

∴ ΔADE ~ ΔABC.

Example 3. In ΔPQR, D is a point on the side QR so that PD ⊥ QR; So ΔPQD ~ ΔRPD.

Solution: False

Since here in ΔPQD and ΔRPD, only ∠PDQ = ∠PDR (each is. a right angle),

The other two angles are not equal.

Solid Geometry Chapter 5 Similarity

Relation Between The Sides Of Two Similar Triangles Fill In The Blanks

Example 1. Two triangles are similar if their _________ sides are proportional.

Solution: corresponding

Example 2.  The perimeters of ΔABC and ΔDEF are 30 cm and 18 cm respectively. ΔABC ~ ΔDEF; BC and EF are corresponding sides. If BC = 9 cm, the EF = ______ cm.

Solution: 5-4 cm since ΔABC ~ ΔDEF,

\(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{DF}}=\frac{\mathrm{AB}+\mathrm{BC}+\mathrm{CA}}{\mathrm{DE}+\mathrm{EF}+\mathrm{FD}}=\frac{30 \mathrm{~cm}}{18 \mathrm{~cm}}=\frac{15}{9}\) \(\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{15}{9} \text { or }, \frac{9 \mathrm{~cm}}{\mathrm{EF}}=\frac{15}{9}\)

or, 15 EF =9 x 9 cm

or, EF = \(\frac{9 \times 9}{15} \mathrm{~cm}\)

EF= \(\frac{81}{15}\) =5.4 cm

Solid Geometry Chapter 5 Similarity

Relation Between The Sides Of Two Similar Triangles Short Answer Type Questions

Example 1. In the following if ∠ACB = ∠BAD, AC = 8 cm, AB = 16 cm and AD = 3 cm, then find the length of BD.

Solution:

Given:

In the following if ∠ACB = ∠BAD, AC = 8 cm, AB = 16 cm and AD = 3 cm

In ΔABC and ΔABD, ∠ACB = ∠BAD, ∠ABC = ∠ABD,

∴ ΔABC ~ ΔABD.

∴ by Thales’ theorem, \(\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{\mathrm{CA}}{\mathrm{AD}}\)

∴ \(\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{CA}}{\mathrm{AD}}\) or, \(\frac{16 \mathrm{~cm}}{\mathrm{BD}}=\frac{8 \mathrm{~cm}}{3 \mathrm{~cm}}\)

[∵ AB = 16cm, AC = 8 cm, AD = 3 cm]

or, 8BD = 48 or, BD = \(\frac{48}{8}\) = 6.

Hence the length of BD = 6 cm.

Example 2. In the adjoining, ∠ABC = 90° and BD ⊥ AC. If AB = 5.7 cm, BD = 3.8 cm, CD =5.4 cm, then find the length of BC.

Solution:

Given:

In the adjoining, ∠ABC = 90° and BD ⊥ AC. If AB = 5.7 cm, BD = 3.8 cm, CD =5.4 cm

∵ ∠ABC = 90° and BD ⊥ AC,

∴ ΔABD ~ ΔBCD and ΔABD ~ ΔABC, ΔBCD ~ ΔABC.

Then ΔABC ~ ΔBCD implies = \(\frac{A B}{B D}=\frac{B C}{C D}\)

or, \(\frac{5 \cdot 7 \mathrm{~cm}}{3 \cdot 8 \mathrm{~cm}}\) = \(\frac{\mathrm{BC}}{5.4 \mathrm{~cm}}\)

[∵ AB = 57 cm, BD = 38 cm, CD = 54 cm.]

or, 38 BC = 5.7 x 5.4 or, BC = \(\frac{5.7 \times 5.4}{3.8}\) = 8.1.

Hence the length of BC = 8.1 cm.

Example 3. In beside, ∠ABC = 90° and BD ⊥ AC. If BD = 8 cm and AD = 4 cm, then find the length of CD.

Solution:

Given:

In beside, ∠ABC = 90° and BD ⊥ AC. If BD = 8 cm and AD = 4 cm,

∵ ∠ABC = 90° and BD ⊥ AC,

∴ ΔABD ∼ ΔBCD

∴ by Thales’ theorem, \(\frac{A D}{B D}=\frac{B D}{C D}\)

∴ from (1) we get, BD² = AD x CD or, (8)² = 4 x CD

or, 64 = 4CD or CD = \(\frac{64}{4}\) = 16.

Hence the length of CD = 16 cm.

“Applications of similarity in Class 10 Maths”

Example 4. In trapezium ABCD, BC || AD and AD = 4 cm. The two diagonals AC and BD intersect at the point O in such a way that \(\frac{A O}{O C}=\frac{D O}{O B}=\frac{1}{2}\) Find the length of BC.

Solution:

Given:

In trapezium ABCD, BC || AD and AD = 4 cm. The two diagonals AC and BD intersect at the point O in such a way that \(\frac{A O}{O C}=\frac{D O}{O B}=\frac{1}{2}\)

ΔAOD ∼ ΔBOC,

∴ \(\frac{A O}{O C}=\frac{D O}{O B}=\frac{A D}{B C}\)

or, \(\frac{A D}{B C}=\frac{1}{2}\) [∵\(\frac{A O}{O C}=\frac{D O}{O B}=\frac{1}{2}\)]

or, \(\frac{4}{B C}=\frac{1}{2}\) or, BC = 8

Hence the length of BC = 8 cm.

Example 5. ΔABC ∼ ΔDEF and in ΔABC and ΔDEF, the corresponding sides of AB, BC and CA are DE, EF and DF respectively. If ∠A = 47° and ∠E = 83°, then find the value of ∠C.

Solution:

Given

ΔABC ∼ ΔDEF and in ΔABC and ΔDEF, the corresponding sides of AB, BC and CA are DE, EF and DF respectively. If ∠A = 47° and ∠E = 83°

ΔABC ~ ΔDEF,

∴ ∠A = ∠D, ∠B = ∠E, ∠C = ∠F

∴ ∠D = 47°, ∠E = 83° (given)

∴ ∠F = 180° – (∠D + ∠E)

= 180° – (47° + 83°) = 180° – 130° = 50°

∴ ∠C = 50° [∵ ∠F = ∠C]

Hence the value of ∠C = 50°.

Solid Geometry Chapter 5 Similarity

Relation Between The Sides Of Two Similar Triangles Long Answer Type Questions

Example 1. ABC is a right angled triangle whose ∠B is right angle and BD ⊥ AC; if AD = 4 cm and CD = 16 cm, then calculate the length of BD and AB. 

Solution:

Given:

ABC is a right angled triangle whose ∠B is right angle and BD ⊥ AC; if AD = 4 cm and CD = 16 cm,

∵ in ΔABC, ∠B = right angle and BD ⊥ AC.

∴ ΔABC ~ ΔABD

ΔABC ~ ΔBCD

ΔABD ~ ΔBCD

Now, ΔABD ~ ΔBCD implies \(\frac{A D}{B D}=\frac{B D}{C D}\)…..(1)

or, AD x CD = BD² or BD² = AD x CD = 4 x 16 = 64

∴ BD = √64 = 8. ∴ BD = 8 cm

Now, AB = vAD2 +BD2 = v42 +82 cm = √80cm \(\sqrt{\mathrm{AD}^2+\mathrm{BD}^2}=\sqrt{4^2+8^2}\)=4√5cm

Hence the length of BD = 8 cm and AB = 4√5 cm

Example 2. AB is a diameter of a circle with centre O, P is any point on the circle, the tangent drawn through the point P intersects the two tangents drawn through the points A and B at the points Q and R respectively. If the radius of the circle be r, then prove that PQ.PR = r².

Solution:

Given:

AB is a diameter of a circle with centre O, P is any point on the circle, the tangent drawn through the point P intersects the two tangents drawn through the points A and B at the points Q and R respectively.

Let AB is a diameter of the circle with centre at O.

P is any point on the circle. AE and BF are two tangents drawn through A and B which intersect the tangent ST drawn at P at the points Q and R respectively.

To prove PQ.PR = r² (where r = radius of the circle)

Construction: Let us join O, Q and O, R.

Proof: ∵ the lengths of the tangents to a circle drawn from an external point of the circle are equal and they subtend equal angles at the centre,

∴ QA = QP, PR = BR and ∠AOQ = ∠POQ, ∠BOR = ∠POR

∴ ∠AOQ = ∠POQ = \(\frac{1}{2}\) ∠AOP and ∠BOR = ∠POR = \(\frac{1}{2}\) ∠BOP.

∴ ∠QOR = ∠POQ + ∠POR = \(\frac{1}{2}\) ∠AOP + \(\frac{1}{2}\) ∠BOP = \(\frac{1}{2}\) (∠AOP + ∠BOP)

= \(\frac{1}{2}\) x ∠AOB = \(\frac{1}{2}\) x 180° =90° [∵ ∠AOB = 1 straight angle = 180°]

Again, OP ⊥ QR [∵ OP is a radius passing through the point of contact.]

∴ ΔPOQ ~ ΔPOR. PQ OP

∴ by Thales’ theorem, \(\frac{P Q}{O P}=\frac{O P}{P R}\) or, PQ.PR = OP²

or, PQ.PR = r² [∵ OP = r = radius of the circle]

Hence PQ.PR = r² (Proved)

Example 3. Modhurima have drawn a semicircle with a diameter AB. A perpendicular is drawn on AB from any point C on AB which intersects the semicircle at the point D. Prove that CD is a mean- proportion of AC and BC.

 

Solution:

Given:

Modhurima have drawn a semicircle with a diameter AB. A perpendicular is drawn on AB from any point C on AB which intersects the semicircle at the point D.

Let ADB is a semicircle with centre at O, of which AB is a diameter. A perpendicular CD is drawn at C on AB which intersects the semicircle at D.

To prove: CD is a mean-proportion of AC and BC, i.e., AC x BC = CD².

Construction: Let us join A, D and B, D.

Proof: ∵ ADB is a semicircle, ∠ADB is a semicircular angle.

∴ ∠ADB = 1 right angle.

Again, CD is the perpendicular drawn from the right angular point D on the hypotenuse AB.

\(\frac{AC}{CD}=\frac{CD}{BC}\) = or AC x BC = CD².

Hence CD is a mean proportion of AC and BC. (Proved).

Example 4.In right angled triangle ABC, ∠A is a right angle. AD is perpendicular on the hypotenuse BC. prove that \(\frac{\triangle A B C}{\triangle A C D}=\frac{B C^2}{A C^2}\)

Solution:

Given:

In right angled triangle ABC, ∠A is a right angle. AD is perpendicular on the hypotenuse BC.

In ΔABC, ∠A = right angle.

AD is the perpendicular drawn from the right angular point A to the hypotenuse BC.

∴ ΔABC ~ ΔACD.

\(\frac{\text { Area of } \triangle \mathrm{ABC}}{\text { Area of } \Delta \mathrm{ACD}}=\frac{\mathrm{BC}^2}{\mathrm{AC}^2}\)

[∵ BC and AC are corresponding sides]

Hence \(\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{ACD}}=\frac{\mathrm{BC}^2}{\mathrm{AC}^2}\) (proved)

Example 5. AB is a diameter of a circle with centre O. A line drawn through the point A intersects the circle at the point C and the tangent through B at the point D. Prove that

1. BD² = AD.DC
2. The area of the rectangle formed by AC and AD for any straight line is always equal. 

Solution:

Given:

AB is a diameter of a circle with centre O. A line drawn through the point A intersects the circle at the point C and the tangent through B at the point D.

Let AB be a diameter of the circle with centre at O. A straight line drawn through A y intersects the circle at C and the tangent BT drawn at B at the point D.

To prove  TBD² = AD

1. DC.
2. The area of the rectangle formed by. AC and AD for any straight line is always equal.

Construction: Let us join B, C

Proof:

1. In ΔABD and ΔBCD, ∠ABD = ∠BCD [∵ BT is a tangent at B and AB is a radius through point of contact,

∴ ∠ABD = 1 right angle; Again, ∠ACB is a semicircular angle. ∴ ∠ACB = 1 right angle.]

Now, ∠BDC = 90° – ∠DBC [∵  ∠BCD =1 right angle]

= ∠ABC [∵ ∠ABD = 1 right angle]

∴ ΔABD ~ ΔBCD.

∴ by Thales’ theorem \(\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AD}}{\mathrm{BD}}\)

∴ BD² = AD x DC. (Proved).

2. Again, in ΔABD and ΔABC, ∠ABD = ∠ACB [∵ each is a right angle]

∠ADB = 90° – ∠DBC = ∠ABC

∴ ΔABD-ΔABC

∴ \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{AD}}{\mathrm{AB}}\)  [by Thals’s theorem] or, AB² =  AC x AD

or, (constant)² = AC x AD [∵ the value of diameter AB is constant]

∴ AC x AD = constant

∴ The area of the rectangle formed by AC and AD for any straight line is always equal. (Proved)

Example 6. The length of the shadow of a stick of length 6 cm is 4 cm. At the same time, if the length of the shadow of a tower be 28 metres, then find the height of the tower.

Solution:

Given:

The length of the shadow of a stick of length 6 cm is 4 cm. At the same time, if the length of the shadow of a tower be 28 metres,

Let CP be the length of the shadow of the stick CD of length 6 cm. CP = 4 cm.

At the same time, AP is the length of the shadow of the tower AB.

As per question, AP = 28 cm

We have to find the height of the tower AB.

Now, in ΔAPB and ΔCPD, ∠PAB = ∠PCD [∵ each is perpendicular to the base, right angles.]

∠P is common to both the triangles,

∴ ΔAPB ~ ΔPCD

Hence by Thales theorem, \(\frac{\mathrm{AB}}{\mathrm{CD}}\) = \(\frac{\mathrm{AP}}{\mathrm{CP}}\)

or, \(\frac{\mathrm{AB}}{6 \mathrm{~cm}}=\frac{28 \mathrm{metres}}{4 \mathrm{~cm}}\)

or, AB= \(\frac{2800 \mathrm{~cm} \times 6 \mathrm{~cm}}{4 \mathrm{~cm}}\) =4200 cm=42

Hence the height of the tower is 42 metres.

Example 7. Prove by the theorem of Thales that the third side of a triangle is parallel to the line segment obtained by joining the mid-points of any two sides of the triangle and is half in length of its third side.

Solution: Let P and Q be the mid-points of the two sides AB and AC of the ΔABC.

Let us join P and Q.

Proof: P and Q are the mid-points of AB and AC respectively.

∴ AP = BP and AQ = CQ.

\(\frac{\mathrm{AP}}{\mathrm{BP}}=\frac{\mathrm{BP}}{\mathrm{BP}}=1\) and \(\frac{\mathrm{AQ}}{\mathrm{CQ}}=\frac{\mathrm{CQ}}{\mathrm{CQ}}=1\)

Then, \(\frac{\mathrm{AP}}{\mathrm{BP}}\) = \(\frac{\mathrm{AQ}}{\mathrm{CQ}}\)

∴ by the conversed theorem of Thales’ theorem, PQ || BC. (Proved)

Again in ΔAPQ and ΔABC,

∠APQ = ∠ABC [∵ similar angles]

∠AQP = ∠ACB [∵ similar angles]

and ∠A is common to both the triangles ΔAPQ ~ ΔABC.

∴ by Thales’ theorem, \(\frac{\mathrm{PQ}}{\mathrm{BC}}=\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{AQ}}{\mathrm{AC}}\)

∴ \(\frac{\mathrm{PQ}}{\mathrm{BC}}=\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{AP}}{\mathrm{2AP}}\)

[∵ P is the mid-point of AB, ∴ AB = 2AP.]

or, \(\frac{\mathrm{PQ}}{\mathrm{BC}}=\frac{\mathrm{AP}}{\mathrm{2 AP}}=\frac{1}{2}\)

or, PQ = \(\frac{1}{2}\) BC

Hence PQ || BC and PQ = \(\frac{1}{2}\) BC (proved)

Example 8. Two parallel straight lines intersect three concurrent straight lines at the points A, B, C and X, Y, Z respectively. Prove that AB : BC = XY : YZ.

Solution:

Given:

Two parallel straight lines intersect three concurrent straight lines at the points A, B, C and X, Y, Z respectively.

Let two parallel straight lines CD and EF intersects three straight lines P₁Q₁, P₂Q₂, P₃Q₃concurrent at O at the points A, B, C and X, Y, Z respectively.

To prove AB : BC = XY : YZ.

Proof: In ΔAOB and ΔXOY, ∠OAB = alternate ∠OXY, [CD || EF and P₁Q₁ is their transversal]

∠OBA = ∠OYX (for similar reason)

∴ ΔAOB ~ ΔXOY

∴ by Thales’ theorem, \(\frac{A B}{X Y}=\frac{O B}{O Y}\)……..(1)

Again, in ΔBOC and ΔYOZ, ∠OBC = alternate ∠OYZ,

[∵ CD || EF and P₂O₂, is their transversal]

∠BOC = ∠YOZ [∵ opposite angles] ΔBOC ~ ΔYOZ

∴ by Thales’ theorem \(\frac{\mathrm{BC}}{\mathrm{YZ}}=\frac{\mathrm{OB}}{\mathrm{OY}}\)……..(2)

∴ From (1) and (2) we get, \(\frac{A B}{X Y}\)=\(\frac{B C}{Y Z}\)

or, \(\frac{A B}{B C}=\frac{X Y}{Y Z}\)

Hence AB : BC = XY : YZ. (Proved)

Example 9. Kamala have drawn a trapezium PQRS of which PQ || SR. If the diagonals PR and QS intersect each other at O, then prove that OP : OR = OQ : OS; If SR = 2PQ, then prove that O is a point of trisection of both the diagonals.

Solution:

Given:

Kamala have drawn a trapezium PQRS of which PQ || SR. If the diagonals PR and QS intersect each other at O, then prove that OP : OR = OQ : OS; If SR = 2PQ,

In the trapezium PQRS, PQ || SR.

The diagonals PR and QS intersect each other at O.

To prove 1. OP : OR = OQ : OS.
2. If SR = 2PQ, then O is a point of trisection of both the diagonals, i.e., OP : OR = 1:2 and OQ : OS = 1 : 2.

Proof: In ΔPOQ and ΔSOR, ∠OPQ = alternate ∠ORS, ∠OQP = alternate ∠OSR.

∴ ΔPOQ ~ ΔSOR.

∴ by Thales’ theorem, \(\frac{\mathrm{OP}}{\mathrm{OR}}=\frac{\mathrm{OQ}}{\mathrm{OS}}=\frac{\mathrm{PQ}}{\mathrm{SR}}\)…….(1)

From (1) we get, \(\frac{\mathrm{OP}}{\mathrm{OR}}=\frac{\mathrm{OQ}}{\mathrm{OS}}\)

or, \(\frac{\mathrm{OP}}{\mathrm{OR}}=\frac{\mathrm{PQ}}{2 \mathrm{PQ}}\) [∵ SR = 2PQ]

or, \(\frac{\mathrm{OP}}{\mathrm{OR}}\) = \(\frac{1}{2}\)

or, OP: OPR = 1: 2

Similarly, \(\frac{\mathrm{OQ}}{\mathrm{OS}}=\frac{\mathrm{PQ}}{\mathrm{SR}}\)

or, \(\frac{\mathrm{OQ}}{\mathrm{OS}}=\frac{\mathrm{PQ}}{2 \mathrm{PQ}}\)

or, \(\frac{\mathrm{OQ}}{\mathrm{OS}}=\frac{1}{2}\)

or, OQ: OS = 1: 2

Hence O is a point of trisection of both the diagonals PR and QS. (Proved) .

Example10. PQRS is a parallelogram. If a straight line EF through S intersects PQ and extended RQ at the points X and Y respectively, then prove that PS : PX = QY : QX = RY : RS

 

Solution:

Given:

PQRS is a parallelogram. If a straight line EF through S intersects PQ and extended RQ at the points X and Y respectively,

Let PQRS is a parallelogram.

Class 10 Maths Wbbse Solutions

The straight line EF drawn through S intersects PQ and extended RQ at the points X and Y respectively.

To prove PS : PX = QY : QX = RY : RS.

Proof: In ΔPXS and ΔQXY, ∠PSX = ∠QYX.

[∵ SP || RY and SY is their transversal, ∠PSX = alternate ∠QYX] and ∠PXS = ∠QXY, [opposite angles]

∴ ΔAPXS ~ ΔQXY

∴ by Thales’ theorem,

\(\frac{P S}{Q Y}=\frac{P X}{Q X}=\frac{S X}{X Y}\)…….(1)

∴ \(\frac{P S}{Q Y}=\frac{P X}{Q X}\)

or, \(\frac{P S}{P X}=\frac{Q Y}{Q X}\)

or, PS : PX = QY: QX…….(2)

Again, in ΔRSY, XQ || SR [∵ PQ || SR]

∴ ∠YXQ = similar ∠YSR, ∠YQX = ∠YRS [similar angles]

∴ in ΕXYQ and ΔYSR, ∠YXQ = ∠YSR and ∠YQX

= ∠YRS, ΔXYQ ~ ΔYSR.

∴ by Thales’ theorem,\(\frac{\mathrm{QY}}{\mathrm{RY}}=\frac{\mathrm{QX}}{\mathrm{RS}}=\frac{\mathrm{XY}}{\mathrm{SY}}\)

∴ \(\frac{S R}{Y R}=\frac{Q X}{Q Y}\)…….(3)

From (2) and (3) we get, \(\frac{\mathrm{PS}}{\mathrm{PX}}=\frac{\mathrm{QY}}{\mathrm{QX}}=\frac{\mathrm{RY}}{\mathrm{RS}}\)

PS : PX = QY : QX = RY : RS. (Proved)

Example 11. ΔABC and ΔPQR are two similar acute triangles. Their circumcentres are X and Y respectively. If BC and QR be two corresponding sides, then prove that BX : QY = BC : QR.

Solution:

Given:

ΔABC and ΔPQR are two similar acute triangles. Their circumcentres are X and Y respectively. If BC and QR be two corresponding sides,

ΔABC and ΔPQR are two similar acute triangles.

The circumcentre of ΔABC is X and the circumcentre of ΔPQR is Y.

To prove BX : QY = BC : QR.

Proof: ΔABC – ΔPQR, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R.

Now, X and Y are the circumcentres of ΔABC and ΔPQR respectively.

∴ ∠BXC = 2 ∠A and ∠QYR = 2∠P, [angles in central is twice the angles in circle.]

But ∠A = ∠P, ∴ ∠BXC = ∠QYR

Again, BX = CX [∵ radii of same circle]

∴ ∠XBC = ∠XCB

Now, in ΔBXC, ∠XBC + ∠BXC + ∠XCB = 180°

[∵ sum of three angles of a traingle is 180°]

or, ∠XBC + ∠XBC + ∠BXC = 180° [∠XCB = ∠XBC]

or, 2∠XBC + ∠BXC = 180° or, 2 ∠XBC = 180° – ∠BXC

or, ∠XBC = \(\frac{1}{2}\) (180°-∠BXC)…….(1)

= \(\frac{1}{2}\) (180° – ∠QYR) [∠BXC = ∠QYR]

= \(\frac{1}{2}\) (∠QYR + ∠YQR + ∠YRQ – ∠QYR]

= \(\frac{1}{2}\) (∠YQR + ∠YRQ)

= \(\frac{1}{2}\) (∠YQR + ∠YQR) [∠Q = YR, ∴ ∠YRQ = ∠YQR]

= \(\frac{1}{2}\) X 2 ∠YQR = ∠YQR

∴ ∠XBC = ∠YQR

Similarly, ∠XCB = ∠YRQ

∴ in ΔXBC and ΔYQR, ∠BXC = ∠QYR, ∠XBC = ∠YQR and ∠XCB and ∠YRQ

∴ ΔXBC ~ ΔYQR

Bby Thales’ theorem, \(\frac{\mathrm{XB}}{\mathrm{YQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}\)

or, BX : QY = BC : QR

Hence BX : QY = BC : QR. (Proved)

Example 12. The two chords PQ and RS of a circle intersects each other at the point X within the circle. By joining P, S and R, Q prove that ΔPXS and ΔRSQ are similar. From this also prove thatPX : XQ = RX : XS. 

or,

If two chords of a circle intersect internally, then the rectangle of two parts of one is equal to the rectangle of two parts of other.

Solution:

Given:

The two chords PQ and RS of a circle intersects each other at the point X within the circle. By joining P, S and R, Q

Let two chords PQ and RS of a circle with centre at 0 intersect internally at the point X.

Let us join P, S, R, Q and S, Q.

To prove ΔPXS ∼ ΔRSQ and hence PX : XQ = RX : XS

Construction: Let us join R, Q.

Proof: In ΔPXS and ΔRXQ, ∠PXS = ∠RXQ Opposite angles] and ∠SPQ = ∠SRQ [each is angles in circle produced by the chord SQ]

∴ ΔPXS ∼ ΔRXQ.

∴ by Thales’ theorem, \(\frac{P X}{X Q}=\frac{R X}{X S}\)

∴ PX : XQ = RX : XS. (Proved)

Example 13. The two points P and Q are on a straight line. At the points P and Q, PR and QS are perpendicular on the straight line. PS and QR intersect each other at the point O. OT is perpendicular on PQ. Prove that \(\frac{1}{O T}=\frac{1}{P R}+\frac{1}{Q S}\)

Solution:

Given:

The two points P and Q are on a straight line. At the points P and Q, PR and QS are perpendicular on the straight line. PS and QR intersect each other at the point O. OT is perpendicular on PQ.

Let P and Q be two points on the straight line EF.

PR and QS are two perpendiculars on EF drawn at P and Q respectively.

PS and QR intersect each other at the point O. OT ⊥ PQ.

To prove \(\frac{1}{\mathrm{OT}}=\frac{1}{\mathrm{PR}}+\frac{1}{\mathrm{QS}}\)

Proof: PR, QS and OT are each perpendiculars to the straight line EF at the points P, Q and T respectively.

∴ PR || TO || QS.

Now, in ΔPQR, TO || PR.

∴ by Thales’ theorem, \(\frac{\mathrm{OT}}{\mathrm{PR}}=\frac{\mathrm{TQ}}{\mathrm{PQ}}\)……(1)

Similarly, in ΔPQS, TO || QS,

∴ by Thales’ theorem, \(\frac{\mathrm{OT}}{\mathrm{QS}}=\frac{\mathrm{PT}}{\mathrm{PQ}}\)……(2)

Then from (1) and (2) we get, \(\frac{\mathrm{OT}}{\mathrm{PR}}+\frac{\mathrm{OT}}{\mathrm{QS}}=\frac{\mathrm{TQ}}{\mathrm{PQ}}+\frac{\mathrm{PT}}{\mathrm{PQ}}\)

or, \(\mathrm{OT}\left(\frac{1}{\mathrm{PR}}+\frac{1}{\mathrm{QS}}\right)=\frac{\mathrm{TQ}+\mathrm{PT}}{\mathrm{PQ}}\)

or, \(\mathrm{OT}\left(\frac{1}{\mathrm{PR}}+\frac{1}{\mathrm{QS}}\right)=\frac{\mathrm{PQ}}{\mathrm{PQ}\)

or, \(\mathrm{OT}\left(\frac{1}{\mathrm{PR}}+\frac{1}{\mathrm{QS}}\right)=1\)

or, \(\frac{1}{\mathrm{PR}}+\frac{1}{\mathrm{QS}}=\frac{1}{\mathrm{OT}}\)

Hence \(\frac{1}{\mathrm{OT}}=\frac{1}{\mathrm{PR}}+\frac{1}{\mathrm{QS}}\) (proved)

Example 14. ΔABC is inscribed in a circle; AD is a diameter of the circle and AE is perpendicular on the side BC, which intersects the side BC at the point E. Prove that ΔAEB and ΔACD are similar. From this also prove that AB x AC = AE x AD.

Solution:

Given:

ΔABC is inscribed in a circle; AD is a diameter of the circle and AE is perpendicular on the side BC, which intersects the side BC at the point E.

ΔABC is inscribed in a circle; AD is a diameter of the circle and AE is perpendicular on the side BC, which intersects the side BC at the point E.

Let ΔABC is inscribed in a circle with centre O, AD is the diameter of that circle. AE ⊥ BC.

To prove ΔAEB ~ ΔACD and AB x AC = AE x AD.

Construction: Let us join C,D.

Proof: AE ⊥ BC, ∴ ∠AEB = 90° ……..(1)

Again, AD is a diameter of the circle and ∠ACD is a semicircular angle.

∴ ∠ACD = 1 right angle or 90°……. (2)

Two angles in circle ∠ABC and ∠ADC are produced by the chord AC.

∴ ∠ABC = ∠ADC or ∠ABE = ∠ADC……..(3)

Then in ΔAEB and ΔACD, ∠AEB = ∠ACD [from (1) and (2)] and ∠ABE = ∠ADC [from (3)]

ΔAEB ~ ΔACD (Proved)

Now, since ΔAEB ~ ΔACD,

∴ by Thales’ theorem, \(\frac{\mathrm{AB}}{\mathrm{AE}}=\frac{\mathrm{AD}}{\mathrm{AC}}\)

or, AB x AC = AE x AD

Hence ΔAEB ~ ΔACD and AB x AC = AE x AD. (Proved)

WBBSE Solutions For Class 10 Maths Solid Geometry Chapter 7 Construction Of Circumcircle And Incircle Of A Triangle

Circumcircle, circumcentre And Circumradius Of A Triangle

Circumcircle of a triangle

Definition: The circle passing through the three vertices of a triangle, is called the circumcircle of the triangle.

The circle having O as the centre and passing through the vertices A, B and C is the circumcircle of the triangle ΔABC.

Circumcentre of a triangle

Definition: The centre of the circumcircle of a triangle is called the circumcentre of the triangle.

∴ O is the circumcentre of the ΔABC. Again, ∵ O is the circumcentre of Δ ABC,

∴ the distances of all the points on the circumcircle, from O are equal. ∵ OA = OB = OC

Read and Learn More WBBSE Solutions for Class 10 Maths

i.e., ΔBOC, ΔCOA and ΔAOB are all isosceles triangles.

We know that the perpendicular drawn from any vertex of an isosceles triangle to its base bisects the base.

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∴ The perpendiculars OD, OE and OF from O to the sides BC, CA and AB respectively of the ΔABC have bisected the corresponding bases and these perpendicular bisectors intersect each other at a point O.

Therefore, we can say that the perpendicular bisectors of the sides of a triangle are concurrent and the point at which the perpendicular bisectors intersect each other, is the circumcentre of the triangle.

Thus, we take the point at which any two perpendicular bisectors meet one another as the circumcentre of the triangle.

If the triangle be

1. an acute-angled triangle, then the circumcentre of the triangle lie inside the triangle.

2. a right-angled triangle, then the circumcentre of the triangle lie on the hypotenuse of the triangle and it bisects the hypotenuse, i.e., the mid-point of the hypotenuse of a right-angled triangle is its circumcentre and circumradius = \(\frac{1}{2}\) x length of the hypotenuse.

3. an obtuse-angled triangle, then the circumcentre of its circum- circle lie on the outside of the triangle.

Circumradius of a triangle

Definition: The radius of the circumcircle of a triangle is called the circumradius of the triangle.

i.e. the distance of any vertex of a triangle from the point at which the perpendicular bisectors of the sides of a triangle meet each other is called the circumradius of the triangle.

OA or OB or OC is called the circumradius of the ΔABC.

Construction of the circumcircle of a given acute triangle.

Let ΔABC be an acute triangle. We have to construct a circumcircle of this triangle.

Principle: To construct the circumcircle we have to take the point of intersection of the two perpendicular bisectors of any two sides of ΔABC as the centre and the distance of any vertex of the triangle from that centre has to be taken as the radius and then the circumcircle is drawn.

Method of construction:

1. Let us draw the perpendicular bisector PQ of the side BC of ΔABC.
2. Let us draw the perpendicular bisector RS of the side AB of ΔABC.
3. Let PQ and RS intersect each other at O.
4. Let us draw the circle with centre at O and radius equal to OA or OB or OC.

Then the circle with centre at O and radius OA or OB or OC is the required circumcircle of ΔABC.

Proof:

Let us join the points O, A; O, B and O, C.

Now, O lie on the perpendicular bisector of AB.

∴ the point O is equidistant from points A and B.

∴ OA = OB

Similarly, it can be proved that OB = OC.

∴ OA = OB = OC.

∴ the circle with a centre at O and radius O A must pass through the vertices A, B and C of ΔABC.

Hence that very circle is the required circumcircle of ΔABC.

From the above construction, we see that the circumcircle of any acute triangle lie within the triangle.

We shall now consider the case when the triangle is an obtuse angle.

“WBBSE Class 10 circumcircle and incircle solved examples”

Construction of the circumcircle of a given obtuse triangle.

Let ΔABC be an obtuse angle of which ∠A = obtuse. We have to construct the circumcircle of ΔABC.

Principle: The circle, drawn with centre at the point of intersection of two perpendicular bisectors of any two sides of ΔABC and the distance of any one of the vertices of ΔABC from that centre being taken as the radius, is the required circumcircle of the obtuse triangle.

Method of construction:

1. Let us draw the perpendicular bisector PQ of the side AC of ΔABC.
2. Let us draw the perpendicular bisector RS of the side AB of ΔABC.
3. Let PQ and RS intersect at point O.
4. Let us draw the circle with the centre at O and a radius equal to OA or OB or OC.

Then that very circle is the required circumcircle of the obtuse triangle.

Proof:

Let us join O, A; O, B and O, C.

Since O lie on the perpendicular bisector of AB, ∴ O is equidistant from the points A and B.

∴ OA = OB.

Similarly, it can be proved that OB = OC.

∴ OA = OB = OC.

∴ each of A, B and C is equidistant from the point O and the circle passes through A, B and C.

Hence the circle with a centre at O and radius equal to OA or OB or OC is the required circumcircle of ΔABC.

From the above construction, we see that the circumcentre of the ΔABC lie outside the triangle.

Hence we can say that the circumcentre of any obtuse triangle lie outside the triangle.

Now, if the triangle be a right-angled triangle, then where the circumcentre of the triangle lie we shall examine that in the following construction.

Construction of the circumcircle of a given right-angled triangle.

Let ΔABC be a right-angled triangle of which ∠A = right angle. We have to construct a circumcircle of ΔABC.

Method of construction:

1. Let us draw the perpendicular bisector EF of the side AB of ΔABC.
2. Let us draw the perpendicular bisector PQ of the side AC.
3. Let EF and PQ intersect each other at point O on the Side BC.
4. Let us draw a circle with a centre at O and radius equal to OA or OB or OC.

Then the circle with centre at O and a radius to OA or OB or OC is the required circumcircle of ΔABC.

Proof:

Let us join O, A.

O lies on the perpendicular bisector of AB.

the points A and B are equidistant from point O. ∴ OA = OB.

Similarly, it can be proved that OB = OC.

OA = OB = OC.

“Construction of circumcircle of a triangle for Class 10 Maths”

∴ the three vertices A, B and C of the AABC are equidistant from the point O, i.e. the circle passes through the vertices A, B and C of ΔABC.

Hence the circle with centre at O and a radius equal to OA or OB or OC is the required circumcircle of the ΔABC.

From the above construction, we see that the circumcentre of the right-angled triangle AABC lies on the hypotenuse BC of ΔABC, Again, since OB = OC,

∴ O is the mid-point of BC.

∴ We can say that the circumcentre of any right-angled triangle lie on the mid-point of its hypotenuse.

In the following examples we have discussed how to construct circumcircles of different types of triangles using the above constructions.

Solid Geometry Chapter 7 Construction Of Circumcircle And Incircle Of A Triangle Examples

“Chapter 7 circumcircle and incircle exercises WBBSE solutions”

Example 1. Construct the following triangles as directed and then construct the circumcircle in each case stating the location of the circumcentres. Also determine the length of the circumradius in each case.

Solution:

1. An equilateral triangle of sides 6 cm each.

we see that the circum-centre of the triangle lies inside the triangle and the length of the circumradius = 3.4 cm.

2. An isosceles triangle, the base of which is 5.2 cm and the length of each of the equal sides is 7cm.

It is clearly seen that the circumcentre of the triangle lies inside the triangle and the length of its circumradius = 3.75 cm (approx.).

3. An right-angled triangle, the lengths of whose two adjacent sides of right angle are 4 cm and 8 cm respectively.

We see that the circumcentre of the triangle lie on the mid-point of the hypotenuse and the circumradius = 4.5 cm (approx.)

4. A right-angled triangle, the hypotenuse of which is 12 cm and the length of any other side is 5 cm.

We see that the circumcentre of the right-angled triangle lies on the mid-point of its hypotenuse and the length of its circumradius = 6 cm.

“Class 10 Maths steps to draw circumcircle and incircle”

5. A triangle, the length of one of whose sides is 6.7 cm and the magnitudes of its two adjacent angles are 75° and 55° respectively.

We see that the circumcentre of the triangle lies inside the triangle and the length of its circumradius = 3.7 cm (approx.)

6. A triangle ABC of which BC = 5 cm, ∠ABC = 100° and AB = 4 cm.

We see that the circumcentre of the triangle lies outside the triangle and the length of the circumradius = 3.6 cm (approx.).

“Understanding circumcircle and incircle in triangles for Class 10”

Example 2. Given that PQ = 7.5 cm, ∠QPR = 45°, ∠PQR = 75°, ∠QPS = 60°, ∠PQS = 60°. Construct the triangle PQR and PQS in such a way that the points R and S lie on the same side of PQ. Then by drawing the circumcircle ofΕPQR, write the position of the point S within, on and outside the circumcircle. Also, explain your answer.

Solution:

Given:

PQ = 7.5 cm, ∠QPR = 45°, ∠PQR = 75°, ∠QPS = 60°, ∠PQS = 60°.

It is seen that the position of point S is on the circumcircle of ΔPQR, i.e., the points S and R are concyclic.

Explanation:

∠S = 180° – (∠SPQ + ∠QPS) = 180° – (60° + 60°)
= 180° – 120° = 60°

and ∠R = 180° – (∠RPQ + ∠RQP) = 180° – (45° + 75°)
= 180° – 120° = 60°.

∴ ∠S = ∠R.

i.e. ∠S and ∠R are the same angles in circle produced by the same arc PQ.

Hence S and R are concyclic.

Example 3.Given that AB = 5 cm, ∠BAC = 30°, ∠ABC = 60°, ∠BAD = 45°, ∠ABD = 45°. Construct two triangles ΔABC and ΔABD in such a way that point C and D lie on opposite sides of AB. By drawing the circumcircle of ΔABC write the position of the point D with respect to the circumcircle. Also state what other characteristics you are observing here.

Solution:

Given:

AB = 5 cm, ∠BAC = 30°, ∠ABC = 60°, ∠BAD = 45°, ∠ABD = 45°.

We see that the position of the point D is on the circumcircle of ΔABC. i.e., points C and D are concyclic.

Because, here ∠BAC = 30°, ∠ABC = 60°.

∠ACB =180° – (∠BAC + ∠ABC)

= 180° – (30° + 60°) = 180° – 90° = 90°.

∴ in ΔABC, ∠C = 90°, i.e., ∠C is a semicircular angle.

Similarly, ∠ABD = 45°, ∠BAD = 45°

∴ ∠ADB = 180° – (∠ABD + ∠BAD)

= 180° – (45° + 45°) = 180° – 90° = 90°

∴ ∠D = 90°, i.e., ∠D is a semicircular angle.

∴ ∠C + ∠D = 90° + 90° = 180°,

i.e., the sum of two opposite angles of the quadrilateral ABCD is 180°.

Hence the points C and D are concyclic.

“Step-by-step solutions for circumcircle and incircle Class 10”

Example 4. Given that AB = 4 cm, BC = 7 cm, CD = 4 cm, ∠ABC = 60°, ∠BCD = 60°. Construct the quadrilateral ABCD. Then construct the circumcircle of ΔABC and also state what other characteristics you observe.

Solution:

Given:

AB = 4 cm, BC = 7 cm, CD = 4 cm, ∠ABC = 60°, ∠BCD = 60°.

We see that point D lie on the circumcircle of ΔABC.

Also the quadrilateral ABCD is an isosceles trapezium of which AD II BC and AB = CD = 4 cm.

Example 5. Given that PQ = 4 cm, QR = 6 cm. Construct the rectangle PQRS. Also draw the diagonals of the rectangle and without drawing the circumcircle write the position of the centre of the circumcircle of ΔPQR and find the length of circumradius. After all by drawing the circumcircle of ΔPQR verify your answer.

Solution:

Given:

PQ = 4 cm, QR = 6 cm.

The centre of the circumcircle of APQR will be the point of intersection of the diagonals of the rectangle PQRS.

The length of the circumradius = \(\frac{\sqrt{4^2+6^2}}{2}\) cm = \(\frac{\sqrt{52}}{2}\) cm = \(\frac{\sqrt{4 \times 13}}{2}\) cm = \(\frac{2 \sqrt{13}}{2}\) cm = √13 cm

Example 6. If any circular picture is given, then how shall you find its centre? Find the centre of the circle in the adjoining figure.

Solution: We can find the centre of any given circular figure in the following manner:

1. At first let us draw any two chords of any length of the given circle. Let PQ and RS be two such chords of the circle.
2. Let us then draw the perpendicular bisectors of the two chords PQ and RS. Let the bisectors are AB and CD respectively.
3. The point of intersection at which the two perpendicular bisectors AB ad CD of the chords PQ and RS respectively intersect will be the required centre of the circle. Let the perpendicular bisectors AB and CD intersect each other at the point O.

Hence O is the centre of the circle.

Example 7. By drawing the following triangles construct their circumcircles

Solution:

1. The lengths of any two sides of the triangle are 5 cm and 6 cm and their internal angle is 60°.

2. The length of one of the sides of the triangle is 5.4 cm and the two adjacent angles of that side are 60° and 45°.

3. The length of one of the sides of a right-angled triangle is 8 cm and the length of its hypotnuse is 10 cm.

4. The lengths of three sides of the triangle are 5.5 cm, 6.6 cm and 7.7 cm.

“WBBSE Mensuration Chapter 7 practice questions on circles”

Example 8. By constructing the circumcircle of an equilateral triangle of sides 6 cm each, determine the position of the circumcentre and the length of circumradius.

Solution:

We see that the circumcentre of the ΔABC entirely lies inside the triangle and the circumradius = 3.5 cm.

Example 9. Construct an angle of measure 120° without any help of the protractor. Then draw the triangle PQR, where ∠P = 120°. PQ = 4 cm and PR = 3 cm. Also construct the circumcircle of ΔPQR.

Solution: ∠ABC = 120° which has been drawn without any help of a protractor.

We see that the circumcentre of the triangle lies outside the triangle and the circumradius = 3.5 cm.

Example 10. Construct a triangle of sides 5 cm, 7.5 cm and 4 cm. By constructing the circumcircle of the triangle determine the circumradius of the triangle.

Solution: we see that the circumradius = 4 cm.

Example 11. Construct a right-angled triangle of which the two adjacent sides of the right angle are 11 cm and 4.5 cm respectively. Also by drawing the circumcircle of this triangle determine the circumradius.

Solution:

We see that the circumcentre of the ΔABC lies on the mid-point of its hypotenuse AC and the circumradius.

= \(\frac{\sqrt{4.5^2+11^2}}{2}\) cm = 5.94 cm (approx)

Solid Geometry Chapter 7 Construction Of Circumcircle And Incircle Of A Triangle Incircle, Incentre And Inradius Of A Triangle

“Examples of triangle circumcircle and incircle constructions for Class 10”

Incircle

Definition: The circle which completely lie inside any triangle and its circumference just touches all three sides of the triangle is called the incircle of the triangle.

The incentre of the incircle of any triangle is usually denoted by I. In the given figure beside, the circle with the centre I.

1. lie completely inside the ΔABC and
2. its circumference have just touched the sides BC, CA and AB at points D, E and F respectively.

∴ the circle DEFD with centre at I is a incircle of the ΔABC.

Incentre

Definition: The centre of the incircle of any triangle is called the incentre of the triangle.

I is the incentre of the ΔABC.

Now, the circle have touched BC at D, ∴ ID ⊥ BC.

Similarly, IE ⊥ CA and IF ⊥ AB.

Since the distances of any point on the circumference of a circle from its centre are all equal, we have, ID = IE = IF ……(1)

Then, in the triangles ΔAEI and ΔAFI, IE = IF  [∵ by (1)]

∠AEI = ∠AFI [∵ each is a right angle] and AI is common to both.]

∴ ΔAEI ≅ ΔAFI [∵ by the condition of R-H-S congruency ]

∴ ∠ IAE = ∠LAF [∵ they are the similar angles of congruent triangles ]

∴ AI, is the bisector of ∠A.

Similarly, it can be proved that BI and Cl are the bisectors of ∠B and ∠C. respectively.

Therefore, the incentre of the incircle of any ΔABC lie on the bisectors of its angles and they are concurrent.

Hence, the point at which the bisectors of the angles of a triangle intersect is called the incentre of the triangle.

The incentre of any type of triangle lie inside it.

Since the incentre of any triangle lie on the bisectors of its angles,

∴ To find the incentre of any triangle we take the point as incentre at which any two bisectors of its angles intersect and to draw the incircle we take the perpendicular distance of any of its side from this incentre as its radius.

Inradius

Definition: The radius of the incircle of any triangle is called its inradius, i.e., the perpendicular distance of any of the three sides of a triangle from the point at which the bisectors of its angles intersect is called the incentre of the triangle.

ID, IE and IF are all the inradii of ΔABC. Clearly, ID = IE = IF and ID ⊥BC, IE ⊥ CA and IF ⊥ AB.

Ex-circle, Ex-centre and Ex-radius of a triangle

1. Ex-circle: Taking the point of intersection of the bisectors of any two exterior angles of any triangle and the bisector of the third interior angle, as centre and the perpendicular distance of any of its sides from that point as the radius we can draw a circle, which is called the ex-circle of the triangle.

Given beside,the circle DHGD with centre E is a excircle of the ΔABC.

2. Ex-centre: The centre of the ex-circle of any triangle is called its ex-centre.

In besides, E is the ex-centre of the ΔABC.

The perpendicular distance of any side of a triangle from the point of intersection at which the external bisectors of any two angles of the triangle and the internal bisector of the third angle intersect is called the ex-radius of the triangle.

Besides, ED or EG or EH are the ex-radius of the ΔABC. Clearly, ED = EG = EH.

From the above discussion, we see that the incircle and incentre of any type of triangle always lie inside the triangle.

We shall now discuss how the incircle of a triangle is constructed.

Construction of incircle of a given acute triangle.

Let ΔABC be an acute triangle. We have to construct the incircle of this triangle.

Principle: The circle with centre at the point of intersection of the two internal bisectors of any two angles of the triangle and radius as the perpendicular distance a of any side from that centre will be the incircle of the triangle.

Method of construction:

1. Let us draw the triangle as per the given measurement.
2. Now, let us draw the internal bisectors of the angles of ΔABC. Let the bisectors of the angles intersect each other at the point I.
3. A perpendicular ID is drawn from point I to the side BC which intersects BC at point D.
4. Now, let us draw a circle with centre at I and a radius equal to ID, which touches the sides BC, CA and AB of the ΔABC at the points D, E and F respectively.

Hence the circle with centre I and radius ID will be required to incircle of the triangle.

∴ If the triangle be an obtuse or a right-angled triangle.

Construction of incircle of a given obtuse triangle.

Let ΔABC be an obtuse angle of which ∠B is obtuse. We have to construct the incircle of this triangle.

Method of construction:

1. Let us construct the triangle as per the given measurement.
2. Let us draw the internal bisectors of ∠BAC and ∠ACB. Let the bisectors intersect each other at the point I.
3. Let us draw ID perpendicular to BC from the point I, which intersects the side BC at D.
4. Now, let us draw a circle with the centre at I and with radius equal to ID.

Hence that very circle is the required incircle of the ΔABC.

Now we shall discuss how the incircle of a given right-angled triangle is constructed.

Construction of incircle of a given right-angled triangle.

Let ΔABC be a right-angled triangle of which ∠B = right angle. We have to construct an incircle of the ΔABC.

Method of construction:

1. Let us construct the triangle ABC of which ∠B = right angle.
2. Let us draw the internal bisectors of ∠BAC and ∠ACB which intersect each other at the point I.
3. Now, let us draw a perpendicular ID on the side AC from the point I, which intersects AC at point D.
4. Let us then draw a circle with centre at I and with radius equal to ID.

Hence that very circle is the required incircle of the ΔABC.

In the following examples how incircles of different triangles are constructed is discussed.

Solid Geometry Chapter 7 Construction Of Circumcircle And Incircle Of A Triangle Incircle, Incentre And Inradius Of A Triangle Examples

“WBBSE Class 10 Maths solved problems on circumcircle and incircle”

Example 1. The lengths of three sides of a triangle are 7 cm, 6 cm and 5.5 cm. Construct the triangle and determine the length of the inradius of that triangle.

Solution:

Given:

The lengths of three sides of a triangle are 7 cm, 6 cm and 5.5 cm.

Let the lengths of the sides AB, BC and CA of ΔABC are 7 cm, 6 cm and 5.5 cm respectively. We have to construct the triangle.

 

The circle with centre at I is the required incircle of ΔABC and the length of its inradius = 1.7 cm.

Example 2. The length of two sides of a triangle are 7.6 cm and 6 cm and the internal angle between these two sides is 75°. Construct the triangle and then by drawing the incircle of this triangle, determine its inradius of it.

Solution:

Given:

The length of two sides of a triangle are 7.6 cm and 6 cm and the internal angle between these two sides is 75°.

The circle with centre I is the required incircle of the ΔABC and the inradius of this triangle = 2 cm.

The inradius of this triangle = 2 cm.

Example 3. The lengths of two adjacent sides of the right angle of a right-angled triangle are 7 cm and 9 cm. Construct the right-angled triangle. Also construct the incircle of this triangle.

Solution:

Given:

The lengths of two adjacent sides of the right angle of a right-angled triangle are 7 cm and 9 cm.

The circle with centre at I is the required incircle of the ΔABC and the inradius of it = 2.3 cm (approx.)

Example 4. Construct a right-angled triangle, the hypotenuse of which is 11.4 cm and the length of another side is 9 cm. Then by constructing the incircle of this triangle find its inradius of it.

Solution:

Let ΔABC be a right-angled triangle of which ∠A = right angle, hypotenuse BC = 11.4 cm and another side AC = 9 cm.

We have to construct the triangle.

The circle with centre I is the required incircle of the ΔABC and the inradius = 2.35 cm (approx.)

Example 5. The base of an isosceles triangle is 10 cm and each of the equal angles is 45°. By drawing this triangle, construct the incircle of it and hence determine its inradius.

Solution:

Given:

The base of an isosceles triangle is 10 cm and each of the equal angles is 45°.

The circle with centre at I is the required incircle of the ΔABC and the inradius of it = 2 cm (approx.)

Example 6. Construct an equilateral triangle of sides 7 cm each. By constructing its circumcircle and incircle, determine whether there is any relation between their circumradius and inradius or not.

Solution:

The circle with centre at O and passing through the vertices A, B and C of ΔABC is the required circumcircle of the ΔABC and its circumradius = 4 cm.

Again, the circle with centre at O and passing through the points D, E and F is the required incircle of the ΔABC and its inradius = 2 cm.

Hence the circumradius of ΔABC is double of its inradius, which is the required relation.

Example 7. The base of an isosceles triangle is 8.4 cm and the length of each of its equal sides is 9 cm. Construct this triangle and then by drawing its incircle, determine its inradius.

Solution:

Given:

The base of an isosceles triangle is 8.4 cm and the length of each of its equal sides is 9 cm.

The circle with centre I is the required incircle of ΔABC and its inradius = 2.5 cm

Example 8. The two adjacent sides of the right angle of a right-angled triangle are 3 cm and 4 cm. Construct the triangle and then by drawing its incircle, determine the inradius of this triangle.

Solution:

Given:

The two adjacent sides of the right angle of a right-angled triangle are 3 cm and 4 cm.

Let ΔABC be a right-angled triangle of which ∠A = right angle. AB and AC are two adjacent sides of its right angle ∠A, the lengths of which one 3 cm and 4 cm respectively. We have to construct this triangle.

The circle with centre I is the required incircle of the ΔABC and its inradius = 1 cm (approx.)

Example 9. The length of the hypotenuse of a right-angled triangle is 13 cm and its another side is 5 cm. Construct the triangle and then by drawing the incircle of this triangle determine its in radius.

Solution:

Given:

The length of the hypotenuse of a right-angled triangle is 13 cm and its another side is 5 cm.

Let the hypotenuse AC = 13 cm of the ΔABC and the other side BC = 5 cm. We have to construct this triangle.

The circle with centre I is the required incircle of the ΔABC and its inradius = 2 cm (approx.)

“Circumradius and inradius calculations for triangles in Class 10 Maths”

Example 10. Three sides of a triangle are 5 cm, 6 cm and 9 cm respectively. Construct the triangle and then by drawing its incircle, determine the inradius of this triangle.

Solution:

Given:

Three sides of a triangle are 5 cm, 6 cm and 9 cm respectively.

Let in the ΔABC, AB = 5 cm, AC = 6 cm and BC = 9 cm. We have to construct the triangle.

The circle with the centre at I is the required incircle of the ΔABC and its radius = 1.5 cm

WBBSE Solutions For Class 10 Maths Solid Geometry Chapter 6 Pythagoras Theorem

You have studied about what a right angle is, which are hypotenuse, perpendicular and base of a right angle, even about Pythagoras’ theorem and its proof and applications.

In the present chapter, we shall discuss Pythagoras’ theorem and its various applications in real problems more detail according to that earlier studies.

Let ABC be a triangle of which ∠A = right angle.

Then the opposite side of ∠A is BC. That very BC is called the hypotenuse.

The triangle ΔABC have more two sides except BC. such as AB and AC.

Then anyone of AB and AC can be taken as perpendicular and the rest other as the base.

In particular, AC is the base and AB is assumed to be perpendicular.

We are eager to know whether there is any relation between AB, BC and CA or not.

Read and Learn More WBBSE Solutions for Class 10 Maths

You have already known that there is a relation among hypotenuse, perpendicular and base, which is true for all right-angled triangles.

The relation is The square drawn on the hypotenuse of a right-angled triangle is equal to the sum of the squares drawn on its perpendicular and base.

It is known as Pythagoras’ theorem. A question may arise now to you who is Pythagoras?

Pythagoras was a famous philosopher and mathematician in ancient Greece. He was born in Saos, a colony of Greece.

His duration of life was from 580 B.C.- 495 B.C.

WBBSE Solutions for Class 10 History	WBBSE Solutions for Class 10 Geography and Environment
WBBSE Class 10 History Long Answer Questions	WBBSE Solutions for Class 10 Life Science And Environment
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WBBSE Class 10 History Very Short Answer Questions	WBBSE Solutions for Class 10 Physical Science and Environment
WBBSE Class 10 History Multiple Choice Questions

 

Another famous mathematician Thales was his teacher. He invented many mathematical hypothesis except the theorem regarding the right-angled triangle.

The hypothesis regarding the sum of the angles of a triangle is also a pythagorian type.

He was fond of songs and had earned a vast knowledge by travelling Egypt and Bharat.

He had to take his livelihood in a colony of south Italy and so many men recognise him as an Italian mathematician.

However, we shall now discuss Pythagoras’ theorem and how it is proved.

WBBSE Solutions For Class 10 Maths Solid Geometry Chapter 6 Pythagoras Theorem Theorems

Pythagoras Theorem: In any right-angled triangle the area of the square drawn on the hypotenuse is equal to the sum of the squares drawn on other two sides.

Given: ABC is a right-angled triangle of which ∠A is a right angle.

To prove BC² = AB² + AC².

Construction: AD is drawn perpendicular on the hypotenuse BC from the right angular point A, which intersects BC at D.

Proof: In the right-angled ΔABC, AD is perpendicular on the hypotenuse BC.

∴ ΔABC ~ΔABD

∴ \(\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{BD}}{\mathrm{AB}}\) or AB² = BC.BD …….(1)

Again, ΔABC ~ ΔACD

∴ \(\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{DC}}{\mathrm{AC}}\) or AB² = BC.BD …….(2)

Now adding (1) and (2) we get,

AB² + AC² = BC.BD + BC.CD = BC (BD + CD) = BC.BC = BC².

∴ BC² = AB² + AC². (Proved)

We shall now prove this theorem by other two methods.

Method 1. Let in ΔABC, ∠C = 90°, BC = a, AC = b and AB = c

To prove c² = a² + b².

Construction: Let us drawn a square PQRS, the side of which is equal to the sum of (a + b).

The parts PD = QE = RF = SG = a are cut off from the sides PQ, QR, RS and SP respectively.

∴ QD = RE = SF = PG = b.

Let us now join points D, E, F, and G.

Proof: The triangles ΔDQE, ΔREF, ΔSFG and ΔPDG are all right-angled triangles and they are congruent to each other.

Since two side of each of these triangles are equal to the corresponding sides of any other and the internal angles of these two sides of each are right angles, i.e., equal.

Again, these triangles and ΔABC are congruent.

∴ DE = EF = FG = GD = AB = c.

Now, since ΔPDG and ΔDQE are congruent, ∴ PDG = DEQ.

∴ ∠PDG + ∠QDE = ∠DEQ + ∠QDE = 1 right angle. [∠Q = right angle]….. (1)

∴ ∠PDG + ∠GDE + ∠QDE = straight angle = 2 right angle

or, 1 right angle + ∠GDE = 2 right angles [from (1)]

or, ∠GDE = 2 right angle – 1 right angle = 1 right angle.

Similarly, it can be proved that each of the angled ∠DEF, ∠EFG and ∠FGD is a right angle.

∴ the sides of the quadrilateral DEFG are all equal and each of its angles is a right angle.

∴ □DEFG is a square and its area = c².

Now, ΔDQE = \(\frac{1}{2}\) x a x b = \(\frac{1}{2}\)ab.

∴ ΔDQE + ΔREF + ΔSGF + ΔPDG = \(\frac{1}{2}\) ab x 4 = 2ab

Then according to (Square DEFG) = (Square PQRS) – (ΔDQE – ΔREF + ΔSGF + ΔPDG)

= (a + b)² – 2ab = a² + b² + 2ab – 2ab – a² + b²

∴ c² = a² + b². (Proved)

Method 2. Let ABC be a right-angled triangle in which ∠A is a right angle.

To prove BC² = AB² + AC².

Construction: Let us draw squares ABDE, ACFG and BCHK on the sides AB. AC and BC respectively.

A straight line AL is drawn from A parallel to BK such that it intersects BC at O and KH at L.

Let us join C, D and A, K.

Proof: Each of the ∠BAC and ∠BAE is right angle and these are adjacent angles.

∴ AC and AE lie on the same straight line.

Now, ∠ABD = ∠CBK [∵ each is right angle]

∴ ∠ABD + ∠ABC = ∠ABC + ∠CBK

or, ∠DBC = ∠ABK.

Now, in ΔDBC and ΔABK.,BD = AB, BC = BK and internal ∠DBC = internal ∠ABK.

∴ ΔDBC ≅ ΔABK

Now, square ABDE and ΔDBC are on the same base BD and lie between the same pair of parallels is BD and CE.

∴ Square ABDE = 2 ΔDBC…….(1)

Again, the rectangle BKLO and ΔABK lie on the same base BK and between the same pair of parallels BK and AL.

∴ rectangle BKLO = 2 ΔABK…….(2)

But ΔDBC ≅ ΔABK.

∴ rectangle BKLO = square ABDE

Similarly, it can be proved that rectangle CHLO = square ACFG

∴ rectangle BKLO + rectangle CHLO = square ABDE + square CFGA.

∴ square BCHK = square ABDE + square CFGA.

∴ BC² = AB² + AC². (Proved)

A question now obviously arises whether the converse of Pythagoras’ theorem is always true or not. We shall now prove this theorem logically by the method of geometry.

Converse of Pythagoras’ Theorem: If in a triangle, the area of a square drawn on one side is equal to the sum of the areas of squares drawn on the other two sides, then the angle opposite to the first side will be right angle.

Given: Let in ΔABC, the area of the square drawn on the side AB is equal to the sum of the areas of the squares drawn on the sides BC and AC, i.e., AB² = BC²+ AC².

To prove ∠ACB = 1 right angle.

Construction: Let us draw the A line segment FF which is equal to CM.

A perpendicular on the side FE at the point F is drawn and cut off FD from that perpendicular which is equal to the side CA and let us join the points D and E.

Proof: Given that AB² = BC² + AC².

= EF² + DF² [∵ by construction BC = EF and AC = DF.]

= DE² [by Pythagoras theorem]

∴ AB² = DE² or AB = DE.

Now, in ΔABC and ΔDEF, AB = DE, BC = EF and AC = DF.

∴ ΔABC ≅ ΔDEF [by the S-S-S condition of congruency]

∴ ∠ACB = ∠DEE = 1 right angle [∵ by construction DF ⊥ EF]

∴ ∠ACB = 1 right angle. (Proved)

By the application of Pythagoras’ theorem, we shall now prove an important theorem. The theorem is the Apollonius theorem.

Apollonius Theorem: The sum of the areas of two squares drawn on any two sides of a triangle is equal to the twice of the sum of the areas of the squares drawn on half of the third side and on the median to this third side.

Or,

ABC is a triangle. D is the mid-point on its side BC. Prove that AB² + AC² = 2 (BD² + AD²)

“WBBSE Class 10 Pythagoras theorem solved examples”

Given: Let in ΔABC, D is a mid-point of BC, i.e., BD = CD.

To prove AB² + AC² = 2 (BD² + AD²)

Construction: Let us draw a perpendicular AE from A on BC, which intersects BC at point E.

Proof: From the right-angled triangle ABE we get,

AB² = AE² + BE² [by Pythagoras’ theorem]

= AE² + (BD – DE)²

= AE² + BD² – 2 BD.DE + DE²

= AE² + DE²+ BD² – 2BD.DE

= AD² + BD² – BC.DE [∵ in ΔAED, ∠E = right angle

∴ by Pythagoras’ therorem, AE² + DE² = AD² and  ∵ 2BD = BC],

∴ AB² – AD² + BD² – BC.DE…….(1)

Again, from the right-angled triangle AEC we get,

AC² = AE² + CE²

= AE²+ (CD + DE)²

= AE² + CD² + 2CD.DE + DE²

= AE² + DE² + CD² + 2CD.DE

= AD² + CD² + 2CD.DE  [∵ AE²+ DE² = AD²]

= AD² + CD² + BC.DE  [∵ 2CD = BC]

∴ AC² = AD² + CD² + BC.DE ……(2)

Now, adding (1) and (2) we get,

AB² + AC² = AD² + BD² – BC.DE + AD² + CD² + BC.DE

= 2AD² + BD² + CD²

= 2AD² + BD² + BD²  [∵ CD = BD]

= 2 AD² + 2BD²

= 2 (AD² + BD²)

∴ AB² + AC² = 2 (BD² + AD²) (Proved)

We shall now discuss about the application of the above theorems.

WBBSE Solutions For Class 10 Maths Solid Geometry Chapter 6 Pythagoras Theorem Multiple Choice Questions

“Pythagorean theorem problems for Class 10 Maths”

Example 1. A person travels from a place firstly 24 m west and then 10 m north. Then the distance of the person from the starting point will be

1. 34 m
2. 17 m
3. 26 m
4. 25 m

Solution:

Given

A person travels from a place firstly 24 m west and then 10 m north.

Let the person firstly travels 24 m West upto a point A from the starting point O.

He then goes to the ultimate point B which is at 10 m north of A.

Then we shall have to find the distance of OB.

Now, since the north is always at a right angle to the west, so ΔOAB is a right-angled triangle, of which ∠A = right angle.

by Pythagoras’ theorem, OB² = OA² + AB².

or, OB² = (24² + 10²) sq-m

= (576 + 100) sq-m.

= 676 sq-m.

∴ OB = 676 m = 26 m

∴ The required distance of the person from the starting point is 26 m.

∴ 3. 26 m is correct.

Example 2. If ΔABC be an equilateral triangle and AD ⊥ BC, then AD² =

1. \(\frac{3}{2}\) DC²
2. 2DC²
3. 3DC²
4. 4DC²

Solution:

Given

If ΔABC be an equilateral triangle and AD ⊥ BC

Let ΔABC be an equilateral triangle, i.e., AB = BC = CA. AD ⊥ BC.

∴ ΔABD and ΔACD are both right-angled triangles.

∴ by Pythagoras’ theorem we get, AB² = AD² + BD²……(1)

or, BC² = AD² + DC²  [∵ ΔABC is equilateral and AD ⊥ BC, D is the mid-point of BC.]

or, (2DC)² = AD² + DC² or, 4DC²= AD²+ DC²

or, AD = 3DC²

∴ 3. 3DC² is correct

“Chapter 6 Pythagoras theorem exercises WBBSE solutions”

Example 3. ABC is an isosceles triangle of which AC = BC and AB² = 2AC². Then value of ∠C will be

1. 30°
2. 90°
3. 45°
4. 60°

Solution:

Given

ABC is an isosceles triangle of which AC = BC and AB² = 2AC².

∵ AB² = 2AC²,  ∴ AB² = AC² + AC² = AC² + BC²  [∵ AC = BC (given)]

∴ by the converse of Pythagoras’ theorem,

ABC is a right-angled triangle, the hypotenuse of which is AB.

∴ the opposite angle of the hypotenuse ZC = a right angle.

∴ ∠C = 90°

∴ 2. 90° is correct.

Example 4. Two rods of 13m length and 7 m length are situated perpendicularly on the ground and the distance between their foots is 8 m. The distance between their top parts is

1. 9 m
2. 10 m
3. 11 m
4. 12 m

Solution:

Given

Two rods of 13m length and 7 m length are situated perpendicularly on the ground and the distance between their foots is 8 m.

Let the lengths of the two rods AB and CD are 13 m and 7 m respectively.

As per question, AC = 8 m, DE ⊥ AB, ∴ AC = DE = 8 m

Now, from the right-angled triangle BDE we get,

BD² = BE² + DE²

= (AB – AE)² + DE²

= (13 – DC)² + DE²  [∵ AE = DC]

= (13 – 7)² + AC² [∵ DC = 7m and DE = AC]

= 6² + AC² = 36 + (8)² = 36 + 64 = 100

∴ BD = √100 = 10.

∴ the required distance between their top parts = 10 m.

∴ 2. 10m is correct.

Example 5. The lengths of two diagonals of a rhombus are 24 cm and 10 cm respectively. Then the perimeter of the rhombus is

1. 13 cm
2. 26 cm
3. 52 cm
4. 25 cm

Solution:

Given

The lengths of two diagonals of a rhombus are 24 cm and 10 cm respectively.

AC and BD are the two diagonals of the rhombus AC and BD, where AC = 10 cm, and BD = 24 cm.

O is the point of intersection of AC and BD.

We know that the diagonals of a rhombus bisect each other at right angles.

∴ AO = CO = \(\frac{10}{2}\) cm = 5cm

BO = DO = \(\frac{24}{2}\) cm = 12 cm

and ∠AOB = ∠BOC = ∠COD = ∠DOA = 1 right angle.

Now, from the right-angled triangle AOB we get,

AB² = AO² + BO²

or, AB² = {(5)² + (12)²} sq-cm = 169 sq-cm.

∴ AB = 13 cm.

∴ the perimeter of the rhombus ABCD = 4 x 13 cm = 52 cm.

∴ the required perimeter = 52 cm.

∴ 3. 52 cm is correct.

WBBSE Solutions For Class 10 Maths Solid Geometry Chapter 6 Pythagoras Theorem True Or False

“Class 10 Maths applications of Pythagoras theorem”

Example 1. If the ratio of the lengths of three sides of a triangle is 3: 4: 5, then the triangle will always be a right-angled triangle.

Solution:

Given:

If the ratio of the lengths of three sides of a triangle is 3: 4: 5

The statement is true since let the sides be 3x cm, 4x cm and 5x cm.

Then (5x)² = (3x)² + (4x)²

or, 25x² = 9x² + 16x²

∴ by the converse of Pythagoras’ theorem, the triangle is always a right-angled triangle.

Example 2. If in a circle of radius 10 cm in length, a chord subtends right-angle at the centre, then the length of the chord will be 5 cm.

Solution:

Given:

If in a circle of radius 10 cm in length, a chord subtends right-angle at the centr

Let the chord AB of the circle with centre O subtend a right angle at the centre i.e, ∠AOB = right angle.

As per the question, OA = OB = 10 cm.

Now, in the right-angled triangle AOB, by Pythagoras’ theorem, we get, AB² = AO² + BO² = (10)² + (10)²= 100 + 100 = 200.

∴ AB = √200 = 10√2

∴ The required length of the chord = 10 √2 cm.

∴ the statement is False.

WBBSE Solutions For Class 10 Maths Solid Geometry Chapter 6 Pythagoras Theorem Fill In The Blanks

“Understanding Pythagorean theorem in Class 10 Maths”

Example 1. In a right-angled triangle, the area of a square drawn on the hypotenuse is equal to the _______ of the areas of the squares drawn on other two sides.

Solution: Sum

Example 2. In an isosceles right-angled triangle if the length of each of two equal sides is 4 √2 cm, then the length of the hypotenuse will be ______ cm

Solution: 8

since in the right-angled isosceles triangle AB = AC and ∠A = right angle.

As per the question,

AB = AC =4 √2 cm

By Pythagoras’ theorem,

BC² = AB + AC² = (4 √2)²+(4 √2)² = 32 + 32 = 64

∴ BC = √64 = 8.

Hence the length of the hypotenuse = 8 cm.

Example 3. In a rectangular figure ABCD, the two diagonals AC and BD intersect each other ait the point O, if AB = 12 cm, AO = 6-5 cm, then the length of BC is _____ cm

Solutions: 5

We know that the diagonals of a rectangle bisects each other.

∴ AO = CO

∴ AC = 2AO = 2 x 6.5 cm  [∵ AO = 6.5 cm] = 13 cm

Now, from the right-angled triangle ABC by Pythagoras theorem we get,

AC² = AB² + BC²

or, (13)² = (12)² + BC²  [∵ AB = 12 cm]

or, 169 = 144 + BC²

or, BC²  = 169 – 144 = 25

or, BC = √25 = 5

∴ the length of BC = 5 cm.

WBBSE Solutions For Class 10 Maths Solid Geometry Chapter 6 Pythagoras Theorem Short Answer Type Questions

“Step-by-step solutions for Pythagorean theorem Class 10”

Example 1. In ΔABC, if AB = (2a – 1) cm, AC = 2√2a cm and BC = (2a + 1) cm, then find the value of ∠BAC.

Solution:

Given:

In ΔABC, if AB = (2a – 1) cm, AC = 2√2a cm and BC = (2a + 1) cm,

Here, BC² = (2a + 1)² sq-cm = 4a- + 4a + 1 sq-cm.

AB²  = (2a – 1)²  sq-cm = 4a² – 4a + 1 sq-cm

AC²  = (2 √2a)²  sq-cm = 8 a sq-cm.

Now, AB² + AC² = (4a² – 4a + 1 + 8a) sq-cm = (4a² + 4a + 1) sq-cm = BC²

∴ BC² = AB² + AC² .

∴ by the converse of Pythagoras’ theorem we get, ΔABC is a right-angled triangle of which BC is the hypotenuse.

ΔABC is a right-angled triangle of which BC is the hypotenuse ∠BAC = 90°.

Example 2. In the point O is situated within the triangle PQR in such a way that, ∠POQ = 90°, OP = 6 cm and OQ = 8 cm. If PR = 24 cm and ∠QPR = 90Q, then find the length of QR.

Solution:

Given:

In the point O is situated within the triangle PQR in such a way that, ∠POQ = 90°, OP = 6 cm and OQ = 8 cm. If PR = 24 cm and ∠QPR = 90Q,

From the right-angled triangle POQ, by Pythagoras’theorem we get,

PQ² = PO² + OQ²

= {(6)² + (8)²} sq-cm  [∵ OP = 6 cm, OQ = 8 cm]

= (36 + 64) sq-cm = 100 sq-cm.

∴ PQ = 100 cm = 10cm.

Again, from the right-angled triangle PQR by Pythagoras’ theorem we get,

RQ² = PR² + PQ²

= {(24)² + (10)²} sq-cm  [∵ PR = 24 cm, PQ = 10 cm]

= (576 + 100) sq-cm = 676 sq-cm.

∴ RQ = √676 cm = 26 cm

Hence the required length of RQ = 26 cm.

Example 3. The point O is situated within the rectangular figure ABCD in such a way that OB = 6 cm, OD = 8 cm and OA = 5 cm. Determine the length of OC.

Solution:

Given:

The point O is situated within the rectangular figure ABCD in such a way that OB = 6 cm, OD = 8 cm and OA = 5 cm.

Let O be any point within the rectangle ABCD. Then OA² + OC² = OB² + OD²

or, 5² + OC² = 6² + 8² [∵OA = 5 cm, OB = 6 cm, OD = 8 cm]

or, 25 + OC² = 36 + 64 or, OC² = 100 – 25

or, OC² = 75 or, OC = √75 or, OC = \(\sqrt{25 \times 3}\) or, OC = 5√3.

∴ the length of OC = 5 √3 cm.

Example 4. In the triangle ABC, the perpendicular AD, from the point A on the side BC meets the side BC at the point D. If BD = 8 cm, DC = 2 cm and AD = 4 cm, then find the measure of ∠BAC.

Solution:

Given:

In the triangle ABC, the perpendicular AD, from the point A on the side BC meets the side BC at the point D. If BD = 8 cm, DC = 2 cm and AD = 4 cm,

In the right-angled triangle ACD we get, AC² = AD² + CD²

= (4² + 2²) sq-cm  [∵ AD = 4 cm, DC = 2 cm] = 20 sq-cm.

Now, from the right-angled ΔABD we get,

AB² = AD² + BD² = (4² + 8²) sq-cm  [∵ AD = 4 cm, BD = 8 cm] = 80 sq-cm.

Again, BC = BD + CD = 8 cm + 2 cm = 10 cm

∴ BC² = (10)² sq-cm =100 sq-cm

∴ BC² = 100 sq-cm = 20 sq-cm + 80 sq-cm

= AC² + AB², BC² = AC² + AB²

By the converse of Pythagoras theorem, ΔABC is a right -angled triangle of which ∠A = right angle and BC is the hypotenuse.

∴ the value of ∠BAC = 90°.

Example 5. In a right-angled triangle ABC, ∠ABC = 90°, AB = 3 cm, BC = 4 cm and the perpendicular BD on the side AC from the point B which meets the side AC B at the point D. Determine the length of BD.

Solution:

Given:

In a right-angled triangle ABC, ∠ABC = 90°, AB = 3 cm, BC = 4 cm and the perpendicular BD on the side AC from the point B which meets the side AC B at the point D.

From the right-angled triangle ABC, we get,

AC² = AB² + BC² = (3)² + (4)² sq-cm = 25 sq-cm.

∴ AC = √25 cm = 5 cm.

Now, from the right-angled triangle ABD we get,

AB² = BD² + AD²…… (1)

Again, from the right-angled triangle BCD we get,

BC² = BD² + CD²……(2)

From (1) we get, BD² = AB² – AD² = (3)² – AD²  [∵ AB = 3 cm] = 9 – AD² …..(3)

From (2) we get, BD² = (BC)² – (CD)²

= (4)² – CD²  [∵ BC = 4 cm]

= 16 – CD² ………(4)

Comparing (3) and (4) we get,

9 – AD² = 16 – CD²

or, CD² – AD² =16-9

or, (CD + AD)(CD – AD) = 7

or, AC (AC – AD – AD) = 7

or, 5 (5- 2 AD) = 7

or, 25 – 10 AD = 7 or, 10 AD = 18

or, AD = \(\frac{18}{10}\) or, AD = \(\frac{9}{5}\) = 1.8

∴ from (3) we get, BD² = 9 – (1.8)² = 9 – 3.24 = 5.76

∴ BD = √5.76 =2.4

Hence the length of BD = 2.4 cm.

WBBSE Solutions For Class 10 Maths Solid Geometry Chapter 6 Pythagoras Theorem Long Answer Type Question

“WBBSE Mensuration Chapter 6 practice questions on Pythagoras”

Example 1. If the followings are the lengths of the three sides of a triangle, then write the cases where the triangles are right-angled triangles :

1. 8 cm, 15 cm, 17 cm
2. 9cm, 11 cm, 6 cm

Solution:

1.  ∵ 8² + 15² = 64 + 225 = 289 = 172, i.e., 8² + 15² = 17²,

∴ The triangle constructed by the given three sides will be a right-angled triangle.

2. ∵ 9² + 6² = 81 + 36 = 117, which is not a perfect square, and not equal to (11)² = 121,

hence the triangle constructed by the given three sides does not form a right-angled triangle.

Example 2. In the road of Laxmi’s locality there is a ladder of 15 m length kept in such a way that it has touched Pujas’ window at a height of 9 m above the ground. Now keeping the foot of the ladder at the same point of that road, the ladder is rotated in such a way that it touched Laxmi’s window situated on the other side of the road. If Laxmi’s window is 12 m above the ground, then determine the breadth of that road.

Solution:

Given:

In the road of Laxmi’s locality there is a ladder of 15 m length kept in such a way that it has touched Pujas’ window at a height of 9 m above the ground. Now keeping the foot of the ladder at the same point of that road, the ladder is rotated in such a way that it touched Laxmi’s window situated on the other side of the road. If Laxmi’s window is 12 m above the ground,

Let the ladder is situated at the point 0 on the road in such a way that it touches the window of Puja situated at the point B.

Also, keeping the foot of the ladder at O, the ladder is rotated in such a way that it touches the window of Laxmi situated at the point A.

We have to determine the breadth of the road, i.e., the value of DC.

Now, from the right-angled triangle BOD we get, OB² = BD² + OD²……(1)

Again, from the right-angled triangle AOC we get, OA² = AC² + OC²

or, OB² = AC² + OC²….. (2) [∵ OA = OB = length of the ladder]

Then from (1) and (2) we get BD² + OD² = AC² + OC²

or, 9² + OD² = (12)² + OC² or, OD² – OC² = 144 – 81

or, (OD + OC) (OD – OC) = 63 or, DC (OD OC) = 63

or, DC = \(\frac{63}{OD – OC}\)…..(3)

Now, from (1) we get OB² = 9² + OD²  [∵ BD = 9 m]

or, (15)² = 9² + OD²  [∵ length of the ladder OB = 15 m]

or, 225 = 81 + OD² or, OD² = 225 – 81

or, OD² = 144 or, OD = √144 = 12.

From (2) we get, OB² = AC² + OC² or, (15)² = (12)² + OC²

or, 225 = 144 + OC² or, OC² = 225 – 144 or, OC² = 81

or, OC = 81 or, OC = 9

∴ from (3) we get, DC = \(\frac{63}{12 – 9}\) = 21  [∵ OD = 12, OC = 9]

Hence the breadth of the road = 21 m.

Example 3. If the length of one diagonal of a rhombus having the side 10 cm length be 12 cm, then calculate the length of other diagonal.

Solution:

Given:

PQR is a triangle whose ∠Q is right angle. If S is any point on QR,

Let the diagonals AC and BD of the rhombus ABCD intersect each other at the point O and let the length of AC be 12 cm.

We know that the diagonals of a rhombus bisects each other at rightangles.

∴ AO = \(\frac{1}{2}\) AC = \(\frac{1}{2}\) x 12cm = 6cm

As per question, AB = 10 cm

Now, from right-angled triangle AOB we get,

AB² = AO² + BO² or, (10)² = 6² + BO²

or, 100 = 36 + BO² or, BO² = 100 – 36 = 64

or, BO = √64 = 8

∴ BD = 2 BO = 2 x 8 cm = 16 cm.

Hence the length of other diagonal = 16 cm.

Example 4. PQR is a triangle whose ∠Q is right angle. If S is any point on QR, then prove that PS² + QR² = PR² + QS².

Solution:

Given:

PQR is a triangle whose ∠Q is right angle. If S is any point on QR,

∵ in ΔPQR, ∠Q = right angle,

∴ from right-angled triangle PQS by Pythagoras’ theorem we get, PS² = PQ² + QS²….. (1)

Again, PR² = PQ² + QR²

or, QR² = PR² – PQ²……..(2)

Now adding (1) and (2) we get,

PS² + QR² = PQ² + QS²+ PR² – PQ²

or, PS² + QR² = PR² + QS²

Hence PS² + QR² = PR² + QS². (Proved)

Example 5. Prove that the sum of squares drawn on the sides of a rhombus is equal to the sum of squares drawn on two diagonals.

Solution:

Let ABCD be a rhombus the diagonals AC and BD of which intersect each other at O.

To prove 4AB² = AC² + BD²

Proof: We know that the diagonals of a rhombus bisects each other at right angles.

∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = 1 right angle.

Then from right-angled triangl∴ e AOB we get, AB² = OA² + OB²

or, \(\mathrm{AB}^2=\left(\frac{1}{2} \mathrm{AC}\right)^2+\left(\frac{1}{2} \mathrm{BD}\right)^2=\frac{1}{4} \mathrm{AC}^2+\frac{1}{4} \mathrm{BD}^2\)

or, \(\mathrm{AB}^2=\frac{1}{4}\left(\mathrm{AC}^2+\mathrm{BD}^2\right) \text { or, } 4 \mathrm{AB}^2=\mathrm{AC}^2+\mathrm{BD}^2\)

or, 4AB² = AC² + BD²

Hence 4AB² = AC² + BD². (Proved)

Example 6. ABC is an equilateral triangle. AD is perpendicular to BC. Prove that AB² + BC²+ CA² = 4AD².

Solution:

Given:

ABC is an equilateral triangle. AD is perpendicular to BC.

∵ ΔABC is an equilateral triangle and AD ⊥ BC,

∴ D is the mid-point of BC.

∴ BD = CD

A Now, from the right-angled triangle ABD we get, AB² = AD² + BD²…….(1)

Again, from the right-angled triangle ACD we get,

AC² = AD²+ CD²…… (2)

Then adding (1) and (2) we get,

AB² + AC² = 2AD² + BD² + CD²

or, AB² + AC² = 2AD²+ \(\left(\frac{1}{2} B C\right)^2+\left(\frac{1}{2} B C\right)^2\) [∵ D is the midpoint of BC.]

or, AB² +AC² = 2AD² + \(\frac{1}{2}\) BC²

or, AB² +AC² = \(\frac{1}{2}\)(4AD² + BC²)

or, 2AB² + 2AC² = 4AD² + BC²

or, 2AB² + AC² + AC² = 4AD² + AB²  [∵ BC² = AB²]

or, 2AB² – AB² + BC² + AC² = 4AD² [∵ AC = BC]

or, AB² + BC² + CA² = 4AD²

Hence AB² + BC² + CA² – 4AD² (Proved)

“Examples of right-angled triangle problems for Class 10”

Example 7. ABC is a right-angled triangle of which ∠A = right angle. P and Q are two points on AB and AC respectively. By joining P, Q; B, Q and C, P prove that BQ² + PC² = BC²+ PQ²

Solution:

Given:

ABC is a right-angled triangle of which ∠A = right angle. P and Q are two points on AB and AC respectively.

∵ In ΔABC, ∠A is right-angled,

∴ ΔPAQ is a right-angled triangle.

∴ AP² + AQ² = PQ²……(1)

Again, from the right-angled triangle APC

we get, AP² + AC² = PC²….. (2)

Again, from the right-angled triangle ABC we get, AB² + AC² = BC²…….(3)

Also from right-angled triangle ABQ we get, AB² + AQ²= BQ²….. (4)

Now, adding (2) and (4) we get,

PC² + BQ² = AP² + AC² + AB² + AQ²

= (AB² + AC²) + (AP² + AQ²)

= BC² + PQ² [from (3) and (1)]

Hence BQ² + PC² = BC² + PQ² (Proved)

Example 8. If the diagonals of the quadrilateral ABCD intersect each other orthogonally, then prove that AB² + CD² = BC² + DA².

Solution:

Given:

If the diagonals of the quadrilateral ABCD intersect each other orthogonally

Let the diagonals AC and BD intersect each other orthogonally at the point O.

∴ ∠AOB = ∠BQC = ∠COD = ∠DQA = 1 right angle.

Now, from the right-angled triangle AOB, we get,

AB² = AO² + BO² …….(1)

From the right-angled triangle COD, we get, CD² = CO² + DO² ……(2)

Similarly, BC² = BO² + CO² ……(3) [from the right-angled triangle BOC]

and DA² = AO² + DO² …….(4) [from the right-angled triangle AOD]

Now, adding (1) and (2) we get,

AB² + CD² = AO² + BO² + CO² + DO²

= (AO² + DO²) + (BO² + CO²)

= DA² + BC² [from (4) and (3)].

Hence AB² + CD² = BC² + DA². (Proved)

Example 9. AD is the height of the triangle ABC. If AB > AC, then prove that AB²– AC² = BD²– CD².

Solution:

Given:

AD is the height of the triangle ABC. If AB > AC

Let in ΔABC, AB > AC and AD ⊥ BC, then AD is the height of ΔABC.

∴ from right-angled ΔABD we get, AB² = AD² + BD²

Again, from the right-angled triangle ACD we get, AC² = AD² + CD²……(1)

Now, subtracting (2) from (1) we get,

AB² – AC² = AD² + BD² – AD² – CD²

or, AB² – AC² = BD² – CD²

Hence AB² – AC² = BD²– CD². (Proved)

Example 10. Two perpendicular BD and CE are drawn from the vertices B and C respectively on the sides AC and AB of the ΔABC, which intersect each other at the point P. Prove that AC² + BP² = AB² + CP².

Solution:

Given:

Two perpendicular BD and CE are drawn from the vertices B and C respectively on the sides AC and AB of the ΔABC, which intersect each other at the point P.

Let two perpendiculars BD and CE are drawn on the sides AC and AB respectively of the ΔABC from its vertices B and C, where BD and CE intersect each other at the point P.

To prove AC² + BP² = AB² + CP².

Construction: Let us join A, P.

Proof: AC² + BP² = AE² + CE² + BE² + PE²

[From right-angled triangles AEC and BEP respectively by Pythagoras’ theorem.]

∴ AC² + BP² = (AE² + PE²) + (CE² + BE²)

= AP² + BC² [From the right-angled triangles ΔAEP and ΔBEC by Pythagoras’ theorem]

= (AD² + PD²) + (BD² + CD²) [From the right-angled triangles ΔAPD and ΔBCD by Pythagoras’ theorem]

= (AD² + BD²) + (PD² + CD²)

= AB² + CP² [From right-angled triangles ΔABD and ΔCPD by Pythagoras theorem]

Hence AC² + BP² = AB² + CP² (Proved)

Example 11. ABC is a right-angled isosceles triangle of which ∠C is a right angle. If D is any point on AB. Then prove that AD² + BD² = 2CD²

Solution:

Given:

ABC is a right-angled isosceles triangle of which ∠C is a right angle. If D is any point on AB.

Let in the right-angled isosceles triangle, ∠C = right angle of which AC = BC.

D is any point on AB.

To prove AD² + BD² = 2CD².

Construction: Let us draw CE ⊥ AB, where CE intersects AB at E.

Proof: From the right-angled triangle ABC we get,

AB² = AC² + BC² = AC² + AC²  [∵ BC = AC] = 2 AC∵

or, (AD + BD)² – 2 AC² or, AD² + BD² + 2AD.BD = 2 AC²

or, AD² + BD² = 2AC² – 2AD.BD

= 2AC² – 2(AE – DE) (BE + DE)

= 2 AC²– 2(AE- DE)(AE + DE)

[∵ BE = AE, since the perpendicular drawn form the right angular vertex of a right-angled isosceles triangle on its hypotenuse bisects the hypotenuse.]

= 2AC² – 2 (AE²– DE²)

= 2AC² – 2 AE² + 2 DE²

= 2 (AC² – AE²) + 2 DE²

= CE² + 2DE² [AC² = AE² + CE²]

= 2 (CE² + DE²) = 2CD² [from the right-angled triangle CED by Pythagoras’ theorem]

Hence AD² + BD² = 2 CD² (Proved)

“WBBSE Class 10 Maths solved problems on Pythagorean theorem”

Example 12. In ΔABC, ∠A = right angle. If CD is a median, then prove that BC² = CD² + 3AD².

Solution:

Given:

In ΔABC, ∠A = right angle. If CD is a median,

Let in ΔABC, ∠A = 90° and CD is a median.

∴ AD = \(\frac{1}{2}\)AB or AB = 2AD.

To prove BC² = CD² + 3AD².

Proof: From the right-angled triangle ABC by Pythagoras’ theorem we get

BC² = AC² + AB² …..(1)

= AC² + (2AD)²  [∵ AB = 2AD]

= AC² + 4AD²

= (AC² + AD²) + 3AD²

= CD² + 3AD²  [Since from the right-angled triangle ΔACD we get, AC² + AD² = CD²]

Hence BC² = CD² + 3AD² (Proved)

Example 13. OX, OY and OZ are the perpendiculars drawn from an internal point O of the AABC on its sides BC, CA and AB respectively. Prove that AZ² + BX² + CY² = AY² + CX² + BZ²

Solution:

Given:

OX, OY and OZ are the perpendiculars drawn from an internal point O of the AABC on its sides BC, CA and AB respectively.

Let O be any internal point in ΔABC, OX, OY and OZ are the perpendiculars drawn from O on the sides BC, CA and AB respectively.

To prove AZ² + BX² + CY² = AY² + CX² + BZ².

Proof: From the right-angled triangle AOZ by Pythagoras’ theorem

we get, OA² = AZ² + OZ².

or, AZ² = OA²– OZ² ….(1)

Similarly, BX²– OB²– OX²….(2)

and CY² = OC² – OY²…. (3)

Now adding (1), (2) and (3) we get,

AZ² + BX² + CY² = (OA² + OB² + OC²) – (OZ² + OX² + OY²)

= (OA² – OY²) + (OC² – OX²) + (OB² – OZ²)

= AY² + CX² + BZ²

[From the right-angled triangles ΔAOY, ΔCOX and ΔBOZ by Pythagoras’ theorem.]

∴ AZ² + BX² + CY² = AY² + CX² + BZ². (Proved)

Example 14. In the ΔRST, ∠S = right angle. X and Y are the midpoints of RS and ST respectively. Prove that RY² + XT² = 5 XY².

Solution:

Given:

In the ΔRST, ∠S = right angle. X and Y are the midpoints of RS and ST respectively.

Let in ΔRST, ∠S = right angle.

X and Y are the midpoints of RS and ST respectively.

To prove RY² + XT² = 5 XY².

Construction: Let us join the points R, Y; X, T; and X, Y.

Proof: ∵ ∠S = right angle,

∴ from the right-angled triangle ΔSXY by Pythagoras’ theorem we get, XY² = SX² + SY²…..(2)

Similarly, from the right-angled triangle ΔRSY by Pythagoras’ theorem we that RY2 = RS2 + SY2…….(2)

Now, RY² + XT² = RS² + SY² + SX² + ST² [from (2) and XT² = SX² + ST²]

= (2SX)² + SY² + SX² + (2SY)²  [∵ RS = 2SX and ST = 2SY]

= 4SX² + SY² + SX² + 4SY²

= 5 SX² + 5SY² = 5 (SX² + SY²)

= 5 XY² [from (1)]

Hence RY² + XT² = 5XY². (Proved)

Example 15. If in ΔABC, AD ⊥ BC, then prove that AB² + CD² = AC² + BD².

Solution:

Given:

If in ΔABC, AD ⊥ BC,

In ΔABC, AD ⊥ BC,

∴ from the right-angled triangle ΔABD by Pythagoras’ theorem we get, AB² = AD² + BD² = AC²= CD² + BD²

[∵ in right-angled triangle ACD, AC² = AD² + CD²]

or, AB² + CD² = AC² + BD²

Hence AB2 + CD² = AC² + BD² (Proved)

Example 16. Prove that the area of the square drawn on the diagonal of a square is twice the area of the given square.

Solution:

Let ABCD be a square and AC is one of its diagonals.

To prove AC² = 2AB².

Proof: ABC is a right-angled triangle of which ∠B = right angle.

∴ by Pythagoras’ theorem we get, AC² = AB² + BC²

= AB² + AB² [∵ in the square ABCD, AB = BC = CD = DA]

∴ AC² = 2AB² (Proved)

[Instead of AC if we take BD as the diagonal, then the statement will also be true, since the diagonals of a square are equal.]

Example 17. O is any point inside a rectangle ABCD. Prove that OA² + OC² = OB² + OD²

Solution:

Given:

O is any point inside a rectangle ABCD.

Let O be any point inside the rectangle ABCD.

To prove OA² + OC² = OB² + OD²

Construction: Let us draw a straight line PQ parallel to AB through point O which intersects AD at P and BC at Q.

Let us join O, A; O, B; O, C and O, D.

Proof: ∵ AB | | PQ, ∴ ∠P = right angle and ∠Q = right angle

[∵ PQ ⊥ AD and PQ ⊥ BC.]

∴ by Pythagoras’ theorem, in the right-angled triangle AOP we get, OA² = OP² + AP²……(1)

Similarly, OC² = OQ² + CQ²…..(2) [in ΔCOQ]

Then adding (1) and (2) we get,

OA² + OC² = OP² + AP² + OQ² + CQ² = AP² + OQ² + CQ² + OP²

= BQ² + OQ² + PD² + OP² [∵ AB || PQ || DC, ∴ AP = BQ and PD = CQ.]

= OB² + OD²

[∵ in right-angled triangle ΔBOQ, BQ² + OQ² = OB² and in right-angled ΔDOP, PD² + OP² = OD²]

Hence OA² + OC² = OB² + OD² (Proved)

“Pythagorean theorem proofs and applications for Class 10”

Example 18. ABCD is a rhombus. Prove that AB² + BC² + CD²+ DA² = AC²+ BD².

Solution:

Given:

ABCD is a rhombus.

Let the diagonals AC and BD of the rhombus ABCD intersect each other at O.

We know that the diagonals of any rhombus bisects each other at right angles.

Then ∠AOB = ∠BOC = ∠COD = ∠DOA – 1 right angle.

To prove AB² + BC² + CD² + DA² = AC² + BD².

Proof: From the right-angled triangle AOB by Pythagoras’ theorem, we get,

AB² = OA² + OB²….(1)

Similarly, BC² = OB² + OC²…..(2)

CD² = OC² + OD²……(3)

DA² = OD² + OA²……..(4)(

Then adding (1), (2), (3) and (4) we get,

AB² + BC² + CD² + DA² = 2 (OA² + OB² + OC² + OD²)

= 2 (OA² + OB² + OA² + OB²)

[∵ O is the midpoint, ∴ OC = OA and OD = OB.]

= 2 (2OA² + 2OB²)

= 4OA² + 4OB² = (2OA)² + (2OB)²

= AC² + BD²  [∵ 2OA = AC and 2OB = BD]

Hence AB² + BC² + CD² + DA² = AC² + BD² (Proved)

Example 19. In ΔABC, AD ⊥ BC. Prove that AB² – BD² = AC² – CD².

Solution:

Given:

In ΔABC, AD ⊥ BC.

Given that AD ⊥ BC.

∴ from the right-angled triangle ABD by Pythagoras’ theorem we get, AB² = AD² + BD²…..(1)

Again, in right-angled triangle ACD, we get, AC² = AD² + CD²….(2)

Now, subtracting (2) from (1) we get,

AB² – AC² = BD² – CD²

Hence AB² – BD² = AC² – CD². (Proved)

Example 20. In ΔABC, AD ⊥ BC which intersects BC at D. If BD = 3 CD, then prove that 2AB² = 2AC²

Solution:

Given:

In ΔABC, AD ⊥ BC which intersects BC at D. If BD = 3 CD

∵ AD ⊥ BC, from right-angled triangle ABD by Pythagoras’ theorem we get,

AB² = AD² + BD²…..(1)

Again, in right-angled triangle ACD we get,

AC² = AD² + CD² ……..(2)

Now, subtracting (2) from (1) we get, AB² – AC² = BD² – CD²

= (3CD)² – CD² = 9CD² – CD² = 8CD²

or, 2AB² – 2AC² = 16 CD²

or, 2AB² – 2AC² = BC² [∵ BC² = (BD + CD)² = (3CD + CD)² = (4CD)² = 16 CD².]

or, 2AB² = 2AC² + BC².

Hence 2AB² = 2AC² + BC². (Proved)

Example 21. In the isosceles triangle ABC, AB = AC and BE is perpendicular to AC from B. Prove that BC² = 2AC x CE.

Solution:

Given:

In the isosceles triangle ABC, AB = AC and BE is perpendicular to AC from B.

Let in the isosceles triangle ABC, AB = AC and BE ⊥ AC.

Then, from the right-angled triangle ABE we get, AB² = BE² + AE²

= BE² + (AC – CE)²

= BE²+ AC² – 2AC x CE + CE²

= BE² + CE² + AC² – 2AC x CE

= BC² + AB² – 2AC x CE

or, 0 = BC² – 2AC x CE

or, BC² = 2AC x CE

Hence BC² = 2AC x CE (Proved)

Example 22. In an isosceles right-angled triangle ABC, ∠B = 90°. The bisector of ∠BAC intersects the side BC at the point D. Prove that CD² = 2BD².

Solution:

Given:

In an isosceles right-angled triangle ABC, ∠B = 90°. The bisector of ∠BAC intersects the side BC at the point D.

In ΔABC, ∠B = 90° and AB = BC.

AD is the bisector of  ∠BAC and it intersects the side BC at point D

To prove CD² = 2BD².

Construction: Let us draw perpendicular DE on AC from D which intersects AC at E.

Proof: ∵ ∠B = 90° and AB = BC, ∠ACB = 45°.

∴ in ΔDEC, ∠E = 90° [by construction] ∠DCE = 45°

∴ ∠CDE = 90° – ∠DCE = 90°- 45° = 45°

∴ DE = CE

Again, in ΔABD and ΔAED, ∠BAD = ∠DAE [AD is the bisector of ∠BAC], ∠ABD = ∠AED [∵ each is right angle] and AD is common to both.

∴ ΔABD ≅ ΔAED [by R-H-S condition of congruency]

∴ BD = DE [∵ corresponding sides of two congruent triangles.]

Now in the right-angled triangle CED by Pythagoras’ theorem, we get,

CD² = DE² + CE² = DE² + DE²  [∵ CE = DE]

= 2DE² = 2BD²  [∵ DE = BD.]

Hence CD² = 2BD². (Proved)

Example 23. P is an external point of the square ABCD. If PA > PB, then prove that PA² – PB² = PD² – PC².

Solution:

Given:

P is an external point of the square ABCD. If PA > PB,

Let P is any point outside the square ABCD.

Given that PA > PB.

Prove that PA² – PB² = PD² – PC².

Construction: Let us draw AE ⊥ PD and BF ⊥ CP. AE intersects PD at E and BF intersects CP at F.

Proof: From the right-angled triangle PEA by Pythagoras’ theorem we get,

PA² = AE² + PE²…..(1)

Similarly, PB² = BF² + PF² …..(2)

Now, subtracting (2) from (l) we get,

PA² – PB² = AE² + PE² – BF² – PF²

= AE² + (PD – DE)² – BF² – (CP – CF)²

= AE² + PD² + DE² – 2PD.DE – BF² – CP² – CF²+ 2CP.CF.

= AE² + DE² + PD² – CP² – (BF² + CF²) – 2PD.DE + 2CP.CF

= AD² + PD² – CP² – BC² – 2PD.DE + 2CP.CF

= BC² + PD² – CP² – BC² – 2PD.DE + 2CP.CF

= AD² – 2PD.DE + 2CP.CF – BC² + PD² – PC²

= AD² – AD² + BC² – BC² + PD² – PC²

[∵ 2PD.DE = AD² and 2CP.CF = BC² ]

= PD² – PC²

Hence PA²– PB² = PD² – PC². (Proved)

Example 24. In the right-angled triangle ABC, ∠A = 1 right angle. BE and CF are two medians of ΔABC. Prove that 4 (BE² + CF²) = 5BC².

Solution:

Given:

In the right-angled triangle ABC, ∠A = 1 right angle. BE and CF are two medians of ΔABC.

In ΔABC, ∠A = 1 right angle.

From the right-angled triangle ΔABE we get, BE² = AB² + AE²…..(1)

Similarly, CF² = AC² + AF² ……(2)

∴ adding (1) and (2) we get, BE² + CF² = AB² + AC² + AE² + AF²

= AB² + AC² + \(\left(\frac{1}{2} \mathrm{AC}\right)^2+\left(\frac{1}{2} \mathrm{AB}\right)^2\)

= AB² + AC² + \(\frac{1}{4}\) AC² + \(\frac{1}{4}\) AB²

= \(\frac{5}{4}\) AB²+ \(\frac{5}{4}\) AC² = \(\frac{5}{4}\)(AB + AC)²

= \(\frac{5}{4}\) BC² [ ∵ in the right-angled triangle ABC, AB² + AC² = BC²]

or, 4 (BE² + CF²) = 5 BC².

Hence 4 (BE² + CF²) = 5 BC². (Proved)

Example 25. A perpendicular AD is drawn on BC from the vertex A of the acute ΔABC. Prove that AC² = AB² + BC² – 2BC.BD.

OR,

Prove that the square drawn on the opposite side of the acute angle of an acute triangle is equal to the areas of the sum of the squares drawn on its other two sides being subtracted by twice the area of the rectangle formed by one of its sides and the projection of the other side to this side.

Solution:

Let ABC be an acute angle. AD ⊥ BC.

To prove AC² = AB² + BC² – 2BC.BD.

Proof: AD ⊥ BC, ΔACD is a right-angled triangle and hence by Pythagoras’ theorem we get,

AC² = AD² + DC²

= AD² + (BC – BD)²

= AD² + BC² + BD² – 2BC.BD

= AD² + BD² + BC² – 2BC.BD

= AB² + BC² – 2BC.BD [∵ in right-angled triangle ΔABD, AD² + BD² = AB²]

Hence AC² = AB² + BC² – 2BC.BD. (Proved)

Example 26. Prove that if equilateral triangles are drawn on the sides of a right-angled triangle then the area of the equilateral triangle, drawn on the hypotenuse is equal to the sum of the areas of the other two equilateral triangles drawn on its other two sides.

Solution:

 

Let ΔABC be a right-angled triangle of which ∠A = right angle.

ΔPAB, ΔQBC and ΔCRA are three equilateral triangles drawn on the sides AB, hypotenuse Q BC and CA respectively.

To prove ΔQBC = ΔPAB + ΔCRA.

Proof: We know that if the length of the sides of an equilateral triangle be a units, then its area = \(\frac{\sqrt{3}}{4}\) a² sq-units…..(1)

∴ Δ QBC = \(\frac{\sqrt{3}}{4}\) x BC² sq- units [side = BC]……(2)

Similarly, Δ PAB = \(\frac{\sqrt{3}}{4}\) x AB² sq- units……(3)

Now, ∵ ΔABC is a right-angled triangle,

∴ by Pythagoras’ theorem, we get, BC² = AB²+ AC²

or, \(\frac{\sqrt{3}}{4}\) BC² = AB² + \(\frac{\sqrt{3}}{4}\) AC²  [multiplying by \(\frac{\sqrt{3}}{4}\)]

or, ΔQBC = ΔPAB + ΔCRA [from (1), (2) and (3)]

Hence ΔQBC = ΔPAB + ΔCRA (Proved)

WBBSE Solutions For Class 10 Maths Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle Secant To A Circle

A straight line may or may not intersect a circle.

If the straight line does not intersect the circle, then we say that there is no common point between the straight line and the circle.

AB is a straight line and PQRS is a circle which have no common point.

So here the straight line AB does not intersect the circle with centre at O.

Again, if the straight line intersect the circle then we shall see that the straight line can intersect the circle atmost two points.

For example the straight line AB intersects the circle with centre.

O in two points P and Q. In this case, we call the line AB a secant of the circle and PQ is a chord of the secant AB.

It is very clear that the secant AB has intersected the circle with centre O atmost two points P and Q.

So, any secant can intersect a circle atmost two points.

Now, let the straight line AB is rotated anti-clockwise fixing it at the point A.

 WBBSE Solutions for Class 10 Maths

Where B₁ and B₂ are the new positions of B and in both the positions of B₁ and B₂, the straight line intersects the circle at two points.

Clearly, the new position of P are P₁ and P₂. But in position B₃, P coincides entirely on the point Q.

In this position, the straight line intersects at one point In such cases, P coincides entirely with Q.

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WBBSE Class 10 History Multiple Choice Questions

 

In this position the straight line intersect the circle with centre at O only at one point Q,

i.e., there is only one common point between straight line and the circle, which is nothing but Q.

In such a case, we say that the straight line touches the circle with centre at O.

AB is the tangent and Q is the point of contact.

Tangent and point of contact 

Definition: If a straight line intersects a circle in two coincident points, then the straight line is called the tangent to the circle and the point at which they coincide is called the point of contact.

AB is the secant, B₃Q is the tangent and Q is the point of contact. PQ is the corresponding chord of the secant AB.

What do we mean by the term, “Two circles touch each other ?”

Two circles touch each other means

Just before we have seen that two circles can intersect each other atmost in two points.

By joining these two points of intersection, we may get a straight line which is the common chord of two intersecting circles.

Now, if the two centres of the circles be taken away continuously from each other, the length of the common chord gradually decreases and in a certain time two end-points of the common chord coincide at a point on both the circles.

In this position, we say that the two circles touch each other.

For example, PQ is the common chord of the two circles with centres at A and B, when two circles are taken away from each other, then after some time P and Q coincide.

In this position, we say that the two circle touch each other.

 

The two endpoints P and Q of the common chord coincide at point P. In this position, we say that the two circles touch each other externally.

However, if the two circles are derived in one another continuously, then at a certain time, one circle will penetrate completely into another circle, but they remain in contact into an endpoint which is common to both circles.

In this situation we say that the two circles touch each other internally. The two circles with centres at A and B intersect other internally at point C.

Characteristics and properties of two circles touching internally or externally each other 

1. If two circles touch externally, then the distance between centres of the circles is equal to the sum of the radii of the circles.
2. If two circles touch internally, then the distance between the centres of the circles is equal to the difference of their radii.
3. In both cases, the number of point of contact is always 1 and it lie on the same line joining the two centres of the circle.
4. All other points except the point of contact on the tangent to the circle will always lie outside the circle.

Condition for two circles to touch each other 

Let two circles with centres at O and O’ are of radii R₁ and R₂ (R₁ > R₂) respectively and the distance between the centres OO’ = d. Then

1. The two circles will touch each other externally, if R₁ + R₂= d.

2. The two circles will touch each other internally if R₁ – R₂ = d (when R₁ > R₂) and R₂ – R₁ = d when (R₂ > R₁).

3. The-two circles will not touch each other if R₁ – R₂ < d.

 

Common tangent 

Definition: If a straight line touches two given circles at one or two points, then the straight line is called the common tangent to the circles.

For example, the straight line PQ touches two circles with centres at A and B at the point C.

So, here PQ is a common tangent to the circles. Notice that here the common tangent touches the circles at only one point.

Again, the common tangent PQ touches the circle with centre A at C₁ and the circle with centre B at C₂. Notice that here the common tangent PQ touches two circles at two different points.

Types of common tangent

Common tangents are of two types. Such as –

1. Direct common tangent ;
2. Transversed common tangent ;

1. Direct common tangent If the points of contact of the common tangent to two given circles lie on the line segment joining the two centres of the circles or lie on the same side of that line, then the common tangent is called direct common tangent. PQ is a direct common tangent to the circles with centres at A and B.

2. Transversed common tangent If the points of contact of the common tangent of two given circles lie on the opposite sides of the line segment joining the centres of the circles, then the common tangent is called the transverse common tangent.

Both PQ and CD are transversed common tangents to the two given circles with centres at A and B.

Relation between tangent at any point of a circle and the radius passing through that point of contact 

Let PT is a tangent to the circle with centre O at the point P and OP is a radius of the circle passing through P.

We shall now try to find out the relation between PT and OP.

Let OP₁, OP₂, OP₃, OP₄, OP, etc. are the distances from O to some points P₁, P₂, P₃, P₄, P on the tangent PT to the circle with centre O.

If we take measures of OP₃, OP₂, OP₃, OP₄, OP etc with the help of a scale, we shall see that among these distances OP is the smallest. So, OP is perpendicular to PT.

Therefore, tangent to a circle and the radius through the point of contact of the tangent are perpendicular to each other. We shall now prove this theorem logically by the method of geometry.

Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle

Secant To A Circle Theorems

“Theorems related to tangents in a circle for Class 10 Maths”

Theorem 1. The tangent to a circle at any point on it is perpendicular to the radius passing through the point of contact.

 

Given: AB is a tangent at P to the circle with the centre at O and OP is a radius of the circle through P.

To prove: OP and AB are perpendicular to each other, i.e., OP ⊥ AB.

Construction: Let us take another point Q on the tangent AB and join the points O, Q.

Proof: Any point on the tangent AB except P lie on the outside of the circle. So, OQ must intersects the circle at any point. Let the point of intersection be T.

∴ OT < OQ [T lies between O and Q]

Again, OT = OP.[radii of same circle] OP < OQ.

∴ OP < OQ

Q is a point on the tangent AB, so OP is the smallest among all the straight lines that can be drawn from O, the centre of the circle, to the tangent AB.

We know that the smallest distance is the perpendicular distance.

∴ OP ⊥ AB. (Proved)

Obviously, there arises a question that whether the converse of this theorem is true or not. We shall verify this logically.

Converse Of Theorem 2. The perpendicular drawn on the radius at the endpoint of the radius of a circle will be a tangent to the circle at the endpoint of the radius. 

Given: CD is a radius of the circle with centre at C and AB is perpendicular to CD at the point D.

To prove: The straight line AB is a tangent to the circle with the centre at C at the point D.

Proof: Let AB is not a tangent to the circle with centre at C at the  point D. Then let us construct another tangent RS at the point D to the R circle with centre at C.

Now, since RS is a tangent at the point D of the circle with the centre at C.

∴ CD ⊥ RS

∴ ∠CDS = 90°, but ∠CDB = 90° (Given) [CD ⊥ AB]

∴ ∠CDS = ∠CDB.

i.e., RS will coincide with the straight line AB.

RS is not a tangent at D on the circle with centre C.

Hence AB is a tangent at the point D of the circle with centre C. (Proved)

Corollary: 1. The point at which any radius of a circle intersects the circumference of the circle, if a perpendicular is drawn on the radius at that point, then the perpendicular is a tangent to the circle at that point.

Corollary: 2. Only one tangent can be drawn at any point on the circumference of a circle.

Corollary: 3. The perpendicular drawn to a tangent at the point of contact passes through the centre of the circle.

In the following examples, applications of the above theorem are discussed thoroughly.

Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle

Secant To A Circle Examples

“Chapter 4 tangents and circles exercises WBBSE solutions”

Example 1. Prove that the two tangents at the extremities of a diameter of any circle are parallel to each other.

Solution:

Given: Let AB be the diameter of the circle with the centre at O. AC and BD are two tangents at the end-points A and B of AB.

To prove AC | | BD.

Proof: In the circle with the centre at O, AC is a tangent at A and OA is a radius through the point of contact.

∴ OA is perpendicular to AC.

∴ ∠OAC = 1 right angle……… (1)

Again, in the circle with centre at O, BD is a tangent at the point B and OB is a radius through point of contact.

∴ OB ⊥ BD

∴ ∠OBD = 1 right angle……(2)

Now AB has intersected two line segments AC and BD and two adjacent angles on the same side of the transversal AB are ∠BAC and ∠ABD.

Now, ∠BAC + ∠ABD = ∠OAC + ∠OBD

= 1 right angle + 1 right angle [from (1) and (2)]

= 2 right angles

∴ AC and BD are paralled to each other, i.e., AC | | BD. (Proved)

Example 2.  Manas has drawn a circle with centre O of which AB is a chord. A tangent is drawn at the point B which intersects extended AO at the point T. If ∠BAT = 21°, find the value of ∠BTA.

Solution:

Given that

Manas has drawn a circle with centre O of which AB is a chord. A tangent is drawn at the point B which intersects extended AO at the point T. If ∠BAT = 21°

∠BAT = 21°

∴ ∠OBA = 21° [OA = OB, radii of same circle]

Again, BT is a tangent at B of the circle with centre O and OB is a radius passing through B.

∴ OB ⊥ BT, ∠OBT = 90°

Now, ∠ABT = ∠ABO + ∠OBT

= 21° + 90° [∠ABO= ∠OBA – 21° and ∠OBT =90°] = 111°

Then, ∠BTA = 180° – (∠BAT + ∠ABT) [sum of three angles of a triangle is 180°]

= 180° – (21° + 111°) = 180° – 132° – 48°

Hence ∠BTA = 48°.

Example 3. XY is a diameter of a circle. PAQ is a tangent to the circle at the point A lying on the circumference. The perpendicular drawn on the tangent to the circle from X intersects PAQ at Z. Prove that XA is a bisector of ∠YXZ.

Solution:

Given:

XY is a diameter of a circle. PAQ is a tangent to the circle at the point A lying on the circumference. The perpendicular drawn on the tangent to the circle from X intersects PAQ at Z.

Let XY be a diameter of the circle with centre at O. PAQ is a tangent to the circle at point A. The perpendicular XZ drawn from X to PAQ intersects
PAQ at Z.

To prove: XA is a bisector of ∠YXZ, i.e., ∠YXA = ∠ZXA

Construction: Let us join O, A.

Proof: PAQ is a tangent to the circle with centre O at the point A and OA is a radius passing through point of contact A.

∴ OA ⊥ PAQ

∴ ∠OAP = 90° or, ∠OAZ = 90°…….. (1)

Again, given that XZ ∠ PAQ,

∴ ∠XZA = 90° …. (2)

From (1) and (2) we get, XZ | | OA and XA is their transversal.

∴ ∠OAX = ∠ZXA…….(3)

or, ∠ZXA = ∠AXO [OX = OA = radii of same circle]

or, ∠ZXA = ∠YXA

Hence XA is the bisector of ∠YXZ(proved)

“Class 10 Maths properties of tangents to a circle”

Example 4. PR is a diameter of a circle. A tangent is drawn at the point P and a point S is taken on the tangent of the circle in such a way that PR = PS. If RS intersects the circle at the point T., prove that ST = PT. 

Solution:

Given:

PR is a diameter of a circle. A tangent is drawn at the point P and a point S is taken on the tangent of the circle in such a way that PR = PS. If RS intersects the circle at the point T.,

Let PR be a diameter of the circle with centre at O. A tangent AB is drawn at P.

The part PS from.AB is cut equal to PR. RS intersects the circle at the point T.

To prove: ST = PT

Proof: AB is a tangent at P on the circle with centre at O and OP is a radius passing through P.

∴ OP ⊥ AB.

∴ ∠OPS = 90° or, ∠RPT + ∠TPS = 90° ……… (1)

Again, ∠PTR is a semicircular angle, ∠PTR = 90°

∴ ∠TPR + ∠TRP = 90°…….(2)

From (1) and (2) we get, ∠RPT + ∠TPS = ∠TPR + ∠TRP

or, ∠TPS = ∠TRP

∠RPT = ∠TPR or, ∠TPS = ∠TSP [PS = PR, ∴ ∠TRP = ∠TSP]

∴ ST = PT [opposite sides of equal angles are equal.]

Hence ST = PT (Proved).

Example 5. Two radii OA and OB of a circle with centre O are perpendicular to each other. If two tangents are drawn at the points A and B intersect each other at the point T, prove that AB = OT and they bisect each other at a right angle. 

Solution:

Given:

Two radii OA and OB of a circle with centre O are perpendicular to each other. If two tangents are drawn at the points A and B intersect each other at the point T

Two radii OA and OB of a circle with centre at O are perpendicular to each other.

Two tangents ST and RT at the points A and B respectively intersect each other at T.

Let us join the points A, B and O, T. Let AB and OT intersect each other at C.

To prove: AB = OT and AB and OT bisect each other at a right angle.

Proof: In the quadrilateral OATB, OA ⊥ OB (given), ∠AOB = 90°, OA ⊥ TS, ∠OAT = 90° and OB ⊥ TR.

∴ ∠OBT = 90°, i.e., three of the angles of OATB are right-angles. So, the fourth angle is also a right angle,

∴ each of the angles of OATB is a right angle.

Again, OA = OB [radii of same circle]

∴ OA = OB = BT = TA.

∴ each of the sides of the quadrilateral OATB are equal and each of the angles is a right angle.

∴ OATB is a square and AB and OT are two of its diagonals.

We know that diagonals of a square are equal and the diagonals bisect each other at right-angles.

∴ AB = OT and AB and OT bisect each other at a right angle. [Proved]

“Understanding tangent theorems in Class 10 Maths”

Example 6. X is a point on the tangent at the point A lies on a circle with centre O. A secant drawn from a point X intersects the circle at the points Y and Z. If P is the mid-point of YZ, prove that XAPO or XAOP is a cyclic quadrilateral.

Solution:

Given:

X is a point on the tangent at the point A lies on a circle with centre O. A secant drawn from a point X intersects the circle at the points Y and Z. If P is the mid-point of YZ

XT is a tangent at A of the circle with centre at O. A secant XZ through X intersects the circle at the point Y and Z. P is the mid-point of YZ.

To prove: XAPO or XAOP is a cyclic quadrilateral.

Proof: YZ is a chord of the circle with centre at O and P is the mid-point of YZ.

∴ OP ⊥ YZ.

∴ ∠OPX = 90°

Again XT is a tangent at A of the circle with centre at O.

∴ OA ⊥ XT, ∠OAX = 90°……. (2)

Now. from (1) and (2) we get,

∠OPX + ∠OAX = 90° + 90° = 180°

∴ Two opposite angles of the quadrilateral XAOP are supplementary.

∴ XAOP is a cyclic quadrilateral. (Proved)

Example 7. P is any point on diameter of a circle with centre O. A perpendicular drawn on diameter at the point O intersects the circle at the point Q. Extended QP intersects the circle at the point R. A tangent drawn at the point R intersects extended OP at the point S. Prove that SP = SR.

Solution:

Given:

P is any point on diameter of a circle with centre O. A perpendicular drawn on diameter at the point O intersects the circle at the point Q. Extended QP intersects the circle at the point R. A tangent drawn at the point R intersects extended OP at the point S.

P is any point on a diameter EF of the circle with .centre O.

OQ ⊥ EF and OQ intersects the circle at the point Q.

Extended QP intersects the circle at the point R.

The tangent TS of the circle at R intersects extended OP at the point S.

To prove: SP = SR.

Construction: Let us join O, R.

Proof : OR is a radius of the circle passing through the point of contact of the tangent SRT. ∴ OR ⊥ ST.

∴ ∠ORS = 90° or, ∠ORQ + ∠QRS = 90°…….(1)

Again, OQ ⊥ ES, ∠QOS = 90°, ∠OQP + ∠OPQ = 90°……..(2)

from (1) and (2) we get, ∠ORQ + ∠QRS = ∠OQP + ∠OPQ….. (3)

Now, OQ = OR [radii of same circle] ∴ ∠ORQ = ∠OQR

∴ from (3) we get, ∠OQR + ∠QRS = ∠OQP + ∠OPQ [∠ORQ = ∠OQR]

or, ∠OQR + ∠QRS = ∠OQR + ∠SPR [∠OPQ = opposite ∠SPR]

or, ∠QRS = ∠SPR or, ∠PRS = ∠SPR

∴ SP = SR [ in ΔSPR, sides opposite of equal angles are equal] .

∴ SP = SR (Proved)

“Step-by-step solutions for tangents in circles Class 10”

Example 8. QR is a chord of the circle with centre O. Two tangents drawn at the points Q and R intersect each other at point P. If QM is a diameter, prove that ∠QPR = 2 ∠RQM

Solution:

Given:

QR is a chord of the circle with centre O. Two tangents drawn at the points Q and R intersect each other at point P.

QR is a chord of the circle with centre O. Two tangents drawn at the point Q and R intersect each other at the point P. QM is a diameter.

To prove: ∠QPR = 2 ∠RQM

Construction: Let us join O, R.

Proof: OQ1 PS,  ∴ ∠OQP = 90°

Again OR ⊥ PT, ∴ ∠ORP = 90°

In quadrilateral OQPR, ∠QOR + ∠ORP + ∠RPQ + ∠PQO = 360°

or, ∠QOR + 90° + ∠RPQ + 90° = 360°

or, ∠QOR + ∠RPQ = 180° or, ∠RPQ = 180° – ∠QOR

or, ∠RPQ = ∠ROM……. (1) [∠QOR + ∠ROM = 180°]

Now, ∠RQM is the angle in circle and ∠ROM is the central angle produced by the chord RM.

∴ ∠ROM = 2 ∠RQM

or, ∠QPR = 2 ∠RQM [by (1)]

Hence ∠QPR – 2 ∠RQM. (Proved)

Example 9. Two chords AC and BD of a circle intersect each other at point O. If two tangents drawn at the points A and B intersect each other at the point P and two tangents drawn at the points C and D intersect at the point Q, prove that ZP + ZQ = 2

Solution:

Given:

Two chords AC and BD of a circle intersect each other at point O. If two tangents drawn at the points A and B intersect each other at the point P and two tangents drawn at the points C and D intersect at the point Q

In the circle with centre O’, two chords AC and BD intersect each other at O.

Two tangents PAS and PBT drawn at the points A and B respectively intersect each other at P.

Again, the tangents QCN and QDM drawn at C and D respectively intersect each other at Q.

To prove: ∠P + ∠Q = 2 ∠BOC

Construction: Let us join O’, A; O’, D; A, B and C, D.

Proof: From example 8 above, we get P = 2 ∠BAO’ and Q = 2 ∠DCO’

∴ ∠P + ∠Q = 2 (∠BAO’ + 2 ∠DCO’)

= 2 (∠BAO’ + ∠DCO’)

= 2 (∠BAC + ∠CAO’ + ∠DCO’) [∠BAO’ = ∠BAC = ∠CAO’]

= 2 (∠BDC + ∠ACO’ + ∠DCO’)

= 2 (∠BDC + ∠ACD) [∠ACO’ + ∠DCO’ = ∠ACD]

= 2 (∠ODC + ∠OCD) [∠BDC = ∠ODC, ∠ACD = ∠OCD]

= 2 ∠BOC [in ΔCOD external ∠BOC = internally opposite (∠ODC + ∠OCD]

Hence ∠P +∠Q = 2 ∠BOC (Proved)

Example 10. Prove that if a quadrilateral is circumscribed about a circle, then the angles subtended at the centre by any two opposite sides are supplementary. 

Solution:

Let the quadrilateral ABCD is circumscribed about the circle with centre O.

Two of its opposite sides AB and CD have subtended the angles ∠AOB and ∠COD at the centre.

To prove: ∠AOB and ∠COD are supplementary to ∠AOB + ∠COD = 180°

Construction: Let the sides AB, BC, CD and DA of the quadrilateral ABCD touch the circle at the points P, Q, R and S respectively.

Let us join the points O, A; O, P; O, B; O, Q; O, C; O, R; O, D and O, S.

Proof: AB is a tangent to the circle with centre O and OP is a radius passing through the point of contact P.

OP ⊥ AB. ∠OPA = ∠OPB = 90°

Similarly, OQ ⊥ BC, OR ⊥ CD, OS ⊥ AD.

∴ ∠OQB = ∠OQC = 90°, ∠ORC = ∠ORD = 90°, ∠OSD = ∠OSA = 90°.

Again, PA || OS, [∠OPA = ∠OSA = 90°] and OA is their transversal,

∴ ∠OAP = alternate ∠AOS.

Similarly, ∠OBP = alternate ∠BOQ; ∠COR = alternate ∠OCQ.

∠DOR = alternate ∠ODS; ∠DOS = alternate ∠ODR.

Now, ∠AOB + ∠COD = ∠AOP + ∠BOP + ∠COR + ∠DOR

= ∠AOS + ∠BOQ + ∠COQ + ∠DOS

= ∠AOS + ∠DOS + ∠BOQ + ∠COQ = ∠AOD + ∠BOC

= 360° – (∠AOB + ∠COD)

or, 2 (∠AOB + ∠COD) = 360°.

or, ∠AOB +∠COD = 180°

Hence ∠AOB + ∠COD = 180° (Proved)

Example 11. PQ is the diameter of the circle with centre O. The tangent drawn at any point R on the circle intersects the tangents drawn at P and Q at two points A and B respectively. Prove that ∠AOB = 1 right angle.

Solution:

Given:

PQ is the diameter of the circle with centre O. The tangent drawn at any point R on the circle intersects the tangents drawn at P and Q at two points A and B respectively.

Let PQ.is a diameter of the circle with centre O.

EF and CD are two of its tangents drawn at P and Q respectively.

R lies on the circle and the tangent AB drawn at R intersects the previous tangents at A and B respectively. Let us join O, A; O, B.

To prove ∠AOB = 1 right angle.

Proof: In ΔAOP and ΔAOR,

∠APO =  ∠ARO [each is right angle]

OR = OP [radii of same circle] and hypotenuse OA is common to both.

∴ ΔAOP ≅ ΔAOR [by the RHS condition of congruency]

∴ ∠OAP = ∠OAR ……(1) [similar angles of two congruent triangles]

Similarly, it can be proved that ∠OBR = ∠OBQ…….(2)

Then, ∠AOB = ∠AOR + ∠BOR

= 90° – ∠OAR + 90° – ∠OBR

= 90° – ∠OAP + 90° – ∠OBQ

= ∠AOP + ∠BOQ

= 180° – ∠AOB [∠AOB + ∠AOP + ∠BOQ = 180°] or, 2 ∠AOB = 180° or, ∠AOB = 90°

Hence ∠AOB = 1 right angle. (Proved)

Example 12. The length of radii of two circles are r₁ unit and r₂ unit respectively where r₁ > r₂. If the distance between the centres of the circles be p unit, then prove that the length of the direct common tangent to -the two circles, PQ = \(P Q=\sqrt{p^2-\left(r_1-r_2\right)^2} \text { unit. }\) unit.

Solution:

Given:

The length of radii of two circles are r₁ unit and r₂ unit respectively where r₁ > r₂. If the distance between the centres of the circles be p unit

Let the length of the radius of the circle with centre C₁ be r₁ unit and the length of the radius of the circle with centre C₂ be r₂ unit. (r₁ > r₂).

RS is a direct common tangent to both the circles which intersects the circle with centre C₁ at P and the circle with centre C₂ at Q respectively.

To prove PQ = \(P Q=\sqrt{p^2-\left(r_1-r_2\right)^2} \text { unit. }\)unit,, where p = distance between the centres of the circles;

Construction: Let us draw a perpendicular C₂M from C₂ to PC₁. Let us join P, C₂.

Then PM = QC₂ = r₂

∴ MC₁ = PC₁ – PM

= r₁ – QC₂

= r₁ – r₂

Again, PQ = MC₂

Proof: In the right-angled ΔMC₁C₂ [∠C₁MC₂ = 90°],

MC₁² + MC₂² = C₁C₂² [by Pythagoras’ theorem]

or, (r₁ – >₂)² + PQ² = p² [MC₁ = 1- r₁ – r₂, MC₂ = PQ and C₁C₂ = p]

or, PQ² = p² – (r₁ – r₂)² or, PQ = \(P Q=\sqrt{p^2-\left(r_1-r_2\right)^2} \text { unit. }\)

Hence PQ = \(P Q=\sqrt{p^2-\left(r_1-r_2\right)^2} \text { unit. }\)  unit. (Proved)

Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle

Tangent To A Circle

The number of tangents to a circle from an external point

In the definition of tangent, we have said earlier that if any straight line intersect a circle at only one point, then the straight line is called a tangent of the circle.

Now, from an external point of a circle, two tangents can be drawn at two points, one point lying above the centre and the other point lying below the centre of the circle.

Such as, let T be an external point of a circle with centre at O. TP and TQ are two tangents drawn from T at the points P and Q respectively on the circle.

Later on, we shall prove it logically.

Number of direct common tangents to a circle 

There can be drawn at most three direct common tangents to two given circles.

Number of transverse common tangents to a circle 

There can be atmost two transverse common tangents to two given circles.

Now, let T be an external point of a circle with centre at O.

Two tangents TP and TQ are drawn from T on the points P and Q on the circle. ∠POT and ∠QOT are two front central angles produced by the tangents TP and TQ.

If we take measures of TP and TQ and the angles ∠POT and ∠QOT with the help of a scale and a protractor, then we shall see that TP and TQ are equal in length and ∠POT = ∠QOT,

i.e., the lengths of the tangents are equal and they produce equal front angles at the centre.

We shall now prove this theorem logically by the method of geometry.

 

Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle

Tangent To A Circle Theorems

“WBBSE Mensuration Chapter 4 practice questions on tangents”

Theorem 1. If two tangents are drawn to a circle from a point outside it, then the line segments joining the point of contact and the exterior point are equal and they subtend equal angles at the centre.

Given: Let PA and PB be two tangents drawn from an external point P of the circle with centre at O and let A and B be their points of contact.

After joining O, A; O, B and O, P, the two line segments PA and PB subtends two angles ∠POA and ∠POB at the centre.

To prove (1) PA = PB and (2) ∠POA = ∠POB.

Proof: PA and PB are two tangents and OA and OB are two radii passing through points of contact.

∴ OA ⊥ PA and OB ⊥ PB .

∠OAP = 90° and ∠OBP = 90°

Now, in the right-angled triangles ΔAOP and ΔBOP, ∠OAP = ∠OBP, [each is a right angle]

OA = OB [radii of same circle] and hypotenuse OP is common to both.

∴ ΔAOP ≅ ΔBOP [by the RHS condition of congruency]

∴  PA = PB [similar sides of congruent triangles] and ∠POA = ∠POB [similar sides of congruent triangles]

Hence PA = PB [(1) is proved]

and ∠POA = ∠POB [(2) is proved]

we have seen that two circles can touch each other in two ways-internally or externally.

In either cases, the point of contact of the two circles always lie on the line segment joining the two centres of the two circles.

We shall now prove this theorem logically by the method of geometry.

Theorem 2. If two circles touch each other, then the point of contact will lie on the line segment joining the two centres.

 

Given: Let two circles with centres A and B touch each other at point P.

To prove: A, P and B are collinear.

Construction: Let us join A, P and B, P.

Proof: Two circles with centres at A and B touch each other at point P.

∴ the two circles have a common tangent at P. Let ST be the common tangent which has touched both the circles at P.

Now,since in the circle with centre A, ST is a tangent to the circle and AP is its radius passing through point of contact P. ∴ AP ⊥ BP.

Again, in the circle with centre at B, ST is a tangent and BP is a radius passing through point of contact P, ∴ BP ⊥ ST.

∴ AP and BP are both perpendicular to ST at the same point P.

AP and BP lie on the same straight line, i.e., the points A, P and B are collinear. (Proved)

In the following examples various applications of the above theorems in practical problems have been discussed throughly.

Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle

Tangent To A Circle Multiple Choice Questions

Example 1. In a circle with centre O, a tangent is drawn from an external point A to the circle which touches the circle at B. If OB = 5 cm, OA = 13 cm, then the length of AB 

1. 1. 12 cm
2. 13 cm
3. 6.5 cm
4. 6 cm

 

Solution: In the circle with centre O, AB is a tangent and OB is a radius passing through point of contact.

∴ OB ⊥ AB, ∠OBA = 90°

∴ ΔAOB is a right-angled triangle* of which OA is the hypotenuse.

∴ OB² + AB² = OA² [by Pythagoras theorem] or, 5² + AB² =13² [OB = 5 cm, OA = 13 cm]

or, 25 + AB² = 169 or, AB² = 169 – 25

or, AB² = 144 or, AB = √144 or, AB = 12

Hence the length of AB = 12 cm

∴ 1. 12 cm is correct.

Example 2. Two circles touch each other externally at the point C. AB is a common tangent to both the circles and touches the circle at the points A and B. Then the value of ∠ACB is

1. 60°
2. 45°
3. 30°
4. 90°

Solution: Let two circles with centres O and O’ touch each other externally at point C.

Let us join OCO’, O,A; O’,B; A, C and B, C.

Now, ∠OCO’ = 1 straight angle =180°

or, ∠OCA + ∠ACB + ∠O’CB = 180°

or, ∠OCA + ∠ACB + ∠O’BC =180°

[∠OAC = ∠OAC, ∠O’CB = ∠O’BC]

or, 90° – ∠BAC + 90° – ∠ABC + ∠ACB = 180°

[∠OAC = 90° – ∠BAC,- ∠O’BC = 90° – ∠ABC]

or, 180° – ∠BAC – ∠ABC + ∠ACB = 180°

or, ∠BAC + ∠ACB + ∠ABC – ∠BAC – ∠ABC + ∠ACB = 180°

[sum of three angles of a triangle is 180°.]

or, 2 ∠ACB = 180° or, ∠ACB = \(\frac{180^{\circ}}{2}\)

or, ∠ACB = 90°

∴ 4. 90° is correct.

Example 3. The length of radius of a circle with centre O is 5 cm. P is a point at a distance of 13 cm from the point O. PQ and PR are two tangents from the point P to the circles: the area of the quadrilaterals PQOR is

1. 60 sq-cm
2. 30 sq-cm
3. 120 sq-cm
4. 150 sq-cm

 

Solution: The length of the radius of a circle with centre O. ∴OQ = OR = 5 cm.

P is a point at a distance of 13 cm from the point O. ∴ OP = 13. cm

Now, PQ and PR are two tangents from P and OP and OR are radii passing through the point of contact.

∴ OQ ⊥ PQ and OR ⊥ PR

∴ ∠OQP = 90° and ∠ORP = 90° i.e., ΔPOQ and ΔPOR are both right-angled triangles.

∴ ΔPOQ = \(\frac{1}{2}\) x PQ x OQ [PQ = base and OQ = height]

= \(\frac{1}{2}\) x 12 x 5 sq-cm PQ \(=\sqrt{\mathrm{OP}^2-\mathrm{OQ}^2}=\sqrt{13^2-5^2} \mathrm{~cm}\)

= 30 sq.cm \(\sqrt{169-25} \mathrm{~cm}=\sqrt{144} \mathrm{~cm}=12 \mathrm{~cm}\)

Similarly ΔPOR = \(\frac{1}{2}\) x PR x OR [PR = base and height = OR]

= \(\frac{1}{2}\) x 12 x 5 sq-cm [ PR = PQ = 12 cm] = 30 sq-cm

∴ Area of the quadrilateral PQOR = ΔPOQ + ΔPOR = 30 sq-cm + 30 sq-cm = 60 sq-cm.

∴ 1. 60 sq-cm is correct.

Example 4. The lengths of radii of two circles are 5 cm and 3 cm. The two circles touch externally. Then the distance between the centres of the circles is 

1. 2 cm
2. 2.5 cm
3. 1.5 cm
4. 8 cm

 

Solution: Let two circles with centres O and O’ touch each other externally at the point C.

The radius of the. first circle is 5 cm ∴ OC =5cm, and the radius of the second circle is 3 cm,

∴ O’C = 3cm.

Now, the distance between the centres, of the circles = OO’ = OC + CO’ = 5 cm + 3 cm = 8 cm

∴ 4. 8 cm is correct.

Example 5. The lengths of radii of two circles are 3.5 cm and 2 cm. They touch each other internally. the distance between the centres of the circle is

1. 5.5 cm
2. 1 cm
3. 1.5 cm
4. None of thsese

 

Solution: Let two circles with centres O and O’ touch each other internally at the point C

The radii of the circles are 3.5 cm and 2 cm ∴ OC = 3.5 cm, O’C = 2cm.

Now, the distance between the centres of the circles = OO’ = OC – O’C = 3.5 cm – 2 cm = 1.5 cm

∴ 3. 1.5 cm is correct.

Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle

Tangent To A Circle True Or False

Example 1. P is a point inside a circle; No tangent of the circle will pass through P.

Solution: True

Example 2. In a circle more than two tangents can he drawn which are parallel to a fixed straight line.

Solution: False

Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle

Tangent To A Circle Fill In The Blanks

Example 1. If a straight line intersects the circle at two points, then the straight line is called a ________ to the circle. 

Solution: Secant

Example 2. If two circles do not intersect or touch each other, then the maximum number of common tangents can be drawn is _______

Solution: 4

Example 3. Two circles touch each other externally at point A. A common tangent drawn to two circles at the point A is a ________ common tangent, (direct/transverse).

Solution: Transverse

Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle

Tangent To A Circle Short Answer Type Questions

Example 1.  In the adjoining O is the centre and BOA is the diameter of the circle. A tangent drawn to the circle at the point P intersects the extended BA at point T. If ∠PBO = 30°, find the value of ∠PTA.

Solution:

Given:

In the adjoining O is the centre and BOA is the diameter of the circle. A tangent drawn to the circle at the point P intersects the extended BA at point T. If ∠PBO = 30°,

Let us join O, P and A. P.

PT is tangent to the circle at P and OP is a radius through the point of contact P.

∴ OP ⊥ PT

∴ ∠OPT = 90°, ∠OPB – ∠OBP = 30° ∠BPT = ∠OPB + ∠OPT

= 30° + 90° [∠OPB = 30° and ∠OPT = 90°] = 120°

Now, in ΔBPT, ∠BTP + ∠TPB + ∠PBT = 180° or, ∠BTP + 120° + 30° = 180°

or, ∠PTA + 150° = 180° or, ∠PTA = 180° – 150° or, ∠PTA = 30°

∴ The value of ∠PTA = 30°.

Example 2. In the adjoining ΔABC circumscribes a circle and touches the circle at the points P. Q, R. If AP = 4 cm, BP = 6cm, AC = 12 cm and BC = x cm, then determine the value of x. 

Solution:

Given:

In the adjoining ΔABC circumscribes a circle and touches the circle at the points P. Q, R. If AP = 4 cm, BP = 6cm, AC = 12 cm and BC = x cm

Let us join O, P; O, Q; O,A, O,B and O,C.

Now ΔAOP = ΔAOR [OP = OR, ∠OPA = ∠ORA = 90° and hypotenuse OA is common to both.]

∴ AP = AR [similar sides of congruent triangles]

Then CR = AC – AR

= 12 cm – AP [AR = AP]

= 12 cm – 4 cm [AC = 12 cm and AP = 4 cm] = 8 cm

Again, ΔCOQ ≅ ΔCOR [by the R-H-S condition of congruency]

CQ = CR [similar sides of congruent triangles]

= 8 cm [CR = 8 cm]…….(1)

Again, ΔBOQ ≅ ΔBOP. [by the R-H-S condition of congruency]

∴ BQ = BP [similar sides of congruent triangles]

= 6 cm [BP = 6 cm]……… (2)

Then, BC = BQ + CQ = 6 cm + 8 cm [from (1) and (2)]

∴ BC = x cm = 14 cm;

The value of x = 14 cm.

“Examples of tangent problems for WBBSE Class 10 Maths”

Example 3. In the adjoining three circles with centres A, B, and C touch one another externally. If AB =  5 cm, BC = 7 cm and CA = 6 cm, then find the length of radius of a circle with centre A. 

Solution:

Given:

In the adjoining three circles with centres A, B, and C touch one another externally. If AB =  5 cm, BC = 7 cm and CA = 6 cm

As per the question, AB = 5 cm.

∴ AP + BP = 5 [two circles with centres A and B touch each other externally at P]………(1)

Similarly, BC = 7 cm

∴ CR + BR = 7 cm ……….. (2)

and CA = 6 cm, ∴ AQ + CQ = 6cm

or, AQ + CR = 6 cm…..(3)

[CQ = CR, radii of same circle.]

Then from (2) and (3) we get,

BR – AQ = 1 cm.

or, BP – AP = 1 cm [BR = BP and AQ = AP] ….. (4)

Now, subtracting (4) from (1) we get,

2 AP = 4 cm or, AP = \(\frac{4}{2}\) cm = 2 cm

∴ the length of the radius of the circle with centre A = 2 cm.

Example 4. In the adjoining two tangents drawn from external point C to a circle with centre O touches the circle at the points P and Q respectively. A tangent drawn at another point R of the circle intersects CP and CQ at the points A and B respectively. If CP = 11 cm and BC = 7 cm, then determine the length of BR.

 

Solution:

Given:

In the adjoining two tangents drawn from external point C to a circle with centre O touches the circle at the points P and Q respectively. A tangent drawn at another point R of the circle intersects CP and CQ at the points A and B respectively. If CP = 11 cm and BC = 7 cm,

Let us join O, P ; O, Q ; O, R and O, B.

Now, in the circle with centre O, CP and CQ are two tangents from an external point C.

∴ CP = CQ

∴ CQ = 11cm [CP = 11 cm (given)]

Again, given that BC = 7 cm

∴ BQ = CQ – BC

= 11 cm – 7cm = 4 cm

Now, in ΔBOQ and ΔBOR, OQ = OR, [radii of same circle]

∴ ∠OQB = ∠ORB [each is right angles] and hypotenuse OB is common to both.

∴ ΔBOQ = ΔBOR [by the R-H-S condition of congruency]

∴ BQ = BR [similar sides of congruent triangles]

But BQ = 4 cm, ∴ BR = 4 cm.

The length of BR = 4 cm

Example 5. The length of radii of two circles are 8 cm and 3 cm and distance between two centres is 13 cm. Find the length of a direct common tangent of two circles.

Solution:

Given:

The length of radii of two circles are 8 cm and 3 cm and distance between two centres is 13 cm.

Let the radii of the circles with centres A and B are 8 cm and 3 cm respectively and . the distance PQ is the direct common tangent of both the circles.

We have to find the length of PQ. Let us join A, P and B, Q and let us draw BM perpendicular to AP from B.

Then PQBM is a rectangle. [∠APQ  – ∠BQP = 90°]

∴ PQ = BM and PM = BQ.

As per question, AP = 8 cm, BQ = 3 cm.

Then AM = AP – PM = 8 cm – BQ = 8 cm – 3 cm = 5 cm

Now, in the right-angled triangle ABM we get by Pythagoras theorem,

BM² + AM² = AB² [∴ AB = hypotenuse] .

or, BM² + 5² = 13² or, BM² = 13² – 5² = 169 – 25 = 144

∴ BM = √144 = 12

∴ PQ = 12 cm [BM = PQ]

∴ the required length of the direct common tangent of two circles is 12 cm.

Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle

Tangent To A Circle Long Answer Type Questions

“WBBSE Class 10 Maths solved problems on circle tangents”

Example 1. An external point is situated at a distance of 17 cm from the centre of a circle having 16 cm diameter. Determine the length of the tangent drawn to the circle from the external point.

Solution:

Given:

An external point is situated at a distance of 17 cm from the centre of a circle having 16 cm diameter.

Since the diameter of the circle with centre O is 16 cm.

∴ radius = \(\frac{16}{2}\) cm = 8 cm.

PT is a tangent to the circle from an external point P which is situated at a distance .of 17 cm from the centre of the circle ∴ OP = 17 cm.

We have to determine the length of PT.

Now, PT is a tangent at T to the circle with centre O and OT is the radius passing through point of contact.

∴ OT ⊥ PT, ∴ ∠PTO = 1 right angle.

So, from the right-angled triangle POT by Pythagoras theorem,

OP² = OT² + PT²

or, 17² = 8² + PT² [OP = 17 cm, OT = 8 cm]

or, 289 = 64 + PT² or, PT² = 289 – 64 .

or, PT² = 225 or, PT = V225 = 15.

∴ the required length of the tangent = 15 cm.

Example 2. The tangent drawn at points P and Q on the circumference of a circle intersect at A. If ∠PAQ = 60°, find the value of ∠APQ. 

Solution:

Given:

The tangent drawn at points P and Q on the circumference of a circle intersect at A. If ∠PAQ = 60°,

Let P and Q be any two points on the circle with centre O. Two tangents AP and AQ drawn at the points P and Q intersect each other at point A.

Let us join P, Q and O, A.

Given that ∠PAQ = 60°.

Now, AP is a tangent to the circle with centre O and OP is the radius passing through the point of contact P.

∴ OP ⊥ AP, ∴  ∠OPA = 90°.

Similarly, ∠OQA = 90°

In Δ’s POA and ΔQOA,

∠OPA = ∠OQA [each is a right-angle]

OP = OQ [radius of same circle] and hypotenuse OA is common to both A POA = AQOA [by the R-H-S condition of congruency]

∴ ∠OAP = ∠OAQ [similar angles of congruent triangles]

Given that ∠PAQ = 60° or, ∠PAO + ∠QAO = 60° or, ∠PAO + ∠PAO = 60° [∠QAO = ∠PAO] = 60°

or, 2 ∠PAO = 60° or, ∠PAO = \(\frac{60^{\circ}}{2}\) = 30°

Then, ∠POA = 90° – ∠OAP = 90° – 30° = 60° [∠OAP = ∠PAO = 30°]

∴ ∠POQ = 2 ∠POA [∠POA = ∠QOA] = 2 x 60° = 120°

∴ ∠OPQ + ∠OPQ – 180° – 120° [∠OQP = ∠OPQ and∠POQ = 120°]

or, 2 ∠OPQ = 60° or, ∠OPQ = \(\frac{60^{\circ}}{2}\)= 30°

∴ ∠APQ = ∠APO – ∠OPQ = 90° – 30° – 60°

∴ value of ∠APQ = 60°.

Example 3. AP and AQ are two tangents drawn from an external point A to a circle with centre O, P and Q are points of contact. If PR is a diameter, prove that OA | | RQ. 

Solution:

Given:

AP and AQ are two tangents drawn from an external point A to a circle with centre O, P and Q are points of contact. If PR is a diameter

AP and AQ are two tangents to the circle with centre O and OP and OQ are two radii of the circle passing through points of contact P and Q respectively.

∴ OP ⊥ AP and OQ ⊥ AQ

∴ ∠OPA = ∠OQA = 90°

Now, ΔAOP = ΔAOQ, [since ∠APO = ∠AQO, OP = OQ and hypotenuse OA is common to both.]

∴ ∠AOP = ∠AOQ

[similar angles of congruent triangles]……… (1)

∴ ∠POQ = ∠AOP + ∠AOQ

= ∠AOQ + ∠AOQ [from (1)] = 2 ∠AOQ ……..(2)

Again, ∠POQ is the central angle and ∠PRQ is the angle in circle produced by the arc PQ,

∴ ∠POQ = 2 ∠PRQ

or, 2 ∠AOQ = 2 ∠PRQ [from (2)] .

or, ∠AOQ = ∠PRQ = ∠OQR [OQ = OR, ∴ ∠ORQ = ∠OQR]

∴ ∠AOQ = ∠OQR.

But these are two alternate angles when OQ intersects the line segments OA and RQ, and they are also equal.

∴ OA | | RQ. (Proved)

Example 4. Prove that a parallelogram circumscribing by a circle is always a rhombus. 

Solution:

Let ABCD is a parallelogram circumscribed by a circle with a centre at O.

We have to prove that ABCD is a rhombus.

Proof: Let the circle touches the parallelogram ABCD at the points P, Q, R and S.

Then AP and AS are two tangents to the circle from the external point A.

∴ AP = AS…… (1)

Similarly, BP = BQ …….(2)

CQ = CR ……… (3) and

DR = DS ……. (4)

Now, ABCD is a parallelogram, AB = DC and AD = BC.

Now, AB + DC = AP + BP + DR + CR

= AS + BQ + DS + CQ

= AS + DS + BQ + CQ

= AD + BC

or, AB + AB = BC + BC [DC = AB and AD = BC]

or, 2AB = 2BC or, AB = BC

Similarly, AD = DC, AB = BC = CD = DA

∴ four sides of the parallelogram ABCD are equal and ABCD is also a parallelogram

∴ ABCD is a rhombus. (Proved)

Example 5. Two circles drawn with centres A and B touch each other externally at C, O is a point on the ṭangents drawn at C, OD and OE are tangents drawn to the circles of centres A and B respectively. If ∠COD = 56°, ∠COE = 40°, ∠ACD = x°, and ∠BCE = y°, then prove that Oc = Od = OE and x-y = 8

Solution:

Given:

Two circles drawn with centres A and B touch each other externally at C, O is a point on the ṭangents drawn at C, OD and OE are tangents drawn to the circles of centres A and B respectively. If ∠COD = 56°, ∠COE = 40°, ∠ACD = x°, and ∠BCE = y°,

OC and OD are two tangents at the points C and D on the circle with centre at A from an external point O.

∴ OC = OD …….(1)

Again, OC and OE are two tangents at the points C and E to the circle with centre B from an external point O.

∴ OC = OE……..(2)

Then from (1) and (2) we get, OC = OD = OE

Now in ΔACD, ∠ADC = ∠ACD = x° [AC = AD]

∴ ∠ODC = ∠OCD = 90° – x° [∠ADO = ∠ACO – 90°]

We know that ∠OCD + ∠ODC + ∠COD = 180°

or, 90° – x° + 90° – x° +56° = 180° [∠COD = 56°] or, 2x° = 56° or, x° = \(\frac{56^{\circ}}{2}\) =28°.

∴ x = 28 ……… (3)

Similarly, ∠BEC = ∠BCE = y° [BC = BE]

∴ ∠OCE = ∠OEC = 90° – y° [∠BCO – ∠BEO = 90°]

We know that in ΔOCE, ∠OCE + ∠OEC + ∠COE = 180°

or, 90° -y°+ 90° – y° + 40° = 180° [∠COE = 40°]

or, 2y° = 40° or, y° = \(\frac{40^{\circ}}{2}\) = 20° .

∴ y = 20 …(4)

Then x – y = 28 – 20 = 8 [from (3) and (4)]

Hence OC = OD = OE and x – y = 8 (Proved)

Example 6. Two circles with centres A and B touch each other internally. Another circle touches the larger circle internally at point X and the smaller circle externally at the point Y. If O be the centre of that circle, prove that (AO + BO) is constant.

Solution:

Given:

Two circles with centres A and B touch each other internally. Another circle touches the larger circle internally at point X and the smaller circle externally at the point Y. If O be the centre of that circle,

Let the circle with centre A touches the circle with centre B internally at the point C.

We know that if two circles touches each other either externally or internally, then the centres of the circles and the point of contact lie on the same straight line.

∴ the points A, O, X; A, B, C and B, Y, O lie on the same straight line each.

Now, AO + BO = AO + BY + OY

= AO + OY + BY

= AO + OX + BY [ OY = OX = radii of same circle]

= AX + BY [AO + OX = AX]

= Radius of the larger circle + Radius of the smaller circle.

= Constant [both the circles are fixed]

Hence (AO + BO) = constant (Proved)

“WBBSE Class 10 tangents to a circle solved examples”

Example 7.  Two circles have been drawn with centres A and B which touch each other externally at the point O. A straight line is drawn passing through the point O and intersects the two circles at P and Q respectively. Prove that AP | | BQ.

Solution:

Given:

Two circles have been drawn with centres A and B which touch each other externally at the point O. A straight line is drawn passing through the point O and intersects the two circles at P and Q respectively.

Let two circles with centres A and B touch each other externally at the point O.

∴ A, O, B will lie on the same straight line

Now, in ΔAOP and ΔBOQ, ∠AOP = ∠BOQ [Opposite angles]

Again, in circle with centre at A, AO = AP [radii of same circle]…..(1)

Also, in circle with, the centre at B,

BO = BQ [Radii of same circle]

∠BOQ = ∠BQO …….. (2)

Since ∠AOP = ∠BOQ, from (1) and (2) we get, ∠APO = ∠BQO

But these are the alternate angles produced when two line segments AP and BQ are intersected by the transversal PQ and also they are equal.

∴ AP | | BQ (Proved)

Example 8. Three equal circles touch one another externally. Prove that the centres of the three circles form an equilateral triangle. 

Solution:

Given:

Three equal circles touch one another externally.

Let three equal circles with centres A, B and C touch one another externally at the points P, Q and R.

To prove: ΔABC is an equilateral triangle.

Proof: In the circle with centre A,

AP = AR [radii of same circle]

Similarly, BP = BQ and CQ = CR.

Again, the circles are equal, so the radii of them are all equal, i.e., AP = BP = BQ = CQ = CR = RA…… (1)

Now, AB = AP + BP = BQ + CQ = BC ……(2)

[AP = BQ and BP = CQ, from (1)]

Again, BC = BQ + CQ = CR + AR [BQ = CR and CQ = AR] = CA… (3)

∴ from (2) and (3) we get, AB = BC = CA.

∴ ΔABC is an equilateral triangle. (Proved)

“Tangent properties and theorems for Class 10 students”

Example 9. Two tangents AB and AC drawn from an external point A of a circle touch the circle at the point B and C. A tangent, drawn to a point X lies on minor arc BC intersects AB and AC at the points D and E respectively. Prove that perimeter of ΔADE = 2AB.

Solution:

Given:

Two tangents AB and AC drawn from an external point A of a circle touch the circle at the point B and C. A tangent, drawn to a point X lies on minor arc BC intersects AB and AC at the points D and E respectively.

Let AB and AC are two tangents drawn from an external point A to the circle with centre O at the points B and C respectively.

Similarly, DB = DX and EX = EC……. (2)

Now, perimeter of ΔADE = AD + DE + EA

= AD + (DX + EX) + EA

= AD + (DB + EC) + EA [from (2)]

= (AD + DB) + (AE + EC)

= AB + AC

= AB + AB [AC = AB] = 2AB.

Hence perimeter of ΔADE = 2AB.(Proved)

Example 10. PQ is a diameter. The tangent drawn at the point R, intersects the two tangents drawn at the points P and Q at the points A and B respectively. Prove that ∠AOB is a right angle.

Solution:

Given:

PQ is a diameter. The tangent drawn at the point R, intersects the two tangents drawn at the points P and Q at the points A and B respectively.

Let us join O, R. Also, let OA intersects the arc PR at E and OB intersects the arc RQ at F.

AP = AR, ∴ arc PE = arc RE

The central angles subtended by them are equal, i.e., ∠POE = ∠ROE or, ∠AOP = ∠AOR

Similarly, ∠BOR = ∠BOQ.

Then ∠AOP + ∠AOR + ∠BOR + ∠BOQ = 180° [straight angle]

or, ∠AOR + ∠AOR + ∠BOR + ∠BOR = 180°

[∠AOP = ∠AOR, ∠BOQ = ∠BOR]

or, 2 ∠AOR + 2 ∠BOR = 180° or, 2 (∠AOR + ∠BOR) = 180°

or, 2 ∠AOB = 180° or, ∠AOB = \(\frac{180^{\circ}}{2}\)= 90°

Hence ∠AOB is a right angle. (Proved)

WBBSE Solutions For Class 10 Maths Solid Geometry Chapter 3 Theorems Related To Cyclic Quadrilateral

Cyclic quadrilateral

If the four vertices of a quadrilateral entirely lie on the circumference of a circle, then it is called a cyclic quadrilateral.

For example, all four vertices of the following quadrilaterals ABCD. PQRS. DEFG. etc. entirely lie on the circumference of a circle. So these are arc all cyclic quadrilaterals.

Characteristics of cyclic quadrilateral 

1.  All the vertices of a cyclic quadrilateral always lie on the circumference of any circle.
2. The centre of the circle may be inside or outside the quadrilateral. Such as besides, the centre of the circle is inside the quadrilateral in the first and outside the quadrilateral in the second.
3. The opposite angles of any cyclic quadrilateral are supplementary.
4. Conversely, if two opposite angles of a quadrilateral arc are supplementary, i.e. if sum of the two opposite angles be 180°, then it is a cyclic quadrilateral.

The most important characteristic of a cyclic quadrilateral is that two of its opposite angles are supplementary to each other.

We shall now prove logically this theorem by the method of geometry.

Solid Geometry Chapter 3 Theorems Related To Cyclic Quadrilateral

Cyclic Quadrilateral Theorem

Theorem 1. The opposite angles of a cyclic quadrilateral are supplementary.

 

Given: Let ABCD is a cyclic quadrilateral in the circle with centre at O.

To prove:

1. ∠ABC + ∠ADC = 2 right-angles
2. ∠BAD + ∠BCD = 2 right-angles

Construction: Let us join A, O and C, O.

Proof: The central angle produced by the circular arc \(\overparen{A B C}\) is ∠AOC and the angle in circle is ∠ADC.

∴ ∠AOC = 2 ∠ADC……. (1)

 WBBSE Solutions for Class 10 Maths

Again, the central angle produced by the arc \(\overparen{A D C}\) is reflex ∠AOC and angle in circle is ∠ABC.

∴ reflex ∠AOC = 2 ∠ABC …..(2)

Now, adding (1) and (2) we get,

∠AOC + reflex ∠AOC = 2 ∠ADC + 2 ∠ABC

or, 4 right angles = 2 (∠ADC + ∠ABC)

or, ∠ADC + ∠ABC = \(\frac{4}{2}\) right angles = 2 right angles.

∴ ∠ABC + ∠ADC = 2 right angles ((1) proved)

WBBSE Solutions for Class 10 History	WBBSE Solutions for Class 10 Geography and Environment
WBBSE Class 10 History Long Answer Questions	WBBSE Solutions for Class 10 Life Science And Environment
WBBSE Class 10 History Short Answer Questions	WBBSE Solutions for Class 10 Maths
WBBSE Class 10 History Very Short Answer Questions	WBBSE Solutions for Class 10 Physical Science and Environment
WBBSE Class 10 History Multiple Choice Questions

 

Similarly, by joining O, B and O, D we can prove that ∠BAD + ∠BCD = 2 right angles.

Hence, 1. ∠ABC + ∠ADC = 2 right angles

2. ∠BAD + ∠BCD = 2 right angles (Proved)

From the above theorem, we can say conversely that if any two opposite angles of a quadrilateral be supplementary to each other, then it is a cyclic quadrilateral.

The question now arises whether this theorem is always true.

We shall now prove it logically by the method of geometry.

Theorem 2. If the opposite angles of any quadrilateral be supplementary, then the vertices of the quadrilateral are concyclic.

Given: Let PQRS be a quadrilateral in which two opposite angles ∠PQR and ∠PSR are supplementary

i.e., ∠PQR + ∠PSR = 2 right angles.

To prove: The vertices P, Q, R, and S of the quadrilateral are concyclic.

Construction: We know that through three non-collinear points P, Q, and R only one circle can be drawn.

Let us draw the circle. Also, let the constructed circle does not pass through S. The circle intersects PS or extended PS at the point T. Let us join T and R.

Proof: Given that ∠PQR + ∠PSR = 2 right angles.

But by construction, ∠PQR + ∠PTR = 2 right angles.

∴ ∠PQR + ∠PSR = ∠PQR + ∠PTR or, ∠PSR = ∠PTR.

But it is impossible since any external angle of any triangle can never be equal to its internally opposite angle.

∴ ∠PSR = ∠PTR is possible only if the points S and T coincide, i.e., if the circle is drawn through three points P, Q, and R passes through point S.

∴ the circle through the points P, Q, and R also passes through S.

Hence the points P, Q, R, and S are concyclic. (Proved)

Corollary Theorem: If one side of a cyclic quadrilateral is produced then the external angle thus obtained is equal to its internally opposite angle.

Given: Let ABCD be a cyclic quadrilateral. Side BC of it is extended to E so that an external angle ∠DCE is produced.

To prove ∠DCE = internally opposite ∠BAD.

Proof: ABCD is a cyclic quadrilateral,

∴ ∠BAD + ∠BCD = 180°……. (1)

Again, ∠BCD + ∠DCE = 1 straight angle =180°.

∴ ∠BAD + ∠BCD = ∠BCD + ∠DCE [from (1) and (2)] or, ∠BAD = ∠DCE

Hence ∠DCE = internally opposite ∠BAD (proved)

Solid Geometry Chapter 3 Theorems Related To Cyclic Quadrilateral

Cyclic Quadrilateral Multiple Choice Questions

“WBBSE Class 10 cyclic quadrilateral solved examples”

Example 1. In the adjoining, O is the centre of the circle and AB is one of its diameters. If ∠ADC = 120°, then the value of ∠BAC is

1. 50°
2. 60°
3. 30°
4. 40°

Solution:

Given:

In the adjoining, O is the centre of the circle and AB is one of its diameters. If ∠ADC = 120°,

ABCD is a cyclic quadrilateral ∠ABC + ∠ADC = 180° or, ∠ABC + 120° = 180° or, ∠ABC = 180° – 120° = 60°

Again, AOB is a diameter. ∠ACB is a semicircular angle, ∠ACB = 90°

Then in ΔABC, ∠BAC + ∠ABC + ∠ACB = 180°

or, ∠BAC + 60° + 90° = 180° [30°ABC = 60° and ∠ACB = 90°]

or, ∠BAC = 180° – 150° = 30°

∴ 3. 30° is correct.

The value of∠BAC is 3. 30°

Example 2. In the adjoining, O is the centre of the circle and AB is the diameter. ABCD is a cyclic quadrilateral. If ∠ABC = 65°, ∠DAC = 40°, then the value of ∠BCD is 

1. 75°
2. 105°
3. 115°
4. 80°

Solution:

Given:

In the adjoining, O is the centre of the circle and AB is the diameter. ABCD is a cyclic quadrilateral. If ∠ABC = 65°, ∠DAC = 40°

AB is the diameter of the circle with centre at O.

∴ ∠ACB is a semicircular angle. ∴ ∠ACB = 90°.

Now, ABCD is a cyclic quadrilateral,

∴ ∠ADC + ∠ABC = 180°

or, ∠ADC + 65° = 180° or, ∠ADC = 180° – 65° or, ∠ADC =115°

∴ in ΔACD, ∠ACD + ∠ADC + ∠CAD = 180°

or ∠ACD + 115° + 40° = 180° or, ∠ACD =180° – 155° or, ∠ACD = 25°

∴ ∠BCD = ∠ACB + ∠ACD = 90° + 25° = 115°

∴ 3. 115°  is correct.

The value of ∠BCD is  3. 115°

Example 3. In the adjoining, O is the centre of the circle and AB is one of its diameters. ABCD is a cyclic quadrilateral in which AB || DC and if ∠BAC = 25°, then the value of ∠DAC is

1. 50°
2. 25°
3. 130°
4. 40°

Solution:

Given:

In the adjoining, O is the centre of the circle and AB is one of its diameters. ABCD is a cyclic quadrilateral in which AB || DC and if ∠BAC = 25°,

AOB is a diameter of the circle,

∴ ACB is a semicircular angle. ∴ ∠ACB = 90°

Now, ∠ABC =180°- (∠BAC + ∠ACB) =180°- (25° + 90°) = 180° – 115° = 65°

Again, ABCD is a cyclic quadrilateral,

∴ ∠ADC + ∠ABC =180°

or, ∠ADC + 65° = 180° or, ∠ADC = 180° -65° =115°

Also, AB || DC and AC is their transversal,

∴ ∠ACD = alternate ∠BAC = 25°

∴ ∠DAC = 180° – (∠ADC + ∠ACD)

= 180° – (115° + 25°) = 180° – 140° = 40°

∴ 4. 40°  is correct.

“Theorems related to cyclic quadrilateral for Class 10 Maths”

Example 4. In the adjoining, ABCD is a cyclic quadrilateral. BA is produced to F. If AE || CD, ∠ABC = 92° and ∠FAE = 20°, then the value of ∠BCD is

1. 20°
2. 88°
3. 108°
4. 72°

Solution:

Given:

In the adjoining, ABCD is a cyclic quadrilateral. BA is produced to F. If AE || CD, ∠ABC = 92° and ∠FAE = 20°

ABCD is a cyclic quadrilateral.

∴ ∠ABC + ∠ADC = 180°

or, 92° + ∠ADC = 180° [∠ABC = 92°]

or, ∠ADC = 180° – 92° = 88°

Also given that AE || CD and AD is their transversal.

∴ ∠CDA = alternate ∠DAE

or, 88° = ∠DAE [∠CDA = 88°] ∴ ∠DAE = 88°

∴ ∠DAF = ∠DAE + ∠EAF = 88° + 20° [∠DAE = 88°] = 108°

∴ 3. 108° is correct.

Solid Geometry Chapter 3 Theorems Related To Cyclic Quadrilateral

Cyclic Quadrilateral True Or False

“Chapter 3 cyclic quadrilateral exercises WBBSE solutions”

Example 1. In the adjoining, AD and BE arc the perpendiculars on side BC and CA respectively of the ΔABC, Then, A, B, D, E are concyclic.

Solution:

Given:

In the adjoining, AD and BE arc the perpendiculars on side BC and CA respectively of the ΔABC

True; since the angles ∠ADB and ∠AEB on the same side of AB are equal for being right angles each.

Hence the four points A, B, D, E are concyclic.

Example 2. In ΔABC, AB = AC, BE and CF are the bisectors of the angles ∠ABC and ∠ACB and they intersect AC and AB at the points E and F respectively. Then four points B, C, E, and F are not concyclic.

Solution:

Given:

In ΔABC, AB = AC, BE and CF are the bisectors of the angles ∠ABC and ∠ACB and they intersect AC and AB at the points E and F respectively.

False; since in ΔABC, AB = AC,

∴ ∠ABC = ∠ACB ⇒ \(\frac{1}{2}\) ∠ABC = \(\frac{1}{2}\)∠ACB

⇒ ∠CBE = ∠BCF [BE and CF are the bisectors of ∠ABC and ∠ACB respectively]

⇒ ∠BEC = ∠BFC [∠ABC = ∠ACB]

Thus, the angles ∠BEC and ∠BFC on the same side of BC are equal, so the four points B, C, E, and F are concyclic.

But given that B, C, E, and F are not concyclic. Hence the statement is false.

Solid Geometry Chapter 3 Theorems Related To Cyclic Quadrilateral

Cyclic Quadrilateral Fill In The Blanks

Example 1. All angles in the same segment are ______

Solution: Equal

Example 2. If the line segment joining two points subtends equal angles at two other points on the same side, then the four points are _______

Solution: Concyclic

Example 3. If two angles on the circle formed by two arcs are equal, then the lengths of arcs are ______

Solution: Equal.

Solid Geometry Chapter 3 Theorems Related To Cyclic Quadrilateral

Cyclic Quadrilateral Short Answer Type Questions

Example 1. In the adjoining figure, if ∠BAD = 60°, ∠ABC = 80°, then find the values of ∠DPC and ∠BQC.

Solution:

Given:

In the adjoining figure, if ∠BAD = 60°, ∠ABC = 80°

∠DPC = ∠APB

= 180° – (∠PAB + ∠PBA)

= 180° – (60° + 80°) = 180° – 140° = 40°

Again, ABCD is a cyclic quadrilateral.

∴ ∠ADC + ∠ABC =180°

or, ∠ADC + 80° = 180° or, ∠ADC = 180° – 80° = 100°

∴ ∠QDA =100°

∴ ∠BQC = ∠AQD

= 180° – (∠QAD + ∠QDA) = 180° – (60° + 100°) [∠QAD = 60° (given), ∠QDA = 100°]

= 180° – 160° = 20°

∴ ∠DPC = 40° and ∠BQC = 20°

“Class 10 Maths properties of cyclic quadrilaterals”

Example 2. In the adjoining, two circles with centres P and Q intersect each other at points B and C, ACD is a line segment. If ∠ARB = 150°, ∠BQD =x°, then find the value of x.

Solution:

Given:

In the adjoining, two circles with centres P and Q intersect each other at points B and C, ACD is a line segment. If ∠ARB = 150°, ∠BQD =x°

ARBC is a cyclic quadrilateral,

∴ ∠ARB + ∠ACB = 180°

or, 150° + ∠ACB = 180° or, ∠ACB = 180° – 150° = 30°

Again, ∠ACD = ∠ACB + ∠BCD = 1 straight angle = 180°

or, 30° + ∠BCD = 180° or, ∠BCD = 180° – 30° = 150°

Now, in the circle with centre Q, ∠BCD is an angle in circle and reflex ∠BQD is its central angle both produced by the circular arc \(\overparen{B D}\)

∴ reflex ∠BQD = 2 ∠BCD

or, 360° – ∠BQD = 2 ∠BCD or, 360° – x° = 2 x 150° or, x° = 360° – 300° or, x° = 60°

Hence the value of x is 60.

Example 3. In the adjoining, two circles intersect each other at the points P and Q. If ∠QAD = 80° and ∠PDA = 84°, then find the value of ∠QBC and ∠BCP.

Solution:

Given:

In the adjoining, two circles intersect each other at the points P and Q. If ∠QAD = 80° and ∠PDA = 84°,

In the circle with centre at O, ADPQ is a cyclic quadrilateral.

∴ ∠ADP + ∠AQP = 180° or, 84° + ∠AQP = 180° [∠ADP = 84°]

or, ∠AQP – 180° – 84° = 96°

Now, ∠BQP + ∠AQP = 1 straight angle or, ∠BQP + 96° = 180° [∠AQP = 96°] or, ∠BQP = 180° – 96° = 84°

Again, BQPC is a cyclic quadrilateral.

∴ ∠BCP + ∠BQP = 180° or, ∠BCP + 84° = 180°

or, ∠BCP = 180° – 84° = 96°

Now, ∠DPQ + ∠DAQ = 180° or, ∠DPQ + 80° =180° or, ∠DPQ = 180° – 80° = 100°

∴ ∠DPQ + ∠QPC = 1 straight angle = 180°

or, 100° + ∠QPC = 180° or, ∠QPC = 180° – 100° = 80°

Again, ∠QPC + ∠QBC = 180° [BQPC is a cyclic quadrilateral]

or, 80° + ∠QBC = 180° or, ∠QBC = 180° – 80° = 100°.

Hence ∠QBC = 100° and ∠BCP = 96°.

Example 4. In the adjoining, O is the centre of the circle and AC is its diameter of it. If DC || EB, ∠AOB = 80° and ∠ACE = 10°, then find the value of ∠BED.

Solution:

Given:

In the adjoining, O is the centre of the circle and AC is its diameter of it. If DC || EB, ∠AOB = 80° and ∠ACE = 10°,

Given that DC || EB and CE is their transversal.

∴ ∠DCE = ∠BEC [alternate angle]……. (1)

Now, ∠AOB = 2 ∠ACB [central angle is twice of angle in circle]

or, 80° = 2 ∠ACB [∠AOB = 80° given]

or, ∠ACB = \(\frac{80^{\circ}}{2}\) = 40°

Also, ∠BOC = 180° – ∠AOB = 180° – 80° = 100°

Again, the central angle produced by the chord \(\overparen{\mathrm{BC}}\) is ∠BOC and angle in circle is ∠BEC.

∴ ∠BEC= \(\frac{1}{2}\) ∠BOC = 100° =50° [∠BOC = 100°]

∴ ∠ECD = 50° [by (1)]

Also, ∠BED + ∠BCD =180° [BCDE is a cyclic quadrilateral]

or, ∠BED = 180° – ∠BCD =180° – 100° = 80°

[∠BCD = ∠ACB + ∠ACE + ∠ECD= 40° + 10° + 50° = 100°] ∴ ∠BED = 80°.

“Understanding cyclic quadrilaterals in Class 10 Maths”

Example 5. In the adjoining, O is the centre of the circle and AB is a diameter. If ∠AOD = 140° and ∠CAB = 50°, then find the value of ∠BED.

Solution:

Given:

In the adjoining, O is the centre of the circle and AB is a diameter. If ∠AOD = 140° and ∠CAB = 50°,

Given that ∠AOD = 140°

∴ reflex ∠AOD = 360° – 140° = 220°

Again, the central angle produced by the arc \(\overparen{\mathrm{ABD}}\) is reflex ∠AOD and angle in circle = ∠ACD.

∴ reflex ∠AOD = 2 ∠ACD

or, 220° = 2 ∠ACD or, ∠ACD = \(\frac{220^{\circ}}{2}\) = 110°

Given that ∠CAB = 50°

Now, in ΔACE, ∠AEC + ∠EAC + ∠ACE = 180° or, ∠AEC = 180° – 50° – 110° = 20°

Hence ∠BED = 20°

Solid Geometry Chapter 3 Theorems Related To Cyclic Quadrilateral

Cyclic Quadrilateral Long Answer Type Questions

Example 1. In the adjoining the diagonals of the cyclic quadrilateral PQRS intersect each other at the point X in such a way that ∠PRS = 65° and ∠RQS = 45°. Find the values of ∠SQP and ∠RSP

Solution:

Given:

In the adjoining the diagonals of the cyclic quadrilateral PQRS intersect each other at the point X in such a way that ∠PRS = 65° and ∠RQS = 45°.

Two angles in circle produced by the arc PS are ∠PRS and ∠PQS.

∴ ∠PQS = ∠PRS – 65° ∴ ∠SQP = 65°

Again, two angles in circle produced by the arc RS are ∠RQS and ∠RPS.

∴ ∠RQS = ∠RPS ⇒∠RPS = 45° [ ∠RQS = 45°]

Then in ΔPRS, ∠RSP + ∠PRS + ∠RPS = 180° or, ∠RSP + 65° + 45° = 180° [ ∠RPS = 45°] or, ∠RSP = 180° – 110° = 70°

∴ ∠SQP = 65° and ∠RSP = 70°

“Step-by-step solutions for cyclic quadrilateral problems Class 10”

Example 2. The side AB of the cyclic-quadrilateral ABCD is extended to X. If ∠XBC = 82° and ∠ADB = 47°. Find the value of ∠BAC.

Solution:

Given:

The side AB of the cyclic-quadrilateral ABCD is extended to X. If ∠XBC = 82° and ∠ADB = 47°.

Given that ∠XBC = 82°

∴ ∠ABC = 180° – ∠XBC = 180° – 82° = 98°

Again, two angles in circle produced by the arc AB are ∠ADB and ∠ACB, ∠ADB = ∠ACB, ∠ACB = 47°.

Now in ΔABC, ∠BAC + ∠ACB + ∠ABC = 180°

or, ∠BAC + 47° + 98° = 180° or, ∠BAC = 180° – 145° = 35°

Hence ∠BAC = 35°

Example 3. Two sides PQ and SR of the cyclic quadrilateral when produced meet at the point T. O is the centre of the circle; If ∠POQ = 110°, ∠QOR = 60°, ∠ROS = 80°, then find the value of ∠RQS and ∠QTR.

Solution:

Given:

Two sides PQ and SR of the cyclic quadrilateral when produced meet at the point T. O is the centre of the circle; If ∠POQ = 110°, ∠QOR = 60°, ∠ROS = 80°,

The angle in circle produced by the arc RS is ∠RQS and the corresponding central angle is ∠ROS = 80° (Given).

∴ ∠RQS = \(\frac{1}{2}\) ∠ROS = \(\frac{1}{2}\) x 80° = 40°.

Now, ∠ROP = ∠ROQ + ∠POQ = 60° + 110° [∠QOR = 60° and ∠POQ =110°] ∠ROP =170° .

Then, ∠RSP = \(\frac{1}{2}\)∠ROP [the angle in circle is ∠RSP and central angle is ∠ROP produced by the arc \(\overparen{\mathrm{RQP}}\)

= \(\frac{1}{2}\) x 170° = 85°

Again, ∠SOQ = ∠QOR + ∠SOR = 60° + 80° = 140°

Then ∠SPQ = \(\frac{1}{2}\)∠SOQ [angle in circle is ∠SPQ and central angle is ∠SOQ produced by the arc \(\overparen{\mathrm{SRQ}}\).

= \(\frac{1}{2}\) x 140° = 70°

∴ in Δ PTS, ∠PTS + ∠SPT + ∠PST = 180° .

or, ∠PTS + 70° + 85° = 180° [∠SPQ = 70° and ∠RSP = 85°] or, ∠PTS = 180° – 70° – 85° = 25° ∴ ∠QTR = 25°

∴ ∠RQS = 40° and ∠QTR = 25°

Example 4. Two circles intersect each other at the points P and Q. Two straight lines through P and Q intersect one circle at the points A and C and the other circle at B and D. Prove that AC 11 BD.

Solution:

Given:

Two circles intersect each other at the points P and Q. Two straight lines through P and Q intersect one circle at the points A and C and the other circle at B and D.

We know that two opposite angles of a cyclic quadrilateral are supplementary.

∴ for the quadrilateral ACQP,

∠CAP + ∠PQC = 180°……(1) and for the quadrilateral BPQD

∠PBD +∠PQD =180°…… (2)

Now adding (1) and (2), we get, ∠CAP + ∠PBD + ∠PQC + ∠PQD = 360°

or, ∠CAP + ∠PBD + ∠CQD = 360° [∠PQC + ∠PQD = ∠CQD = 1 straight angle = 180°]

or, ∠CAP + ∠PBD +180° = 360° or, ∠CAP + ∠PBD = 360° – 180° or, ∠CAP + ∠PBD =180°

But the common transversal of the straight lines AC and BD is AB and two adjacent angles on the same side of AB are ∠CAB and ∠DBA, the sum of which is 180°.

AC || BD [the sum of two adjacent angles on the same side of the common transversal of two straight lines is 180°]

Hence AC || BD. (Proved)

“WBBSE Mensuration Chapter 3 practice questions on cyclic quadrilaterals”

Example 5. Two straight lines are drawn through any point X exterior to a circle to intersect the circle at points A, B and points C, D respectively. Prove that ΔXAC and ΔXBD are equiangular.

Solution:

Given:

Two straight lines are drawn through any point X exterior to a circle to intersect the circle at points A, B and points C, D respectively.

 

ABCD is a cyclic quadrilateral.

∴ ∠ABD + ∠ACD =180° …… (1)

Again, ∠XCA + ∠ACD = 1 straight angle = 180°…..(2)

Then, by subtracting (1) from (2) we get,

∠XCA – ∠ABD = 0 or, ∠XCA = ∠ABD.

Again in cyclic quadrilateral ABDC, ∠BAC + ∠BDC = 180°……(3)

and ∠XAC + ∠BAC = 1 straight angle =180°……(4)

Then subtracting (3) from (4) we get,

∠XAC – ∠BDC = 0 or, ∠XAC = ∠BDC

∴ in triangles ΔXAC and ΔXBD,

∠XCA = ∠XBD [∠ABD = ∠XBD] and ∠XAC = ∠XDB

∴ two angles of each of ΔXAC and ΔXBD are equal.

Hence ΔXAC and ΔXBD are equiangular. (Proved)

Example 6. Two circles intersect each other at the points G and H. A straight line is drawn through the point G which intersect two circles at points P and Q and the straight line through the point H parallel to PQ intersects the two circles at the points R and S. Prove that PQ = RS. 

Solution:

Given:

Two circles intersect each other at the points G and H. A straight line is drawn through the point G which intersect two circles at points P and Q and the straight line through the point H parallel to PQ intersects the two circles at the points R and S.

Let us join P, R and Q, S.

As per the question, PQ || RS and PS is their transversal.

∴ ∠SPQ = ∠RSP…… (1) [alternate angles]

Similarly, PQ || RS and GH is their transversal.

∴ ∠PGH = ∠SHG ……(2) [alternate angles]

Again, PRHG is a cyclic quadrilateral

∴ ∠PGH + ∠PRH = 180° ….(3)

Similarly, QSHG is a cyclic quadrilateral.

∴ ∠SQG + ∠SHG =180°……(4)

From (3) and (4) we get, ∠PGH + ∠PRH = ∠SQG + ∠SHG or, ∠PRH = ∠SQG [from (2)] or, ∠PRS = ∠SQP…..(5)

Now, in ΔPRS and ΔPQS, ∠PSR = ∠QPS [from (1)] ∠PRS = ∠SQP [from (5)] and PS is common to both.

∴ ΔPRS = ΔPQS [by the A-A-S condition of congruency]

∴ RS = PQ [similar sides of congruent triangles]

Hence PQ = RS (proved)

Example 7. In triangle ABC, AB = AC and E is any point on the extended BC. If the circumcircle of ΔABC intersect AE at the point D, then prove that ∠ACD = ∠AEC.

Solution:

Given:

In triangle ABC, AB = AC and E is any point on the extended BC. If the circumcircle of ΔABC intersect AE at the point D,

In ΔABC, AB = AC,

∴ ∠ABC = ∠ACB………..(1)

Now, ABCD is a cyclic quadrilateral,

∴ ∠ABC + ∠ADC = 180°

or, ∠ACB + ∠ADC = 180° …… (2) [from (1)]

Again, ∠ACB + ∠ACE = 1 straight angle = 180°…….(3)

From (2) and (3) we get, ∠ACB + ∠ADC = ∠ACB + ∠ACE

or, ∠ADC = ∠ACE or, ∠DCE + ∠DEC = ∠ACD + ∠DCE or, ∠DEC = ∠ACD

∴ ∠ACD = ∠DEC.

Hence ∠ACD = ∠AEC (Proved)

Example 8. ABCD is a cyclic quadrilateral. The chord DE is the external bisector of ZBDC. Prove that AE (or extended AE), is the external bisector of ∠BAC.

Solution:

Given:

ABCD is a cyclic quadrilateral. The chord DE is the external bisector of ZBDC.

Two angles in circle produced by the arc BC are ∠BAC and ∠BDC.

∴ ∠BAC = ∠BDC……(1)

Again, two angles in circle produced by the are CF arc ∠CAE and ∠CDE.

∴ ∠CAE = ∠CDE….(2)

Now, ∠FDE + ∠CDE + ∠BDC = 1 straight angle = 180°.

or, ∠CDE + ∠CDE + ∠BDC = 180° [DE is bisector of ∠FDC  ∴ ∠FDE = ∠CDE]

or, 2 ∠CDE + ∠BDC = 180°…… (3)

Again, ∠GAE + ∠CAE + ∠BAC = 1 straight angle = 180°…… (4) .

From (3) and (4) we get, 2 ∠CDE + ∠BDC = ∠GAE + ∠CAE + ∠BAC

or, 2 ∠CDE = ∠GAE + ∠CAE [from (1)]

or, 2 ∠CAE = ∠GAC

or, ∠CAE = \(\frac{1}{2}\) ∠GAC

∴ AE is the bisector of ∠GAC.

Hence AE is the external bisector of ∠BAC. (Proved)

“Examples of cyclic quadrilateral calculations for Class 10”

Example 9. BE and CF are perpendicular on sides AC and AB of triangle ABC respectively. Prove that four points B, C, E, F are concyclic. Also prove that two angles of each of ΔAEF and ΔABC are equal.

Solution:

Given:

BE and CF are perpendicular on sides AC and AB of triangle ABC respectively.

BE ⊥ AC, ∠BEC = 90°

Also, CF ⊥ AB, ∠BFC = 90°, i.e., at two points E and F on the same side of BC, the two produced angles are equal.

∴ B, C, E, F are concyclic. (Proved)

Now, in ΔAEF, ∠AEF = ∠AEB – ∠BEF

= 90° – ∠BEF [∠AEB = 90°]

= 90° – ∠BCF [∠BEF = ∠BCF same angles in circle produced by the arc BF]

= ∠CBF [∠BFC = 90°]

Again, ∠AFE = ∠AFC – ∠CFE = 90° – ∠CFE [∠AFC = 90°]

= 90° – ∠CBE [∠CFE = ∠CBE, same angle in circle produced by the arc CE] = ∠BCE [∠BEC = 90°]

∴ in Δ’s AEF and ΔABC, ∠AEF = ∠ABC [∠CBF = ∠ABC] and ∠AFE = ∠ACB [∠BCE = ∠ACB]

Hence two angles of each of ΔAEF and ΔABC are equal. (Proved).

Example 10. ABCD is a parallelogram. A circle passing through points A and B intersect the sides AD and BC at points E and F respectively. Prove that the four points E, F, C, and D are concyclic.

Solution:

Given:

ABCD is a parallelogram. A circle passing through points A and B intersect the sides AD and BC at points E and F respectively.

Given: ABCD is a parallelogram. A circle with centre at O passing through the points A and B intersect the sides AD and BC at the points E and F respectively.

To prove: E, F, C, and D are concyclic.

Proof: As per ∠AEF + ∠DEF = 1 straight angle = 180°…..(1)

Again, ∠ABC + ∠BCD = 180° adjacent angles of a parallelogram]

[ABFE is a cyclic quadrilateral, ∴ ∠ABf + ∠AEF = 180° or, ∠ABF = 180° – ∠AEF or, ∠ABC – 180° – ∠AEF]

or, 180° – ∠AEF + ∠BCD = 180° or, ∠DEF + ∠BCD = 180° [180° – ∠AEF = ∠DEF] or, ∠DEF + ∠FCD = 180°

i.e.., two opposite angles of the quadrilateral EFCD are supplementary to each other.

Hence E, F, C, and D are concyclic. (Proved)

Example 11. ABCD is a cyclic quadrilateral. The two sides AB and DC are produced to meet in the point P and other two sides AD and BC are produced to meet in the point R. The two circumcircles of ABCP and ACDR intersect at the point T. Prove that the points P, T, and R are collinear.

Solution:

Given:

ABCD is a cyclic quadrilateral. The two sides AB and DC are produced to meet in the point P and other two sides AD and BC are produced to meet in the point R. The two circumcircles of ABCP and ACDR intersect at the point T.

ABCD is a cyclic quadrilateral. The two sides AB and DC are produced to meet in the point P and other two sides AD and BC are produced to meet in the point R.

The two circumcircles of ΔBCP and ΔCDR intersect at point T.

Proof: Two angles in circle produced by the arc CP are ∠CBP and ∠CTP,

∴ ∠CBP = ∠CTP…..(1)

Again, ∠CDR and ∠CTR are two angles in circle produced by the chord CR.

∴ ∠CDR = ∠CTR ….(2)

Now, ABCD is a cyclic quadrilateral,

∴ ∠ABC + ∠ADC = 180°

or, 180°- ∠CBP + 180°- ∠CDR = 180° [∠ABC + ZCBP = 180° and ∠ADC + ∠CDR =. 180°]

or, ∠CTP + ∠CTR = 180° [from (1) and (2)]

∴ ∠PTR = 1 straight angle

Hence the three points P, T, R are collinear (Proved).

“WBBSE Class 10 Maths solved problems on cyclic quadrilaterals”

Example 12. O is the orthocentre of the AABC. Prove that O is also the incentre of its pedal triangle.

Solution:

Given:

O is the orthocentre of the AABC.

Let AD, BE and CF be three perpendiculars drawn from the vertices A, B and C of the ΔABC to their opposite sides BC, CA and AB respectively which intersect each other at O.

Then O is the orthocentre of the ΔABC. Let us join D, E; E, F and F, D.

Then ΔDEF is the pedal triangle of ΔABC.

To prove: O is the incentre of ΔDEF.

Proof: The four points A, C, D, and F are concyclic,

[since the two angles ∠DCF and ∠DAF on the same side of FD are equal.

∴ ∠DCF = 90° – ∠ABD = ∠BAD = ∠FAD]

Now, the two angles in circle produced by the arc AF are ∠ADF and ∠ACF,

∴ ∠ADF = ∠ACF

= 90° – ∠EAF [∠AFC = 90°, ∴ ∠ACF + ∠CAF = 90°]

= ∠ABE [∠AEB = 90°, ∴ ∠ABE + ∠EAB = 90°]

= ∠ADE [A, B, D, E are concyclic and ∠ABE and ∠ADE are two angles in circle produced by arc AE.]

i.e., ∠ODF = ∠ODE ∴ OD is the bisector of ∠EDF.

Similarly, it can be proved that OE and OF are the bisectors of ∠DEF and ∠DFE.

So, O lies on the bisectors of the angles of ADEF, i.e., O is the point of intersection of the bisectors of the angles of the ΔDEF.

Hence O is the incentre of ADEF. (Proved)

Example 13. ABCD is a cyclic quadrilateral such that AC bisects ∠BAD. AD is produced to E in such a way that DE = AB. Prove that CE = CA.

Solution:

Given:

ABCD is a cyclic quadrilateral such that AC bisects ∠BAD. AD is produced to E in such a way that DE = AB.

Given: ABCD is a cyclic quadrilateral. AC bisects ∠BAD. AD is produced to E in such a way that DE = AB.

To prove: CE = CA

Proof: ABCD is a cyclic quadrilateral.

∴ ∠ABC + ∠ADC = 180° . . . (1)

Again, ∠ADC + ∠CDE = 1 straight angle = 180° ……. (2)

From (1) and (2) we get, ∠ABC + ∠ADC – ∠ADC + ∠CDE

⇒ ∠ABC = ∠CDE……(3)

Now, in ΔABC and ΔCDE, DE = AB (given),

BC = CD [the angle in circle ∠BAC produced by the chord BC and the angle in circle ∠CAD produced by the chord CD are equal, ∴BC = CD.]

and included ∠ABC = included ∠CDE [from (3)]

∴ ΔABC ≅ ΔCDE [by the S-A-S condition of congruency]

∴ CA = CE [similar sides of congruent triangles]

Hence CA = CE. (Proved)

Example 14. In two circles, one circle passes through the centre O of the other circle and they intersect each other at the points A and B. A straight line passing through A intersect the circle passing through O at the point P and the circle with centre at O at the point R. By joining P, B and R. B prove that PR = PB.

Solution:

Given:

In two circles, one circle passes through the centre O of the other circle and they intersect each other at the points A and B. A straight line passing through A intersect the circle passing through O at the point P and the circle with centre at O at the point R.

Given: Let the circle with centre at C passes through the centre of other circle with centre at O.

The two circles intersect each other at the points A and B. A straight line PAR intersects the circle passing through O at the point P and also intersect the circle with centre at O at the point R.

The points P, B and B, R are joined.

To prove: PR = PB.

Construction: Let us join O, A; O, B and O, R

Proof: In the circle with centre at O, OB = OR [radii of same circle]

∴ ∠OBR = ∠ORB…… (1)

Again, PAOB is a cyclic quadrilateral,

∴ ∠PAO + ∠PBO = 180°  ……… (2)

Now, ∠PAO + ∠OAR = 1 straight angle =180°….(3)

Then from (2) and (3) we get, ∠PAO + ∠PBO = ∠PAO + ∠OAR

or, ∠PBO = ∠OAR or, ∠PBO = ∠ORA [OA = OR, ∠OAR = ∠ORA]……(4)

Now, ∠PBR = ∠PBO + ∠OBR  = ∠ORA + ∠ORB [from (4) and (1)] = ∠PRB.

∴ in ΔPBR , ∠PBR = ∠PRB, PR = PB. (Proved)

Example 15. Prove that cyclic parallelogram must be a rectangle.

Solution:

Let ABCD be a parallelogram inscribed in the circle with centre at O.

To prove: ABCD is a rectangle.

Proof: ABCD is a cyclic quadrilateral,

∴ ∠ABC + ∠ADC = 180° …….(1)

Again, ABCD is a parallelogram.

∴ ∠ABC = ∠ADC [opposite angles of a parallelogram are equal.]

∴  from (1) we get, ∠ADC + ∠ADC = 180° or, 2 ∠ADC = 180° or, ∠ADC = 90°

Similarly, it can be proved that ∠ABC = ∠BCD = ∠CDA = ∠DAB = 90°

Also, since ABCD is a parallelogram,

∴ AB = DC and AD = BC.

Hence ABCD is a rectangle. (Proved)

“Cyclic quadrilateral angle relationships for Class 10 students”

Example 16. Prove that any four vertices of a regular pentagon are concyclic.

Solution:

Let ABCDE he a regular pentagon.

∴ AB = BC = CD = DE = EA.

To prove: Any four let A, B, C, E points are concyclic.

Construction: Let us join A, C and B, E.

Proof: In triangles ABC and ABE, BC = AE, AB is common to both and ∠ABC = ∠BAE [angles of a regular pentagon are equal.]

∴ ΔABC = ΔABE [by the S-A-S condition of congruency]

∴ ∠ACB = ∠AEB,

i.e. the angles produced at two points C and E on the same side of AB are equal.

A, B, C, E are concyclic.

Similarly, by selecting any other four points, it can be proved that the selected four points are concyclic.

Hence any four vertices of a regular pentagon are concyclic. (Proved)

Example 17. ABCD is a cyclic quadrilateral. The side BC of it is extended to E. Prove that the two bisectors of ∠BAD and ∠DCE meet on the circumference of the circle.

Solution:

Given:

ABCD is a cyclic quadrilateral. The side BC of it is extended to E.

Let ABCD is a cyclic quadrilateral in the circle with centre at O. The side BC of it is extended to E.

To prove: The two bisectors of ∠BAD and ∠DCE meet on the circumference of the circle.

Construction: Let us construct the bisector AP of ∠BAD, in which A, intersect the circle at P.

Let us join P, C and PC is extended to Q.

Proof: ∠BCP = ∠QCE [opposite angles] ……….. (1)

Now, ADCP is a cyclic quadrilateral.

∴ ∠PAD + ∠PCD = 180°……(2)

But ∠PCD + ∠DCQ =180°…… (3)

∴ from (2) and (3) we get, ∠PAD + ∠PCD = ∠PCD + ∠DCQ or, ∠PAD = ∠DCQ

or, ∠PAB = ∠DCQ…….. (4) [AP is the bisector of ∠BAD]

or, ∠PCB = ∠DCQ [both ∠PAB and ∠PCB are angles in circle produced by the arc BP.]

or, ∠QCE = ∠DCQ [from (1)] or, ∠ECQ = ∠DCQ

∴ CQ is the bisector of ∠DCE.

i.e…., PQ is the bisector of ∠DCE and intersects the circle at P.

Hence the two bisectors of ∠BAD and ∠DCE meet on the circumference of the circle. (Proved)

Example 18. AB is a diameter of a circle. PQ is such a chord of the circle that it is neither a diameter of the circle nor an interceptor of AB. By joining the points A, P and B, Q it is found that ΔBQP is a quadrilateral of which ∠BAP = ∠ABQ. Prove that ΔBQP is a cyclic trapezium.

Solution:

Given:

AB is a diameter of a circle. PQ is such a chord of the circle that it is neither a diameter of the circle nor an interceptor of AB. By joining the points A, P and B, Q it is found that ΔBQP is a quadrilateral of which ∠BAP = ∠ABQ.

Let AB be the diameter of the circle with centre at O and PQ is such a chord of the circle that is neither a diameter of the circle nor an interceptor of AB.

ABQP is a quadrilateral of which ∠ZBAP = ∠ABQ.

To prove: ABQP is a cyclic trapezium.

Proof: ABQP is a cyclic quadrilateral.

∴ ∠BAP + ∠BQP = 180°……..(1)

and ∠ABQ + ∠APQ =180°…….(2)

From (1) and (2) we get, ∠BAP + ∠BQP = ∠ABQ + ∠APQ

or, ∠BAP + ∠BQP = ∠BAP + ∠APQ [∠ABQ = ∠BAP]

or, ∠BQP = ∠APQ

from (1) we get, ∠BAP + ∠APQ = 180° [∠BQP = ∠APQ]

Thus, the sum of two adjacent angles on the same side of the transversal AP of two line segments AB and PQ is 180°.

∴ AB and PQ are parallel to each other, i.e., two opposite sides AB and PQ of the cyclic quadrilateral ABQP are parallel to each other.

Hence ABQP is a cyclic trapezium. (Proved)

Example 19. ABCD is a cyclic quadrilateral; The bisectors of ∠A and ∠C intersect the circle at points E and F respectively. Prove that EF is the diameter of the circle.

Solution:

Given:

ABCD is a cyclic quadrilateral; The bisectors of ∠A and ∠C intersect the circle at points E and F respectively.

Let ABCD is a cyclic quadrilateral in the circle with centre at O.

The bisector AE of the ∠A intersects the circle at E and the bisector of the ∠C intersects the circle at F. Let us join E, F.

To prove: EF is a diameter of the circle.

Construction: Let us join C, E.

Proof: ABCD is a cyclic quadrilateral.

∴ ∠BAD + ∠BCD = 180°

or, \(\frac{1}{2}\) ∠BAD + \(\frac{1}{2}\) ∠BCD = 90° [dividing by 2]

or, ∠DAE + ∠DCF = 90° [BE and CF are the bisectors of ∠A and ∠C respectively]

But two angles in circle produced by the arc DE are ∠DAE and ∠DCE.

∴ ∠DAE = ∠DCE.

∴ from (1) we get, ∠DCE + ∠DCF = 90° or, ∠ECF = 90°

∴ ∠ECF is a semicircular angle.

Hence EF is the diameter of the circle. [Proved]

Example 20. ΔABC is an acute angled triangle inscribed in a circle. AD is a diameter of the circle. Two perpendiculars BE and CF are drawn from B and C to AC and AB respectively, which intersect each other at the point G. Prove that BDCG is a parallelogram.

Solution:

Given:

ΔABC is an acute angled triangle inscribed in a circle. AD is a diameter of the circle. Two perpendiculars BE and CF are drawn from B and C to AC and AB respectively, which intersect each other at the point G.

Let ΔABC is an acute angled triangle, inscribed in a circle with centre at O.

Two perpendiculars BE and CF are drawn from B and C to the sides AC and AB which intersect each other at the point G.

AD is a diameter of the circle.

To prove: BDCG is a parallelogram.

Proof: AD is a diameter, ∴ ∠ABD = 90° [semicircular angle]

Again two angles in circle produced by the chord BD are ∠BCD and ∠BAD.

∴ ∠BCD = ∠BAD [angles in the same segment of circle are equal.]

= 90° – ∠BDA [∠ABD = 90°, ∠BAD + ∠BDA = 90°]

= 90° – ∠ACB [∠BDA and ∠ACB are angles in circle produced by the arc AB.]

= ∠CBE [∠BEC = 90°, ∴ ∠CBE + ∠BCE = 90°] = ∠CBG .

i.e., ∠BCD = ∠CBG.

But these are alternate angles, BG || DC.

Similarly, it can be proved that BD || GC. So, two opposite sides of the quadrilateral BDCG are parallel.

Hence BDCG is a parallelogram. (Proved)

WBBSE Solutions For Class 10 Maths Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

Angles In A Circle

Let ABCD is a. circle and O is its centre. Points A, B, C and D are on the circle.

If A, D and B, D are joined, then there is produced an angle ∠ADB. Then ∠ADB is called the angle in the circle produced by the arc \(\overparen{\mathrm{AB}}\).

Since D lies on the circumference, the angle ∠ADB is also called angle in circumference.

Similarly, ∠BAC, ∠ACB, ∠BDC, ∠CAD are all angles in a circle, which are produced by the arcs \(\overparen{\mathrm{BC}}, \overparen{\mathrm{AB}}, \overparen{\mathrm{BC}}, \overparen{\mathrm{CD}}\) respectively.

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Hence, any angle on the circle or on the circumference of a circle which is produced by any arc of the circle is called the angle in a circle.

Observe that by joining the two end points of an arc with a point situated on the opposite side of the arc and also on the circumference of the circle, we can get an angle in a circle.

 WBBSE Solutions for Class 10 Maths

However, since there exists infinitely many points on the opposite side of the fixed arc, we can draw an infinite number of angles’ in a circle.

Thus the number of angles in a circle produced by a fixed arc is infinite.

Again, the chord produced by joining the two end points of an arc, we can construct more than one angle in a circle on both the sides of this chord.

Such as, in the given two angles in a circle are ∠ACB (on the upper side) and ADB (on the lower side) produced by the arc \(A \overparen{D D} \text { and }  \overparen{A C B}\) respectively.

Characteristics of angles in circle 

1. The vertex of angle in circle always lies on the circumference of the circle.
2. The centre of the circle may or may not within the region of the angle in circle.
3. Any angle in circle must be produced by an arc of the circle.
4. Infinite number of angles in circle can be made by an arc of the circle.
5. Angle in circle may be either acute, obtuse or right angle.

Let ABC be a circle with centre at O. \(\overparen{\mathrm{AB}}\) is any arc of the circle.

Now, joining A, O and B, O by straight lines we get an angle ∠AOB at O. This angle ∠AOB is called the central angle.

Hence the front angle produced by an arc of the circle is called the central angle.

In the given ∠AOB, ∠BOC, ∠BOD are all central angles, which are produced by the arcs \(\overparen{\mathrm{AB}}, \overparen{\mathrm{BC}} \text { and } \overparen{\mathrm{BD}}\) respectively.

It must be noticed that both ∠AOB and reflex ∠AOB are produced by the arc AB.

Hence two and only two central angles can be constructed by a certain arc one of which is a reflex angle.

Characteristics of central angles 

1. The vertex of any central angle always lie on the centre of the circle.
2. Two and only two central angles can.be constructed by a fixed arc of a circle, one of which is a reflex angle.
3. The value of a central angle may be either an acute or an obtuse or a right-angle or a reflex angle.
4. The range of a central angle is from 0° to 360° including the both.
5. The central angle produced by the entire circumference of a circle is 360° and that produced by the circumference of a semi-circle is 180°.

Relation between angle in circle and its central angle produced by any arc of a circle 

In the above discussion, we have seen that by any arc of a circle both angle in circle and central angle can be constructed, So, it is the question that is their any relation between them?

Let the central angle produced by an arc AB of a circle with centre at O be ∠AOB and ∠ACB be its angle in circle.

If we measure these two angles ∠AOB and ∠ACB by a protractor, we shall see that ∠AOB = 2 ∠ACB, i.e., the angle which an arc of a circle subtends at the centre is double the angle which it subtends at any point on the remaining part of the circumference.

We shall now logically prove this theorem by geometric method.

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

Angles In A Circle Theorem

Theorem: Prove that the front angle formed at the centre of a circle by an arc, is the double of the angle formed by the same arc at any point on the circle.

Given: ∠AOB is the central angle of the circle with centre at O, produced by the arc APB and ∠ACB is the angle at any point on the circle formed by the same arc APB.

To prove: We have to prove that ∠AOB = 2 ∠ACB.

According to the length of the arc APB, these may be of three types, In Number (1) and (4), APB is a minor arc, In number, (2) APB is a semi circle and in number (3), APB is a major arc.

We shall have to prove that theorem for all the cases.

Construction: Let us join C. O and CO is produced upto D.

Proof: In ΔAOC. OA = OC (radii of same circle)

∴ ∠OCA = ∠OAC [in all the mages]

Again, tor the ΔAOC, in every cases, external ∠AOD = internally opposite (∠OAC + ∠OCA)

= ∠OCA + ∠OCA [∠OAC = ∠OCA] = 2 ∠OCA …….. (1)

Again, in ΔBQC, OB = OC [radii of same circle]

Similar is the case, when CO of ΔBOC is produced upto D, external ∠BOD = internally opposite (∠OCB + ∠OBC)

= ∠OCB + ∠OCB [∠OBC = ∠OCB]

= 2 ∠OCB……….(2)

Now in the cases of (1) arid (2), ∠AOD + ∠BOD = 2 ∠OCA + 2 ∠OCB

or, ∠AOB = 2 (∠OCA + ∠OCB) or, ∠AOB = 2 ∠ACB

In the (3), ∠AOD + ∠BOD – 2 ∠OCA + 2 ∠OCB

or, reflex ∠AOB = 2 (∠OCA + ∠OCB)

or, reflex ∠AOB = 2 ∠ACB

In the (4), let us subtract (1) from (2) to get,

∠BOD + ∠AOD = 2 ∠OCB + 2 ∠OCA

or, ∠AOB = 2 (∠OCB – ∠OCA) or, ∠AOB – 2 ∠ACB

∴ in every cases, ∠AOB = 2 ∠ACB (proved)

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle

Angles In A Circle Multiple Choice Questions

“WBBSE Class 10 angles in a circle solved examples”

Example 1. In the given, O is the centre of the circle and PQ is one of its diameter, then the value of x is

1. 140
2. 40
3. 80
4. 20

Solution: In the given, ∠POR = 140° (Given),

∴ ∠ROQ = 180° – 140° = 40°

Now, the angle in circle produced by the arc \(\overparen{Q R}\)

= ∠QSR and the central angle = ∠ROQ.

∴ ∠QSR = \(\frac{1}{2}\)∠ROQ [by theorem] = \(\frac{1}{2}\) x 40° = 20°

∴ x° = ∠QSR = 20°, ∴ x = 20,

∴ 4. 20 is correct.

The value of x is 4. 20.

Example 2. In the given, if O is the centre of the circle, then the value of x is

1. 70
2. 60
3. 40
4. 200

Solution: In the given, ∠POQ = 140°, ∠POR = 80°

∴ ∠QOR = 360° – (∠POQ + ∠POR)

= 360° – (140° + 80°) = 360° – 220° = 140°

Again, the angle in circle produced by the arc \(\overparen{Q R}\) = ∠QPR = x° and the central angle = ∠QOR = 140°

∴ ∠QPR = \(\frac{1}{2}\)∠QOR [by theorem]

= \(\frac{1}{2}\) x 140° =70°, x° = 70° ∴ x = 70

∴ 1. 70 is correct.

The value of x is 1. 70.

Example 3. In the given, if O is the centre of the circle and BC be any diameter of it, then the value of x is

1. 60
2. 50
3. 100
4. 80

Solution: In the given, ∠OAB = 50°, ∠ADC = x°,

Now, OA = OB (radii of same circle)

∴ ∠OAB = ∠OBA – 50° [∠OAB – 50°]

∴ ∠AOC = internally opposite (∠OAB + ∠OBA) = 50° + 50° = 100°

Again, the angle in circle produced by the circular arc \(\overparen{A C}\) = ∠ADC = x° and central angle = ∠AOC = 100°

∴ ∠ADC =  \(\frac{1}{2}\) ∠AOC [by theorem]

or, ∠ADC = \(\frac{1}{2}\) x 100° [∠AOC = 100°] .

. or, x° = 50°, x = 50,

∴ 2. 50 is correct.

The value of x is 2. 50.

Example 4. O is the circumcentre of triangle ABC. If ∠OAB = 50°, then the value of ∠ACB is

1. 50°
2. 100°
3. 40°
4. 80°

Solution:

Given

O is the circumcentre of triangle ABC. If ∠OAB = 50°

Let us join O, B.

Now, OA = OB (radii of same circle)

∴ ∠OBA – ∠OAB = 50° (Given)

∴ ∠AOB = 180° – (∠OBA + ∠OAB) = 180° – (50° + 50°) = 180° – 100° = 80°

Now, the angle in circle produced by the arc \(\overparen{Q R}\) = ∠ACB and the central angle = ∠AOB = 80°

∴ ∠ACB = \(\frac{1}{2}\) ∠AOB [by theorem] = \(\frac{1}{2}\)  x 80° = 40°

∴ 3. 40° is correct.

The value of x is 3. 40° .

Example 5. In the given, if O is the centre of the circle, then the value of ∠POR is 

1. 20°
2. 40°
3. 60°
4. 80°

Solution: In the given, ∠OPQ = 10°, ∠ORQ = 40°

∴ OP = OQ [radii of same circle]

∴ ∠OQP = ∠OPQ = 10° (Given)

Again, OQ = OR (radii of same circle)

∴ ∠OQR = ∠ORQ = 40° (Given)

Now, ∠PQR = ∠OQR – ∠OQP [according to the image]

= 40° – 10° = 30° [∠OQR = 40° and ∠OQP = 10°]

Then, the angle in circle produced by the arc \(\overparen{P R}\) = ∠PQR = 30° and the central angle = ∠POR

∴ ∠POR = 2 ∠PQR [by theorem]

. = 2 x 30° [∠PQR = 30°] .

= 60°

∴ 3. 60°  is correct.

The value of x is 3. 60°.

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle Angles In A Circle True Or False

Example 1. The number of angles in a circle produced by an fixed arc is infinite.

Solution: True

Example 2. The vertex of angle in circle does not always lie on the circumference of the circle.

Solution: False.

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle Angles In A Circle Fill In The Blanks

Example 1. On the same circular arc, the angle in circle is _______ of its central angle.

Solution: Half [by theorem]

Example 2. The lengths of the chords AB and AC of the circle with centre at O are equal. If ∠APB and ∠AQC are two of its angle in circle, the values of the angles are ______

Solution: Equal

Example 3. If O be the circum-centre of an equilateral triangle, then the value of the front central angle produced by any one of its sides is equal to ______.

Solution: 120°, since, let ABC be an equilateral triangle.

∴ ∠BAC = ∠ABC = ∠ACB = 60°.

Now, the front central angle produced by BC = ∠BOC.

The angle in circle produced by BC = ∠BAC.

∴ by theorem, ∠BOC = 2 ∠BAC = 2 x 60° = 120° [∠BAC = 60°]

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle Angles In A Circle Short Answer Type Questions

“Theorems related to angles in a circle for Class 10 Maths”

Example 1. In the adjacent O is the centre of the circle. If ∠OAB = 30°, ∠ABC = 120°, ∠BCO = y° and ∠COA = x°, then find the values of x and y.

Solution:

Given :

In the adjacent O is the centre of the circle. If ∠OAB = 30°, ∠ABC = 120°, ∠BCO = y° and ∠COA = x°

According to the given, ∠AOC = x°, ∠ABC = 120°, ∠OCB = y°, ∠BAC = 30°

Now, the central angle produced by the arc AC = reflex ∠AOC and angle in circle = ∠ABC =120°

∴ reflex ∠AOC = 2 x ∠ABC [by theorem] = 2 x 120° – 240°

Then ∠COA – 360° – reflex ∠AOC = 360° – 240° – 120°

∴ x° = 120° ⇒ x= 120

Again, ∠AOC + ∠OCB + ∠OAB + ∠ABC = 360 [the sum of 4 angles of a quadrilateral is 360°]

or, 120° + y° + 30° + 120° = 360°

or, y° + 270° = 360° or, y° = 360° – 270° or y° = 90°

∴ x = 120 and y = 90.

The values of x and y 120 and 90.

Example 2. O is the circumcentre of AABC and D is the mid-point of BC. If ∠BAC = 40°, then find the value of ∠BOD.

Solution:

Given :

O is the circumcentre of AABC and D is the mid-point of BC. If ∠BAC = 40°

D is the mid-point of BC, BD = CD

Now, in triangles ΔBOD and ΔCOD, OB = OC [radii of tire same circle], BD = CD and ∠OBD = ∠OCD [OB = OC]

∴ Δ BOD = Δ COD [by the condition of S-A-S congruent]

∴ ∠BOD = ∠COD [similar angles of congruent triangles]

∴ ∠BOC = 2 ∠BAC [by theorem]

= 2 x 40° [∠BAC – 40°] = 80°

Then from(1) we get, ∠BOD = \(\frac{1}{2}\)∠BOC = \(\frac{1}{2}\) x 80° [∠BOC = 80°] = 40°

The value of ∠BOD = 40°

Example 3. Three points A, B and C lie on a circle with centre at O in such a way that AOCB is a parallelogram. Find the value of ∠AOC. 

Solution:

Given :

Three points A, B and C lie on a circle with centre at O in such a way that AOCB is a parallelogram.

We know that sum of any two adjacent angles of a parallelogram is 180°.

∴ ∠AOC + ∠OAB = 180° ……(1)

Again, the central angle produced by the arc \(\overparen{A C}\) = reflex ∠AOC and angle in circle = ∠ABC

∴ by the theorem, reflex ∠AOC = 2 ∠ABC or, 360° – ∠AOC = 2 ∠AOC [∠ABC = ∠AOC, since opposite angles of any parallelogram are equal.]

or, 360° = 3 ∠AOC or, ∠AOC = \(\frac{360^{\circ}}{3}\) or, ∠AOC = 120°

∴ ∠AOC = 120°

“Chapter 2 angles in a circle exercises WBBSE solutions”

Example 4. The circumcentre of the isosceles triangle ABC is O and ∠ABC = 120°. It the radius of the circle be 5 cm, then determine the length of AB. 

Solution:

Given :

The circumcentre of the isosceles triangle ABC is O and ∠ABC = 120°. It the radius of the circle be 5 cm

Let us join O, A; O, B and O, C.

The central angle produced by the arc AC = reflex ∠AOC and angle in circle = ∠ABC = 120°

By theorem, reflex ∠AOC = 2 ∠ABC.

or, 360° – ∠AOC = 2 x 120°

or, 360° – ∠AOC = 240°

or, ∠AOC =360° – 240° = 120° ……..(1)

Again, in ΔABC, AB = BC [ABC is isosceles]

Now, in ΔOAB and ΔOCB, OA = OC [radii of same circle] A

B = BC and OB is common to both.

∴ ΔOAB = ΔOCB [by the condition of S-S-S congruent]

∴ ∠AOB = ∠BOC [similar angles of congruent triangles]

Now, ∠ABO + ∠CBO = ∠ABC = 120°

or, ∠ABO + ∠ABO = 120° [∠OBA = ∠OBC]

or, 2 ∠ABO = 120° or, ∠ABO = \(\frac{120^{\circ}}{2}\) = 60°

Again, ∠AOB + ∠COB = ∠AOC = 120° [by (1)]

or, ∠AOB + ∠AOB = 120° [∠COB = ∠AOB]

or, 2 ∠AOB = 120° or, ∠AOB = \(\frac{120^{\circ}}{2}\) =60°.

∴ in ΔAOB, ΔAOB = ∠ABO = 60°

∴ ΔAOB is equilateral.

∴ AB = AO = 5 cm [AO = radius = 5 cm]

Hence the required length of AB = 5 cm.

Example 5. Two circles with centres A and B intersect each other at points C and D. The centre B of the other circle lies on the circle with centre A. If ∠CQD = 70°, then find the value of ∠CPD.

Solution:

Given :

Two circles with centres A and B intersect each other at points C and D. The centre B of the other circle lies on the circle with centre A. If ∠CQD = 70°

Let us join A, C; B, C; A, D and B, D.

Now, in the circle with centre at B, angle in circle produced by the arc \(\overparen{C D}\) = ∠CQD and central angle = ∠CBD.

∴ by theorem, ∠CBD = 2 ∠CQD = 2 x 70° [∠CQD = 70°] =140°

Again, in the circle with centre A, the central angle produced by p the arc \(\overparen{C P D}\)= reflex ∠CAD and angle in circle = ∠CBD.

∴ by theorem, reflex ∠CAD = 2 ∠CBD

or, 360° – ∠CAD = 2. x 140° [∠CBD = 140°] or, ∠CAD = 360° – 280° = 80°

Again, in the circle with centre A, the central angle produced by the arc \(\overparen{C D}\) = ∠CAD and its angle in circle = ∠CPD.

By theorem, ∠CAD = 2 ∠CPD or, 80° – 2 ∠CPD [∠CAD = 80°]

or, ∠CPD = \(\frac{80^{\circ}}{2}\) = 40°

Hence the value of ∠CPD = 40°

“Class 10 Maths angle theorems in circles explained”

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle Angles In A Circle Long Answer Type Questions

Example 1. In the isosceles ΔABC, AB = AC. The circumcentre of ΔABC is O and the centre O lies on the opposite side of BC in which A lies. If ∠BOC = 100°, then find the value of ∠ABC and ∠ABO.

Solution:

Given :

In the isosceles ΔABC, AB = AC. The circumcentre of ΔABC is O and the centre O lies on the opposite side of BC in which A lies. If ∠BOC = 100°,

In ΔBOC, OB = OC [radii of same circle]

∴ ∠OBC = ∠OCB

or, ∠OBC + ∠OBC = ∠OCB + ∠OBC

or, 2 ∠OBC = 180° – ∠BOC or, 2 ∠OBC = 180° – 100°

or, 2 ∠OBC = 80° or, ∠OBC = \(\frac{80^{\circ}}{2}\) = 40°.

Again, the central angle produced by the arc BC = ∠BOC and the angle in circle = ∠BAC.

By theorem, ∠BOC = 2 ∠BAC.

or ∠BAC = \(\frac{1}{2}\) ∠BOC = \(\frac{1}{2}\) x 100° =50°

Now, ∠ABC = 180° – (∠ACB + ∠BAC)

= 180° – ∠ABC – 50° [∠ACB = ∠ABC and ∠BAC = 50°]

= 130°-∠ABC

or, 2 ∠ABC = 130° or, ∠ABC = \(\frac{130^{\circ}}{2}\) = 65°

Then ∠ABO – ∠ABC – ∠OBC – 65° – 40° = 25°

Hence ∠ABC = 65 ° and ∠ABO = 25°

Example 2. In the adjoining, if O is the centre of the circumcircle of ΔABC and ∠AOC = 110°, find the value of ∠ABC

Solution:

Given :

In the adjoining, if O is the centre of the circumcircle of ΔABC and ∠AOC = 110°,

The angle in circle produced by the major arc \(\overparen{A B}\) = ∠ABC and the centre angle = reflex ∠AOC.

∴ by theorem, reflex ∠AOC = 2 ∠ABC or, 360° – ∠AOC = 2 ∠ABC

or, 360° – 110° = 2 ∠ABC [∠AOC = 110°]

or, 250° = 2 ∠ABC 250°

or, ∠ABC = \(\frac{250^{\circ}}{2}\) = 125°

∴ ∠ABC = 125°

Example 3. O is the centre of the circle. If ∠AOD = 40° and ∠ACD = 35° then find the values of ∠BCO and ∠BOD.

Solution:

Given :

O is the centre of the circle. If ∠AOD = 40° and ∠ACD = 35°

The central angle produced by the arc \(\overparen{A B}\) = ∠AOB and the angle in circle = ∠ACB = 35° (Given)

∴ by theorem, ∠AOB = 2 ∠ACB = 2 x 35° = 70°

Again, the central angle produced by \(\overparen{A D}\) = ∠AOD and angle in circle = ∠ACD or ∠ACO.

∴ by theorem, ∠AOD = 2 ∠ACO

or, 40° = 2 ∠ACO [∠AOD = 40° (Given)]

or, ∠ACO = \(\frac{40^{\circ}}{2}\) = 20°

Now, ∠BCO = ∠ACB = 35° + 20° = 55° [∠ACB = 35° (Given) and ∠ACO = 20°]

and ∠BOD = ∠AOD + ∠AOB = 40° + 70° = 110° [∠AOD = 40° (Given) and ∠AOB = 70°] ∠BCO = 55° and ∠BOD = 110°

Example 4. Like the adjoining figure, draw two circles with centres C and D which intersect each other at the points A and B. Draw a straight line through A which intersects the circle with centre C at the point P and the circle with centre D at the point Q.

Prove that (1) ∠PBQ = ∠CAD; (2) ∠BPC = ∠BQD.

Solution:

Given :

Like the adjoining figure, draw two circles with centres C and D which intersect each other at the points A and B.

In the circle with centre C, the central angle produced by the chord AP = ∠ACP and angle in circle = ∠ABP.

By theorem, ∠ACP = 2 ∠ABP or ∠ABP = \(\frac{1}{2}\) ∠ACP….(1)

Again, in the circle with centre at D, the central angle produced by the chord AQ = ∠ADQ and angle in circle ∠ABQ.

∴ by theorem, ∠ADQ = 2 ∠ABQ or, ∠ABQ = \(\frac{1}{2}\) ∠ADQ …(2)

Then, adding (1) and (2) we get,

∠ABP + ∠ABQ = \(\frac{1}{2}\) ∠ACP+ \(\frac{1}{2}\) ∠ADQ

or, ∠PBQ = \(\frac{1}{2}\)(∠ACP + ∠ADQ) = \(\frac{1}{2}\) (180° – 2 ∠PAC +180°- 2 ∠QAD)

[∠ACP + ∠PAC + ∠APC – 180°]

or, ∠ACP + ∠PAC + ∠PAC – 180° or, ∠ACP – 180° – 2 ∠PAC

Similarly, ∠ADQ = 180° – 2 ∠QAD

or, ∠PBQ = \(\frac{1}{2}\)[360° -2 (∠PAC + ∠QAD)] = \(\frac{1}{2}\) [360° -2 (180° – ∠CAD)]

= \(\frac{1}{2}\) [360° – 360° + 2 ∠CAD] = \(\frac{1}{2}\) x 2 ∠CAD = ∠CAD

∴ ∠PBQ = ∠CAD. [Proved (1)]

Now, let us join B, C and B, D

CP = CB (radii of same circle), ∠BPC = ∠PBC ……. (3)

BD = DQ (radii of same circle), ∠DBQ = ∠DQB……. (4)

Now, from (1) we get, ∠PBQ = ∠CAD.

or, ∠PBA + ∠QBA = ∠CAB + ∠DAB .

or, ∠PBC + ∠CBA + ∠DBA – ∠DBQ – ∠CAB + ∠DAB

or, ∠PBC + ∠CAB + ∠DAB – ∠DBQ = ∠CAB + ∠DAB

[CB = CA, ∴ ∠CBA = ∠CAB]

DB = DA, ∴ ∠DBA = ∠DAB

or, ∠PBC – ∠DBQ = 0 or, ∠PBC = ∠DBQ or, ∠BPC = ∠BQD [∠CBP = ∠BPC and ∠DBQ = ∠BQD]

∴ ∠BPC = ∠BQD. [Proved (2)]

Hence (1) ∠PBQ = ∠CAD and (2) ∠BPC = ∠BQD (Proved)

“Understanding angles in a circle for Class 10”

Example 5. Each of two equal circles passes through the centre of the other and the two circles intersect each other at the points A and B. If a straight line through the point A intersects the two circles at points C and D, prove that ABCD is an equilateral triangle.

Solution:

Given :

Each of two equal circles passes through the centre of the other and the two circles intersect each other at the points A and B. If a straight line through the point A intersects the two circles at points C and D

Let the two circles with centres at P and Q respectively are equal.

The two circles intersect each other at A and B.

The straight line CD passes through A and intersects the circle with centre P at C and with centre Q at D.

To prove: We have to prove that ΔBCD is equilateral.

Construction: Let us join P, Q; A, P; B, P; A, Q and B, Q. Let PQ intersects AB at S.

Proof: In triangles ΔAPS and ΔBPS, AP = BP [radii of same circle]

AS = BS [S is the mid-point of AB] and PS is common to both.

∴ ΔAPS ≅ ΔBPS, ∴ ∠APS = ∠BPS [similar angles of congruent triangles]……. (1)

Now, in the circle with centre at P, the angle in circle produced by the arc AB = ∠ACB and the central angle = ∠APB.

∴ ∠APB = 2 ∠ACB [by theorem]

or, ∠APS + ∠BPS = 2 ∠ACB

or, ∠APS + ∠APS = 2 ∠ACB

or, 2 ∠APS = 2 ∠ACB or, ∠APS = ∠ACB…… (2)

Also in the circle with centre at Q, the central angle produced by the arc \(\overparen{A D B}\) = reflex ∠AQB and angle in circle = ∠APB.

∴ by theorem, reflex ∠AQB = 2 ∠APB

or, 360° – ∠AQB – 2 ∠APB

or, 360° – ∠APB -2 ∠APB

[∠AQB = ∠APB,

Since ΔAPS = ΔAQS ⇒  ∠APS = ∠AQS,

Similarly, ∠BPS = ∠BQS]

or, 360° = 3 ∠APB or, ∠APB = \(\frac{360^{\circ}}{3}\) = 120°

or, 2 ∠APS =120° [∠APB = 2 ∠APS]

or, ∠APS = \(\frac{120^{\circ}}{2}\) = 60°

∴ ∠ACB = 60° [from (2)]

Similarly, it can be proved that ∠BDC = 60°

∴ the other angle of ΔBCD is 60°.

i.e., in mangle BCD, ∠BCD = ∠BDC = ∠CBD = 60° .

Hence ΔBCD is an equilateral triangle. (Proved).

Example 6. S is the centre of the circumcircle of ΔABC and if AD ⊥ BC, prove that ∠BAD = ∠SAC.

Solution:

Given :

S is the centre of the circumcircle of ΔABC and if AD ⊥ BC

S is the centre of the circum-circle of ΔABC,

∴ SA = SB = SC.

Also, AD ⊥ BC, ∠ADB = 90°, ∠ABD + ∠BAD = 90°

or, ∠ABD = 90° – ∠BAD……(1)

Now, in the circle with centre S, the central angle produced by the chord AC = ∠ASC and angle in circle = ∠ABC

∴ ∠ASC = 2 ∠ABC…… (2)

But SA = SC [radii of same circle]

∴ ∠SAC = ∠SCA….. (3)

Then, ∠ASC + ∠SAC + ∠SCA = 180°

or, ∠ASC + ∠SAC + ∠SAC = 180° [from (3)] .

or, ∠ASC + 2 ∠SAC = 180°

or, 2 ∠ABC + 2 ∠SAC – 180° [by (2)]

or, ∠ABC + ∠SAC = 90° [dividing by 2]

or, ∠ABD + ∠SAC = 90° [∠ABC and ∠ABD are same angle]

or, 90° – ∠BAD + ∠SAC = 90° [from (1)]

or, ∠SAC – ∠BAD = 0 or, ∠SAC = ∠BAD

Hence ∠BAD = ∠SAC (Proved)

Example 7. Two chords AB and CD of a circle with centre O intersect each other at point P. Prove that ∠AOD + ∠BOC = 2 ∠BPC. If ∠AOD and ∠BOC are supplementary to each other, then prove that the two chords are perpendicular to each other.

Solution:

Given :

Two chords AB and CD of a circle with centre O intersect each other at point P. Prove that ∠AOD + ∠BOC = 2 ∠BPC. If ∠AOD and ∠BOC are supplementary to each other

In the circle with centre at O, the two chords AB and CD intersect each other at P.

To prove: We have to prove that ∠AOD + ∠BOC = 2 ∠BPC and AB ⊥ CD when ∠AOD + ∠BOC = 180°.

Construction: Let us join O, A; O,B; O, C; O, D and B, D.

Proof: The central angle produced by the arc \(\overparen{\mathrm{AD}}\) = ∠AOD and angle in circle = ∠ABD.

∴ ∠AOD = 2 ∠ABD [by theorem] ……….(1)

Again, the central angle produced by the arc BC = ∠BOC and angle in circle = ∠BDC.

∴ ∠BOC = 2 ∠BDC (2) [by theorem]

Now, adding (1) and (2) we get, . ‘ .

∠AOD + ∠BOC = 2 ∠ABD + 2 ∠BDC = 2 (∠ABD + ∠BDC)…….(3)

But by producing DP of ΔPBD upto C, the produced external ∠BPC = internally opposite (∠PBD + ∠BDP) = internally opposite (∠ABD + ∠BDC)

[they are same angles]

∠ABD + ∠BDC = ∠BPC

∴ from (3) we get, ∠AOD + ∠BOC = 2 ∠BPC.

∴ ∠AOD + ∠BOC = 2 ∠BPC. (Proved)

Now, if ∠AOD + ∠BOC = 180°, then we get, 180° = 2 ∠BPC or, ∠BPC = = 90°

Hence AB ⊥ CD [∠BPC = 90°] [Proved]

“Step-by-step solutions for angles in circles Class 10”

Example 8. If two chords AB and CD of a circle with centre O, when produced intersect each other at the point P, prove that ∠AOC – ∠BOD = 2 ∠BPC.

Solution:

Given :

If two chords AB and CD of a circle with centre O, when produced intersect each other at the point P,

Let AB and CD be two chords of the circle with centre at O. Produced AB and CD intersect at P.

To prove: We have to prove that ∠AOC – ∠BOD = 2 ∠BPC

Construction: Let us join B, C.

Proof: The central angle produced by the arc \(\overparen{\mathrm{AC}}\) = ∠AOC and angle in circle = ∠ABC.

∠AOC = 2 ∠ABC…. (1) [by theorem]

Again, the central angle produced by the arc BD = ∠BOD and angle in circle = ∠BCD

∴ ∠BOD = 2 ∠BCD…… (2) [by theorem]

Now, subtracting (2) from (1) we get,

∠AOC – ∠BOD – 2 ∠ABC – 2 ∠BCD ….(3)

Again, by producing PB of ABPC we get, external ∠ABC.

∴ ∠ABC = internally opposite (∠BPC + ∠BCP)

or, ∠ABC = ∠BPC + ∠BCD.

or, 2 ∠ABC = 2 ∠BPC + 2 ∠BCD [multiplying by 2] or, 2 ∠ABC – 2 ∠BCD = 2 ∠BPC

∴ from (3) we get, ∠AOC – ∠BOD = 2 ∠BPC.

Hence, ∠AOC – ∠BOD = 2 ∠BPC. (Proved)

Example 9. Draw a circle with the point A of quadrilateral ABCD as centre which passes through the points B, C and D. Prove that ∠CBD + ∠CDB = \(\frac{1}{2}\) ∠BAD.

Solution:

Let a circle be drawn with centre at A of the quadrilateral ABCD, which passes through B, C and D.

To prove: We have to prove that ∠CBD + ∠CDB = \(\frac{1}{2}\)∠BAD.

Construction: Let us join A, B; B, C; C, D; D, A; B, D and C, A.

Proof: The central angle produced by the arc \(\overparen{\mathrm{BC}}\) = ∠BAC and angle in circle = ∠BDC .

∴ ∠BDC = \([\frac{1}{2}\)∠BAC…… (1) [by theorem]

Again, the central angle produced by \(\overparen{\mathrm{CD}}\) = ∠CAD and angle in circle = ∠CBD

∴ ∠CBD = \([\frac{1}{2}\) ∠CAD… ….(2) [by theorem]

Then, adding (1) and (2) we get,

∠BDC + ∠CBD = \(\frac{1}{2}\)∠BAC + \(\frac{1}{2}\) ∠CAD = \(\frac{1}{2}\) (∠BAC + ∠CAD) = \(\frac{1}{2}\) ∠BAD.

Hence ∠CBD + ∠CDB \(/frac{1}{2}\) = \(/frac{1}{2}\) ∠BAD.

Example 10. O is the circumcentre of ΔABC and OD is perpendicular on the side BC prove that ∠BOD = ∠BAC

Solution:

Given :

O is the circumcentre of ΔABC and OD is perpendicular on the side BC

The circumcentre of ΔABC is O and OD ⊥ BC.

To prove: We have to prove that ∠BOD = ∠BAC

Construction: Let us join O, A; O, B; and O, C.

Proof: In ΔBOD and ΔCOD, OB = OC [radii of same circle],

∠OBD = ∠OCD [OB = OC] and ∠BDO = ∠CDO [each is right angle]

∴ ΔBOD ≅ ΔCOD [by the condition of A-A-S congruence]

∴ ∠BOD = ∠COD [similar angles of congruent triangles]

∴ ∠BOC = ∠BOD + ∠COD = ∠BOD + ∠BOD [∠COD = ∠BOD] = 2 ∠BOD……. (1)

Now, the central angle produced by the chord BC = ∠BOC and angle in the circle = ∠BAC.

∴ ∠BOC = 2 ∠BAC [by theorem]……(2)

Then adding (1) and (2) we get,

2 ∠BOD = 2 ∠BAC or, ∠BOD = ∠BAC

Hence ∠BOD = ∠BAC (Proved)

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle Central Angle

We have seen earlier that the number of angles in circle produced by the same circular arc in the same circle are infinity.

Now the question arises is there any relation among these infinite number of angles in circle?

Let in a circle with centre at O, ∠ACB and ∠ADB are two angles in circle produced by the arc APB.

We shall see that if these two angles are measured by a protractor, then the two angles are equal.

Hence angles in the same segment of a circle are equal.

We shall now prove this theorem logically by the geometric method.

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle Central Angle Theorem

Theorem: Prove that angles in the same segment of a circle are equal.

Given: Let ∠ACB and ∠ADB are two angles in a circle with centre at O, produced by the circular arc \(\overparen{\mathrm{APB}}\)

To prove: ∠ACB = ∠ADB

Construction: Let us join O, A and O, B

Proof: ∠ACB is an angle in circle and ∠AOB is it’s central angle both produced by the same arc \(\overparen{\mathrm{APB}}\) of the circle.

∴ ∠AOB = 2 ∠ACB….. (1)

Similarly, ∠ADB is an angle in circle and ∠AOB is the central angle both produced by the circular arc APB of the circle.

∴ ∠AOB = 2 ∠ADB …… (2)

Now, from (1) and (2) we get, 2 ∠ACB = 2 ∠ADB, or, ∠ACB = ∠ADB

Hence angles in the same segment of a circle are equal (Proved).

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle Concylic Points

Three or more than three points are said to be concyclic if they lie on the same circle.

Such as, in the adjoining, the points A, B, C, and D lie on the circle with centre O. So, the points A, B, C, D are concyclic.

There are some properties and characteristics of concyclic points.

Such as, the opposite angles of the quadrilateral which is obtained by joining the four concyclic points are supplementary to each other.

Again, if a line segment joining two points subtends equal angles at two other points on the same side of it, then the four points are concyclic.

We shall now logically prove this theorem by the geometric method.

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle Concylic Points Theorem

“Examples of angle theorems for WBBSE Class 10 Maths”

Theorem: Prove that if a line segment joining two points subtends equal angles at two other points on the same side of it, then four points are concyclic.

Given: Let the line segment joining two points A and B subtend equal angles ∠ACB and ∠ADB, i.e., ∠ACB = ∠ADB.

To prove: We have to prove that A, B, C, D are concyclic.

Construction: Let us draw the circle passing through three non-collinear points A, B and C.

Proof: If the circle constructed above passes through the fourth point D, then the theorem is proved.

But if the circle does not pass through D, it will intersect AD or produced AD at a point.

Let the circle does not pass through D and intersect AD at E. Let us join E, B.

Now, ∠ACB = ∠AEB [angles in the same segment of a circle]

But given that ∠ACB = ∠ADB = ∠AEB = ∠ADB

But it is impossible since any external angle of a triangle can never be equal to its internally opposite angles.

So the circle passing through the three non-collinear points A, B, C must pass through point D.

Hence the four points A, B, C and D are concyclic. (Proved)

In the following examples, the various application of the above theorems are discussed thoroughly.

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle Concylic Points Multiple Choice Questions

Example 1. In the adjoining figure, O is the centre of the circle; if ∠ACB = 30°, ∠ABC = 60°, ∠DAB = 35° and ∠DBC = x°, then the value of x is

1. 35
2. 70
3. 65
4. 55

Solution:

Given

In the adjoining figure, O is the centre of the circle; if ∠ACB = 30°, ∠ABC = 60°, ∠DAB = 35° and ∠DBC = x°

∠DBC and ∠CAD are two angles in circle produced by the same arc \(\overparen{\mathrm{CD}}\) with centre at O.

∴ ∠DBC = ∠CAD……(1)

Now, in ΔABC, ∠ACB = 30° and ∠ABC = 60°

∴ ∠BAC = 180° – (∠ACB + ∠ABC)

= 180° – (30° + 60°) = 180° – 90° = 90°

Again, ∠DAB = 35° (Given),

∴ ∠CAD = ∠BAC – ∠DAB = 90° – 35° = 55°

∴ from (1) we get, ∠DBC = ∠CAD = 55°, x° = 55°

Hence x = 55

∴ 4. 55 is correct.

Example 2. In the adjoining, O is the centre of the circle; if ∠B AD = 65°, ∠BDC = 45°, then the value of ∠CBD is

1. 65°
2. 45°
3. 40°
4. 20°

Solution:

Given

In the adjoining, O is the centre of the circle; if ∠B AD = 65°, ∠BDC = 45°,

∠BAC = ∠BDC = 45° [∠BDC = 45° (Given)]

Again, ∠BAD = 65° (Given), ∠CAD = ∠BAD – ∠BAC = 65° – 45° = 20°

Now, both ∠CAD and ∠CBD are angles in circle produced by the same arc \(\overparen{\mathrm{CD}}\) of the circle.

∴ ∠CAD = ∠CBD

But ∠CAD = 20°, ∠CBD = 20°

∴ 4. 20° is correct.

Example 3. In the adjoining, O is the centre of the circle. If ∠AEB = 110°, ∠CBE = 30°, then the value of ∠ADE is

1. 70°
2. 60°
3. 80°
4. 90°

Solution:

Given

In the adjoining, O is the centre of the circle. If ∠AEB = 110°, ∠CBE = 30°,

In the circle with centre at O, the angles in circle produced by the arc AB are ∠ADB and ∠ACB.

∴ ∠ADB = ∠ACB…….(1)

Now, if the side CE of ABCE be produced to A, then the external angle ∠AEB produced is equal to the internally opposite (∠CBE + ∠BCE),

i.e., ∠AEB = ∠CBE + ∠BCE

or, 110° = ∠CBE + ∠BCE [∠AEB =110° (Given)]

or, ∠BCE = 110° – ∠CBE or, ∠BCE = 110° – 30° [∠CBE = 30° (Given)]

or, ∠ACB = 80° [∠BCE and ∠ACB are same angles]

∴ ∠ADB = ∠ACB = 80° [from (1)] or, ∠ADB = 80°

∴ 3. 80° is correct.

Example 4. In the adjoining, O is the centre of circle. If ∠BCD = 28°, ∠AEC = 38°, then the value of ∠AXB is 

1. 56°
2. 94°
3. 38°
4. 28°

Solution:

Given

In the adjoining, O is the centre of circle. If ∠BCD = 28°, ∠AEC = 38°

In the circle with centre at O, the angles ∠BAD and ∠BCD are both angles in circle produced by the arc \(\overparen{\mathrm{BD}}\).

∴ ∠BAD = ∠BCD = 28° (Given)

Again, ∠AEC = 38°, ∴ ∠ADC = (∠AEC + ∠DAB)

[external ∠ADC = internally opposite (∠AEC + ∠DAE)]

∴ ADC = 38° + ∠BAD [∠DAB and ∠BAD are same angles] – 38° + 28° = 66°

Again, ∠AXB = internally opposite (∠XDC + ∠XCD)

= ∠ADC + ∠BCD = 66° + 28° = 94°

∴ 2. 94°  is correct.

Example 5. In the adjoining, O is the centre of the circle and AB is its a diameter. If AB ∥ CD, ∠ABC = 25°, then the value of ∠CED is

1. 80°
2. 50°
3. 25°
4. 40°

Solution:

Given

In the adjoining, O is the centre of the circle and AB is its a diameter. If AB ∥ CD, ∠ABC = 25

Let us join B, D and A, D.

AB ∥ CD, ∴ ∠ABD + ∠CDB = 180°……..(1)

Again, in the circle with centre at O, ∠CED and ∠CBD are two angles in circle produced by the arc \(\overparen{\mathrm{AD}}\),

∴ ∠CED = ∠CBD…. (1)

Now, ∠ABC = 25° (Given), ∴ ∠BCD = 25° [AB ∥ CD, alternate angle]

Again, ∠ADC = ∠ABC [angles in the same segment of the circle]

AB is a diameter,

∴ ∠AOB is the central angle and angles in circle is ∠ADB both produced by the arc \(\overparen{\mathrm{AEB}}\).

∴ ∠ADB = \(\frac{1}{2}\)∠AOB = \(\frac{1}{2}\) x 180° = 90° [∴ ∠AOB is a straight angle]

Then ∠BDC = ∠ADC + ∠ADB = 25° + 90° = 115°

∴ in ΔBCD, ∠CBD = 180° – (∠BCD + ∠BDC) = 180° – (25° + 115°) = 180° – 140° = 40°

from (1) we get, ∠CED = ∠CBD = 40°

∴ 4. 40° is correct.

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle Concylic Points True Or False

Example 1. In the adjoining AD and BE are perpendicular to BC and AC respectively of the ΔABC. The four points A, B, D, E are concyclic.

Solution: True,

since the angles ∠AEB and ∠ADB are on the same side of AB.

Now, ∠AEB and ∠ADB are both right angles, ∠AEB = ∠ADB.

Hence the four points A, B, D, E are concyclic.

Example 2. In ΔABC, AB = AC; BE and CF are respectively the bisectors of ∠ABC and ∠ACB and intersect the sides AC and AB at the points E and F respectively. Then the four points B, C, E, F are not concyclic.

Solution: False

since in ΔABC, AB = AC.

∴ ∠ABC = ∠ACB or, \(\frac{1}{2}\)∠ABC = \(\frac{1}{2}\)∠ACB

or, \(\frac{1}{2}\)∠EBC = ∠BCF [BE and CF are the bisectors of ∠ABC and ∠ACB respectively.]

Now, in ΔBEC and ΔBFC we get,

∠EBC = ∠BCF, ∠ECB = ∠FBC and BC is common to both.

∴ ΔBEC ≅ ΔBFC [by the A-A-S condition of congruency]

∴ ∠BEC = ∠BFC.

But they are two such angles on the same side of BC at the points E and F that they are equal.

∴ B, C, E, F are concyclic.

But given that B, C, E, F are not concyclic.

Hence the given statement is false.

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle Concylic Points Fill In the Blanks

Example 1. All angles in the same segment of a circle are _____

Solution: Equal

Example 2. If the line segment joining two points subtends equal angle at two other points on the same side, then the four points are ______

Solution: Concyclic

Example 3. If two angles on the circle formed by two arcs are equal, then the lengths of arcs are ______

Solution: Equal.

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle Concylic Points Short Answer Type Questions

Example 1. In the adjoining, O is the centre of the circle AC is the diameter and chord DE is parallel to the diameter AC. If ∠CBD = 60°, find the value of ∠CDE.

Solution:

Given :

In the adjoining, O is the centre of the circle AC is the diameter and chord DE is parallel to the diameter AC. If ∠CBD = 60°,

Let us join D, O. Then the central angle produced by the chord CD is ∠COD. and the angle in circle by the same chord CD is ∠CBD.

∴ ∠COD = 2 ∠CBD or, ∠COD = 2 x 60° or, ∠COD =120°

∴ in ΔCOD, ∠OCD + ∠ODC = 180° – ∠COD = 180° – 120°

[the sum of three angles of a triangle is 180°]

∴ ∠OCD + ∠OCD = 60° [∠ODC = ∠OCD]

or, 2 ∠OCD = 60° or, ∠OCD = \(\frac{60^{\circ}}{2}\) = 30°

Again, AC | | DE, ∴ ∠ACD = ∠CDE [alternate angles]

or, ∠CDE = ∠OCD [∠ACD and ∠OCD are same angles]

Hence the value of ∠CDE – 30°.

Example 2. In the adjoining, QS is the bisector of an angle ∠PQR, if ∠SQR = 35° and ∠PRQ = 32°, then find the value of ∠QSR.

Solution:

Given :

In the adjoining, QS is the bisector of an angle ∠PQR, if ∠SQR = 35° and ∠PRQ = 32°,

Let us join O, Q and O, R. .

Now, ∠QOR is the central angle produced by the arc QR and ∠QPR and ∠QSR are two angles in circle produced by the same arc QR.

∴ ∠QOR = 2 ∠QPR ……. (1) and ∠QOR – 2 ∠QSR …… (2)

∴ from (1) and (2) we get, 2 ∠QPR = 2 ∠QSR or, ∠QPR = ∠QSR.

Now, ∠PQR = 2 ∠SQR [QS is the bisector of ∠PQR]

= 2 x 35° = 70° and ∠PRQ – 32° [Given]

∴ ∠QPR = 180° – (∠PQR + ∠PRQ)

= 180° – (70° + 32°) [∠PQR = 70° and ∠PRQ = 32°]

= 180° – 102° = 78°

Hence ∠QSR = 78°.

“WBBSE Mensuration Chapter 2 practice questions on angles”

Example 3. In the adjoining, O is the centre of the circle and AB is the diameter. If AB and CD are mutually perpendicular to each other and ∠ADC = 50°, then find the value of ∠CAD.

Solution:

Given :

In the adjoining, O is the centre of the circle and AB is the diameter. If AB and CD are mutually perpendicular to each other and ∠ADC = 50°,

The angles in circle produced by the chord AC are ∠ADC and ∠ABC.

∴ ∠ABC = ∠ADC = 50°

Again, the central angle = ∠AOB and angle in circle = ∠ACB both produced by the arc \(\overparen{A D B}\).

∴ ∠AOB = 2 ∠ACB or, 180° = 2 ∠ACB [∠AOB = straight angle = 180°]

or, ∠ACB = \(\frac{180^{\circ}}{2}\) = 90°

Then ∠CAB = 180° – (∠ACB + ∠ABC) [sum of three angles of a triangle is 180°]

= 180° – (90° + 50°) = 180° – 140° = 40°

Again, AB and CD are perpendiculars to each other. Let AB and CD intersect each other at E.

∴ ∠EAD = 180° – (∠ADE + ∠AED)

= 180° – (50° + 90°) [∠ADE = ∠ADC = 50° and ∠AED = 90°]

= 180° – 140° = 40°

Now, ∠CAD = ∠EAC + ∠EAD

= 40° + 40° = 80° [∠EAC = ∠CAB = 40° and ∠EAD = 40°] .

∴ ∠CAD = 80°

Example 4. In the adjoining, O is the centre of the circle and AB = AC; if ∠ABC = 32°, then find the value of ∠BDC. 

Solution:

Given :

In the adjoining, O is the centre of the circle and AB = AC; if ∠ABC = 32°,

In ΔABC, AB = AC; ∠ABC = ∠ACB = 32°

[∠ABC = 32°]

Now, ∠ACB and ∠ADB are angles in circle, produced by the chord AB.

∴ ∠ADB = ∠ACB = 32°….(1)

Again, ∠ABC and ∠ADC are two angles in circle, produced by the chord AC,

∴ ∠ADC = ∠ABC = 32°……(2)

Also, ∠BDC = ∠ADB + ∠ADC = 32° + 32° = 64° [from (1) and (2)]

∴ ∠BDC = 64°.

Example 5. In the adjoining, BX and CY are the bisectors of ∠ABC and ∠ACB respectively. If AB = AC and BY = 4 cm, then find the length of AX. 

Solution:

Given :

In the adjoining, BX and CY are the bisectors of ∠ABC and ∠ACB respectively. If AB = AC and BY = 4 cm

In ΔABC, AB = AC, ∴ ∠ABC = ∠ACB

or, \(\)∠ABC = ∠ACB

or, ∠ABX = ∠BCY [BX and CY are the bisectors of ∠ABC and ∠ACB respectively.]

∴ by the converse theorem of theorem 35, arc AX = arc BY = 4 cm [BY = 4cm (Given)]

Hence the length of AX = 4cm.

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle Concylic Points Long Answer Type Questions

Example 1. O is the orthocentre of ΔABC and AD ⊥ BC. If AD is produced it intersects the circumcircle of ΔABC at the point G. Prove that OD = DG.

Solution:

Given :

O is the orthocentre of ΔABC and AD ⊥ BC. If AD is produced it intersects the circumcircle of ΔABC at the point G.

The perpendiculars AD, BE and CF on the sides BC, CA and AB respectively of ΔABC intersect each other at O.

So, O is the orthocentre of ΔABC. When AD is extended it intersects the circumcircle with centre P at point G.

To prove: We have to prove that OD = DG.

Construction: Let us join C, G.

Proof: ∠BAG and ∠BCG are two angles in circle, produced by the arc \(\overparen{B G}\).

∴ ∠BAG = ∠BCG.

or, 90° – ∠ABD = ∠BCG [AD ⊥ BC, ∴ ∠ADB = 90°]

or, 90° – (90° – ∠BCF) = ∠BCG [CF ⊥ AB, ∴ ∠BFC = 90°] or, ∠BCF = ∠BCG

or, ∠OCD = ∠GCD [∠BCF and ∠OCD also ∠BCG and ∠GCD are same angles]

∴ ∠OCD = ∠GCD …(1)

Now, in ΔCOD and ΔCGD,

∠CDO = ∠CDG [each is right angle],

∠OCD = ∠GCD [from (1)] and CD is common to both.

∴ ΔCOD ≅ ΔCGD [by the A-A-S condition of congruency]

∴ OD = DG [similar sides of congruent triangles]

Hence OD = DG (Proved)

Example 2. I is the centre of the incircle of ΔABC; produced AI intersects the circumcircle of that triangle at the point P. Prove that PB = PC = PI.

Solution:

Given :

I is the centre of the incircle of ΔABC; produced AI intersects the circumcircle of that triangle at the point P.

I is the incentre of ΔABC.

∴ ∠IBC = ∠IBA, ∠IAB = ∠IAC and ∠ICB = ∠ICA

∴ ∠IAB = ∠IAC, ∴ ∠PAB = ∠PAC

∴ PB = PC [by the converse theorem]…..(1)

Again, ∠BCP and ∠BAP are two angles in circle produced by the arc \(\overparen{P B}\).

∴ ∠BCP = ∠BAP

or, ∠PBC = ∠BAP [PB = PC, ∴ ∠BCP = ∠PBC]

Now, external ∠BIP of ΔAIB = internally opposite

(∠IAB + ∠IBA) = ∠IAB + ∠IBC

= ∠BAP + ∠IBC [∠IAB and ∠BAP are same angles]

= ∠PBC + ∠IBC

[∠PBC = ∠BAP] = ∠IBP

∴ ∠BIP + ∠IBP

∴ PB = BI or, [PB = PI, radius of same circle]……(2)

∴from (1) and (2) we get, PB = PC = PI. (Proved)

Example 3. Ankita drew two circles which intersect each other at the points P and Q. Through the point P two straight lines are drawn so that they intersect one of the circles at the points A and B and the other circle at the points C and D respectively. Prove that ∠AQC = ∠BQD.

Solution:

Given :

Ankita drew two circles which intersect each other at the points P and Q. Through the point P two straight lines are drawn so that they intersect one of the circles at the points A and B and the other circle at the points C and D respectively.

 

Given: Two circles with centres at O and O’ respectively intersect each other at the points P and Q.

Through the point P two straight lines APC and BPD are drawn which intersect the first circle at A and B and the second circle at C and D respectively.

To prove: We have to prove that ∠AQC = ∠BQD

Construction: Let us join A. Q; B, Q; C, Q and D, Q.

Proof: ∠APB and ∠AQB are two angles in circle produced by the arc AB in the circle with centre at O.

∴ ∠APB = ∠AQB ……. (1)

Again ∠CQD and ∠CPD are two angles in circle produced by the arc \(\overparen{C D}\) in the circle with centre at O’.

∴  ∠CQD = ∠CPD…..(2) .

But ∠APB = ∠CPD [opposite angles]

∴ ∠AQB = ∠CQD [from (1) and (2)]

Now, ∠AQD + ∠AQB = ∠AQD + ∠CQD [∠AQB = ∠CQD]

or, ∠BQD = ∠AQC

∴ ∠AQC = ∠BQD (Proved)

Example 4. Two chords AB and CD of a circle are perpendicular to each other. If a perpendicular drawn to AD from the point of intersection of those two chords AB and CD is produced to meet BC at the point E, prove that the point E is the mid-point of BC. 

Solution:

Given :

Two chords AB and CD of a circle are perpendicular to each other. If a perpendicular drawn to AD from the point of intersection of those two chords AB and CD is produced to meet BC at the point E

AB ⊥ CD, ∴ ∠APD = ∠BPC = 90°

∴ ∠ADP = 90° – ∠DAP……. (1)

Now, ∠ADC and ∠ABC are two angles in circle, produced by the arc \(\overparen{A C}\).

∴ ∠ADC = ∠ABC…….(2)

Again, ∠BPE = ∠APF [opposite angles]…… (3)

Now, ∠APF – 90° – ∠PAF [∠AFP = 90°]

= ∠ADP [∠APD = 90°]……..(4)

From (3) and (4) we get, ∠BPE = ∠ADP = ∠ADC = ∠ABC = ∠PBE

∴ in ΔBPE, ∠BPE = ∠PBE. ∴BE = PE……….. (5)

Similarly, it can be proved that CE = PE…….(6)

Then from (5) and (6) we get, BE = CE

Hence E is the mid-point of BC. (Proved)

Example 5. If in a cyclic quadrilateral ABCD, AB = DC, then prove that AC = BD. [GP-X]

Solution:

Given :

If in a cyclic quadrilateral ABCD, AB = DC

In the cyclic quadrilateral ABCD, AB = DC

∴ the angles in circle produced by the chords AB and CD are equal.

∴ ∠ACB = ∠ADB ……. (1) and

∠CBD = ∠CAD…….(2)

Moreover ACB = ∠CBD [AB = DC] in ΔEBC, ∠ECB = ∠EBC, ∴ BE = CE

Again, ∠ADB = ∠CAD [AB = DC]

∴ in ΔEAD, ∠EAD = ∠EDA  ∴DE = AE

Now, AE + CE = DE + BE [AE = DE and CE = BE]

or, AC = BD (Proved)

Example 6. OA is the radius of a circle with centre at O, AQ is its chord and C is any point on the circle. A circle passes through the points O, A, C intersects the chord AQ at the point P. Prove that CP = PQ.

Solution:

Given :

OA is the radius of a circle with centre at O, AQ is its chord and C is any point on the circle. A circle passes through the points O, A, C intersects the chord AQ at the point P.

Given: OA is the radius of the circle with centre at O and AQ is its chord. C is any point on the circle.

A circle passes through the points O, A, C intersects the chord AQ at the point P. Let us join C, P.

To prove: We have to prove that CP= PQ

Construction: Let us join O, C; O, Q and O, P.

Proof: In the circle APOC, ∠OAP and ∠OCP are two angles in circle produced by the chord OP,

∴ ∠OAP = ∠OCP……. (1)

Again, in ΔOAQ, OA = OQ [radii of the same circle]

∴ ∠OAQ = ∠OQA……..(2)

from (1) and (2) we get, ∠OCP = ∠OQA …… (3)

∴ CP = PQ [in ΔCPQ, ∠QCP = ∠CQP]

Hence CP = PQ (Proved).

Example 7. The triangle ABC is inscribed in a circle, the bisectors AX, BY and CZ of the angles ∠BAC, ∠ABC and ∠ACB intersect at the points X, Y, Z, on the circle respectively. Prove that AX is perpendicular to YZ. 

Solution:

Given :

The triangle ABC is inscribed in a circle, the bisectors AX, BY and CZ of the angles ∠BAC, ∠ABC and ∠ACB intersect at the points X, Y, Z, on the circle respectively.

Let ΔABC is inscribed in the circle with centre at O. AX, BY and CZ are the bisectors of ∠BAC, ∠ABC and ∠ACB respectively and intersect the circle at X, Y and Z respectively.

Let AX, BY and CZ intersect each other at O. Then O is the incentre of the circle. Again, let AX intersect YZ at the point E.

To prove: We have to prove that AX ⊥ YZ, i.e., AE ⊥ YZ

Proof: ∠ABY = ∠CBY (BY is the bisector of ∠ABC]

∴ arc AY = arc CY [by the converse]

∴ Similarly, arc BX = arc CX and arc AZ = arc BZ.

[angles in the same segment of a circle are equal.]

Again, arc AZ and arc BZ produced two equal central angles.

∴ ∠AOZ = ∠BOZ…..(1)

Similarly, ∠AOY = ∠COY [arc AY = arc CY]

or, ∠AOY = ∠BOZ [opposite angles] …… (2)

∴ ∠AOZ = ∠AOY or, ∠EOZ = ∠EOY …..(3)

Now, in ΔEOY and ΔEOZ, OY = OZ (radii of same circle),

∠EOY = ∠EOZ [from (3)] and OE is common to both.

∴ ΔEOY ≅ ΔEOZ [by the S-A-S condition of congruency]

∴ ∠OEY = ∠OEZ [similar angles of congruent triangles]

But these two angles are adjacent obtained when the line segment OE stands upon the line segment YZ and they are equal.

∴ each is right angle, i.e., ∠OEY = ∠OEZ = right angle.

∴ OE ⊥ YZ or AX ⊥ YZ.

∴ AX is perpendicular to YZ. (Proved)

Example 8. ΔABC is inscribed in a circle,, the bisectors of the angles ∠BAC, ∠ABC and ∠ACB intersect at the points X, Y, Z on the circle respectively. Prove that in ΔXYZ, ∠YXZ = 90 – \(\frac{1}{2}\) ∠BAC

Solution:

Given :

ΔABC is inscribed in a circle,, the bisectors of the angles ∠BAC, ∠ABC and ∠ACB intersect at the points X, Y, Z on the circle respectively.

Given: ΔABC is inscribed in the circle with centre at O.

The bisectors of ∠BAC, ∠ABC, and ∠ACB intersect the circle at the points X, Y and Z respectively. Let us join the points X, Y; Y, Z and Z, X.

To prove: ∠YXZ = 90° – \(\frac{1}{2}\)∠BAC.

Proof: ∠AXZ and ∠ACZ are two angles in circle produced by the arc AZ,

∴ ∠AXZ = ∠ACZ……..(1)

Similarly, ∠AXY and ∠ABY are two angles in circle produced by the arc AY.

∴ ∠AXY = ∠ABY…….. (2)

Now, adding (1) and (2) we get,

∠AXZ + ∠AXY = ∠ACZ + ∠ABY

or, ∠YXZ=\(\frac{1}{2}\)∠C+\(\frac{1}{2}\)∠B=\(\frac{1}{2}\) (∠C + ∠B)

= \(\frac{1}{2}\) (180° -∠A) [∠A + ∠B + ∠C = 180°]

= 90° –\(\frac{1}{2}\)∠A =90°-\(\frac{1}{2}\)∠BAC

Hence ∠YXZ = 90° –\(\frac{1}{2}\)∠BAC. [Proved]

Example 9. The isosceles triangle ABC is inscribed in the circle with centre at O. If AP be a diameter passsing through A, then show that AP is the internal bisector of ∠BPC.

Solution:

Given :

The isosceles triangle ABC is inscribed in the circle with centre at O. If AP be a diameter passsing through A

Given: ΔABC is an isosceles triangle of which AB = AC.

O is the centre of the circle in which ΔABC is inscribed and AP is a diameter of the circle.

To prove: AP is the internal bisector of ∠BPC, i.e., ∠APB = ∠APC.

Construction: Let us join the points O, B; O, C; B,P and C, P.

Proof: In ΔABC, AB = AC, ∴ ∠AOB = ∠AOC [equal segments of a circle produce equal front angles at the centre.]

or, 2 ∠APB = 2 ∠APC or, ∠APB = ∠APC.

Hence AP is the internal bisector of ∠BPC. (Proved)

Example 10. In ΔABC, AB = AC and E is any point on the extended BC. The circumcircle of ΔABC intersects AE at the point D. Prove that ∠ACD = ∠AEC.

Solution:

Given :

In ΔABC, AB = AC and E is any point on the extended BC. The circumcircle of ΔABC intersects AE at the point D.

Given: In ΔABC, AB = AC and E is any point on the produced BC.

The circumcircle of ΔABC intersects AE at point D.

To prove: We have to prove that ∠ACD = ∠AEC

Construction: Let us join B. D.

Class 10 Maths Wbbse Solutions

Proof: External ∠ACB of AΔACE = internally opposite (∠AEC + ∠CAE) = ∠AEC + ∠CBD [the angles produced by CD, ∠CAD = ∠CBD]

or, ∠ACB – ∠CBD = ∠AEC or, ∠ABC – ∠CBD = ∠AEC [AB = AC. ∴ ∠ACB = ∠ABC]

or, ∠ABD =∠AEC

or, ∠ACD = ∠AEC [the angles in circle produced by AD. ∠ABD = ∠ACD]

∴ ∠ACD = ∠AEC (Proved)

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle Angle In Semi Circle

In your previous classes you have learnt about what a semi-circle is. If a circle is divided into two equal parts by cutting it through any of its diameters, then each part of the two is called a semi-circle.

If we take any point P on the circumference of a semi-circle, then the angle obtained by joining two end points of the diameter of the semi-circle with the point P is called an angle in the semi-circle.

For example, in the following figure. AB is the diameter of the semi-circle and O is its centre. P₁, P₂, P₃, P₄ are four points on its circumference.

If we join the endpoints A and B of the diameter AB with the points P₁, P₂, P₃, P₄ respectively by some straight lines, we get the angles ∠AP₁B, ∠AP₂B, ∠AP₃B and ∠AP₄B.

Then these angles are called angles in a semi-circle. Now it is the question that what the value of these angles are and is there any relation among them.

If we measure these angles with hands-on trial, then we shall see that all these angles are equal and the value of each of them is 1 right angle, i.e..

∠AP₁B = ∠AP₂B = ∠AP₃B = ∠AP₄B= 1 right angle.

Hence angle in a semi-circle is a right angle.

We shall now prove this theorem logically by the method of geometry.

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle Angle In Semi Circle Theorem

Theorem: Prove that angle in a semi-circle is a right angle.

Given: ∠ACB is an angle in a semi-circle in a circle with centre at O.

To prove ∠ACB = 1 right angle or 90°.

Proof : In the circle with centre at O, ∠AOB is the front angle and ∠ACB is the front angle in circle both produced by the same arc \(\overparen{A P B}\).

∴ ∠AOB = 2 ∠ACB ……. (1)

Now, ∠AOB = 1 straight angle [AOB is a line segment]

= 2 right angles or 180°.

∴ from (1) we get, 2 ∠ACB = 2 right angles or 180° or, ∠ACB = 1 right angle or 90°

∴ ∠ACB = 1 right angle or 90°.

Hence angle in a semi-circle is a right angle. (Proved)

In a similar way, by taking any angle in a semicircle we can prove that an angle in a semi-circle is a right angle.

We have discussed much more of the application of this theorem in the following examples.

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle Angle In Semi Circle Multiple Choice Questions

Example 1. If PQ be a diameter of a circle with centre at O and if PR = RQ, then the value of ∠RPQ is

1. 30°
2. 90°
3. 60°
4. 45°

Solution: PQ is a diameter of the circle with centre at O and R is any point on its circumference.

∴ ∠PRQ = 90° [by theorem]

Again, PR = RQ, ∴ ∠RPQ = ∠RQP…… (1) [opposite angles of equal sides of a triangle are equal]

Now, in ΔPQR, ∠PQR + ∠QRP + ∠ZRPQ = 180° [sum of three angles of a triangle is 180°]

or, ∠PQR + ∠RPQ + 90° = 180° [by (1) and ∠PRQ = 90°]

or, ∠RPQ + ∠RPQ + 90° = 180° [by (1) and ∠PRQ = 90°]

or, 2 ∠RPQ = 180° – 90° or, 2 ∠RPQ = 90°

or, ∠RPQ = \(\frac{90^{\circ}}{2}\)= 45° ∴ ∠RPQ = 45°

∴ 4. 45° is correct.

Example 2. QR is a chord of the circle with centre at O and POR is a diameter of it. OD is perpendicular to QR. If OD = 4 cm, then the length of PQ is

1. 4 cm
2. 2 cm
3. 8 cm
4. None of these

Solution: OD ⊥ QR, ∴ ∠ODR = 1 right angle

Again, POR is a diameter of the circle with centre at O.

∴ ∠PQR is an angle is semicircle.

∴ ∠PQR = 1 right angle.

∴OD || PQ and ∠PRQ is common to both ΔPQR and ΔODR. ΔPQR and ΔODR are similar triangles.

∴ \(\frac{O D}{P Q}=\frac{O R}{P R}\)

or, [/latex]\frac{4}{P Q}=\frac{O R}{2 O R}[because \mathrm{PR}=2 O R][/latex]

or, \(\frac{4}{P Q}=\frac{1}{2} \text { or, } P \mathrm{PQ}=8\)

∴ the length of PQ = 8 cm.

∴ 3. 8 cm is correct

Aliter: POR is a diameter of the circle with centre at O.

∴ ∠PQR is an angle in semicircle.

∴ ∠PQR = 1 right angle or 90°

∴ ΔPQR is a right-angled triangle of which PR is the hypotenuse.

∴ PR² = PQ² + QR² …….(1) [by Pythagoras’ theorem]

Again, OD ⊥ QR, where is a chord of the circle.

∴ D is the mid-point of QR,

∴ DR = \(\frac{1}{2}\) QR.

∴ ΔODR is a right-angled triangle of which OR is the hypotenuse.

∴ OR² = OD² + DR²

or, \(\left(\frac{1}{2} \mathrm{PR}\right)^2=4^2+\left(\frac{1}{2} \mathrm{QR}\right)^2\)

\(\cdot\left[\mathrm{OR}=\frac{1}{2} \mathrm{PR} \text { and } \mathrm{DR}=\frac{1}{2} \mathrm{QR} \text { and } \mathrm{OD}=4 \mathrm{~cm}\right]\) \(\frac{\mathrm{PR}^2}{4}=16+\frac{\mathrm{QR}^2}{4}\)

or, PR² = 64 + QR² ……(2)

Now, from (1) and (2), we get,

PQ²+ QR² = 64 + QR²

or, PQ² = 64 or, PQ = √64 = 8.

∴ the length of PQ = 8 cm.

∴ 3. 8cm is correct.

Example 3. AOB is a diameter of a circle. When the chords AC and BD are extended they meet at the point E. If ∠COD = 40°, then the value of ∠CED is 

1. 40°
2. 80°
3. 20°
4. 70°

Solution:

Given

AOB is a diameter of a circle. When the chords AC and BD are extended they meet at the point E. If ∠COD = 40°

Let us join B, C and C, D. AOB is the diameter of the circle With centre at O.

∴ ∠ACB is an angle in semicircle.

∴ ∠ACB = 90°…..(1)

Again, ∠BCE = 180° – ∠ACB = 180° – 90° [from (1)] = 90°…..(2)

Now, the front central angle produced by the chord CD is ∠COD = 40° and the front angle in circle is ∠CBD.

∴ ∠CBD = \(\frac{1}{2}\)∠COD

or,∠CBE = \(\frac{1}{2}\) x 40° = 20°…..(3) [∠CBD and ∠CBE are same angles]

So from ΔBCE we get, ∠BEC + ∠BCE + ∠CBE = 180°

or, ∠BEC + 90° + 20° = 180° [from (2) and (3)] or, ∠BEC = 180° – 110° = 70°

∴ ∠BEC = 70°

∴ 4. 70° is correct.

Example 4. AOB is a diameter of a circle, If AC = 3 cm and BC = 4 cm, then the length of AB is

1. 3 cm
2. 4 cm
3. 5 cm
4. 8 cm

Solution: AOB is a diameter of the circle with centre at O. ∠ACB is an angle in a semicircle

∴ ∠ACB = 90°

∴ from the right-angled triangle ABC by Pythagoras theorem we get,

AB² = AC² + BC² = 3² + 4² = 25 [AC = 3 cm and BC = 4cm]

∴ AB = √25 = 5 the length of AB = 5cm

∴ 3. 5 cm is correct.

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle Angle In Semi Circle True Or False

Example 1. The angle in circle produced by a minor axis is an obtuse angle.

Solution: False since it will be an acute angle.

Example 2. O is the mid-point of AB in ΔABC and OA = OB – OC; If a circle is drawn by taking AB as diameter, then the circle will pass through point C.

Solution: True, since OA = OB = OC is also a radius of the circle with centre at O, so the circle will pass through the point C.

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle Angle In Semi Circle Fill In The Blanks

Example 1. Semicircular angle is a ______ angle.

Solution: Right

Example 2. The angle in the segment of a circle which is less than a semi-circle is an _________ angle

Solution: Obtuse

Example 3. The circle drawn with hypotenuse of a right angled triangle as diameter passes through the _______

Solution: Angular angle

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle Angle In Semi Circle Short Answer Type Questions

Example 1. In the adjoining, O is the centre of the circle and AB is the diameter. If ∠BCE = 20°, ∠CAE = 25°, then find the value of ∠AEC.

Solution:

Given :

In the adjoining, O is the centre of the circle and AB is the diameter. If ∠BCE = 20°, ∠CAE = 25°

The angles in circle produced by the chord CA are ∠AEC and ∠ABC.

∴ ∠AEC = ∠ABC…… (1)

Again, the angles in circle produced by the chord BE are ∠BAE and ∠BCE.

∴ ∠BAE = ∠BCE – 20° [∠BCE = 20°]

Similarly, the angles in circle produced by the chord CE are ∠CBE and ∠CAE,

∴ ∠CBE = ∠CAE = 25° [∠CAE = 25°]

Again, ∠AEB is a semicircular angle, ∠AEB = 90°

Then, in ΔABE, ∠ABE = 180° – (∠BAE + ∠AEB) = 180° – (20°+ 90°) = 70°

∴ ∠ABC = ∠ABE – ∠CBE = 70°-25° = 45°

∴ from (1) we get, ∠AEC = ∠ABC = 45° ∴ ∠AEC = 45°

Example 2. In isosceles triangle ABB = AC; if a circle is drawn with the side AB as diameter, then the circle intersects the side BC it the point D. When BD = 4 cm. find the length of CD.

Solution:

Given :

In isosceles triangle ABB = AC; if a circle is drawn with the side AB as diameter, then the circle intersects the side BC it the point D. When BD = 4 cm.

Let O is the mid-point of AB.

The circle drawn with centre at O intersects the side BC at the point D.

Let us join A. D. Then ∠ADB is a semi-circular angle.

∴ ∠ADB = 90°, ∴ ∠ADC = 90°

Now in ΔABD and ΔACD, we have AB = AC, ∠ABD = ∠ACD and ∠ADB = ∠ADC.

∴ ΔABD = ΔACD [by die A-A-S condition of congruency ]

∴ BD = CD [similar sides of congruent triangles]

But BD = 4 cm (Given); CD = 4 cm.

Example 3. Two chords AB and AC of a circle are perpendicular to each other. If AB = 4 cm and AC = 3cm, then find the radius of the circle. 

Solution:

Given :

Two chords AB and AC of a circle are perpendicular to each other. If AB = 4 cm and AC = 3cm,

AB and AC are perpendicular to each other.

∠BAC = 90°,

∴ ΔABC is a right-angled triangle of which BC is the hypotenuse.

∴ BC² = AB² + AC² [by Pythagoras theorem] = 4²+ 3² = 16 + 9 = 25

∴ BC = √25 = 5

Now, BC is one of the diameters of the circle.

∴ radius of the circle = \(\frac{5}{3}\) cm = 2.5 cm

Example 4. Two chords PQ and PR of a circle are perpendicular to each other. If the radius of the circle be r cm. then find the length of the chord QR. 

Solution:

Given :

Two chords PQ and PR of a circle are perpendicular to each other. If the radius of the circle be r cm.

PQ and PR are perpendicular to each other, ∠QPR = 90°

∴ ∠QPR is a semicircular angle.

∴ QR is a diameter.

Since QR is a diameter of the circle,

radius of the circle = r cm (Given)

∴ QR = 2r cm.

Example 5. AOB is a diameter of a circle. C is a point on the circle. If ∠OBC = 60°, then find the value of ∠OCA

Solution:

Given :

AOB is a diameter of a circle. C is a point on the circle. If ∠OBC = 60°

AOB is a diameter of a circle.

C is a pont on circle

∴ ∠ACB = 1 right angle or 90°

Again, in ΔOBC, OB = OC (radii of same circle)

∴ ∠OCB = ∠OBC = 60° [∠OBC = 60° (Given)]

∴ ∠OCA= ∠ACB – ∠OCB OCB = 90° – 60° = 30°

∴ ∠OCA = 30°

 Example 6. In the adjoining, O is the centre of a circle and AB is one of its diameters. The length of the chord CD is equal to the radius of the circle. When AC and BD are extended, they intersect each other at a point P. Find the value of ∠APB.

Solution:

Given :

In the adjoining, O is the centre of a circle and AB is one of its diameters. The length of the chord CD is equal to the radius of the circle. When AC and BD are extended, they intersect each other at a point P.

Let us join B, C.

Now, AB is the diameter of the circle,

∴ ∠ACB = 90° [semi-circular angle]

Again in ΔCOD, OC = OD = CD (Given)

∴ ΔCOD is an equilateral triangle.

∴ each angle of the triangle = 60° ∴ ∠COD = 60°

Now, the central angle produced by the chord CD is ∠COD and angle in circle is ∠CBD.

∴ ∠COD = 2 ∠CBD or, 60° = 2 ∠CBD or, ∠CBD = 30°

∴ in ΔPBC, ∠PCB = 90° [∠ACB = 90°], ∠PBC = ∠CBD = 30°.

∴ ∠BPC = 180° – (∠PCB + ∠PBC) = 180° – (90° + 30°) = 180° – 120° = 60°

∴ ∠APB = 60° [∠BPC = ∠APB]

Solid Geometry Chapter 2 Theorems Related To Angles In A Circle Angle In Semi Circle Long Answer Type Questions

Example 1. The angle B of the AABC is a right angle. If a circle is drawn with AC as diameter, then it intersects AB at a point D. Then which one of the following is correct?

1. AB > AD
2. AB = AD
3. AB < AD
4. AB ≠ AD

Solution:

Given

The angle B of the AABC is a right angle. If a circle is drawn with AC as diameter, then it intersects AB at a point D.

The angle B of ΔABC is a right angle.

∴ ∠ABC is a semicircular angle.

∴ AC is a diameter of the circle.

∴ The circle drawn with AC as diameter intersects AB at a point B, i.e., B and D are the same points.

∴ AB = AD

∴ 2. AB = AD is correct.

Example 2. Prove that the circle drawn with any one of the equal sides of an isosceles triangle as diameter bisects the unequal side.

Solution:

Let in ΔABC, AB = AC. The circle drawn with AB as diameter intersects BC at a point D.

To prove: D is the mid-point of BC, i.e., BD = CD.

Proof: The circle intersects BC at a point D.

∴ ∠ADB is a semi-circular angle.

∴ ∠ADB = 90° ⇒ ∠ADC = 90°

Again, in ΔABC, AB = AC

∴ ∠ABC = ∠ACB or, ∠ABD = ∠ACD

So, in ΔABD and ΔACD, ∠ADB = ∠ADC [each is 90°]

∠ABD = ∠ACD and AD is common to both.

ΔABD = ΔACD [by the A-A-S condition of congruency]

∴ BD = CD. [Similar sides of congruent triangle]

Hence D is the mid-point of BC, i.e., the circle bisects the unequal side. [Proved]

Example 3. Parama drew two circles which intersect each other at the points P and Q. If the diameters of the two circles are PA and PB respectively, then prove that A, Q, B are collinear.

Solution:

Given :

Parama drew two circles which intersect each other at the points P and Q. If the diameters of the two circles are PA and PB respectively,

Let two circles with centres O and O’ intersect each other at P and Q.

PA is a diameter of the circle with centre at O and PB is also a diameter of the circle with centre at O.

To prove: A, Q and B are collinear.

Construction: Let us join P, Q.

Proof: In the circle with centre at O, ∠AQP is a semi-circular angle. ∴ ∠PQA = 1 right angle.

Again, in the circle with centre at O’, ∠BQP is a semicircular angle. ∴ ∠PQB = 1 right angle.

Now, ∠PQA + ∠PQB = 1 right angle + 1 right angle.

or, ∠AQB = 2 right angle or, ∠AQB = 180° or, ∠AQB =1 straight angle.

∴ ∠AQB – 1 straight angle.

Hence the points A, Q, B are collinear. (Proved)

Example 4. Debanjan drew a line segment PQ of which mid-point is R and two circles are drawn with PR and PQ as diameter. Debaratee drew a straight line through the point P which intersects the first circle at the point S and the second circle at the point T. Prove that PS = ST. 

Solution:

Given :

Debanjan drew a line segment PQ of which mid-point is R and two circles are drawn with PR and PQ as diameter. Debaratee drew a straight line through the point P which intersects the first circle at the point S and the second circle at the point T.

Let O be the centre of the circle drawn with PR as the diameter.

Let us join S,R and T, R. Since PR is a diameter. ∴ ∠PSR = 90° [Semicircular angle]

Obviously, ∠TSR = 90°, [Since PST is a straight line, so that ∠PST = 1 straight line = 180°]

Now in ΔPSR and ΔTSR, PR = TR [radii of same circle with centre at R]

∠PSR = ∠TSR [ each is right angle] and SR is common to both.

∴ Δ PSR = Δ TSR [by the R-H-S condition of congruency]

∴ PS = ST [similar sides of two congruent triangles]

Hence PS = ST. (Proved)

Example 5. Three points P, Q, R lie on a circle. The two perpendiculars PQ and PR at the point P intersect the circle at the points S and T respectively. Prove that RQ = ST.

Solution:

Given :

Three points P, Q, R lie on a circle. The two perpendiculars PQ and PR at the point P intersect the circle at the points S and T respectively.

Three points P, Q, R are on the circle with centre at O. PS ⊥ PQ and PT ⊥ PR.

To prove RQ = ST.

Construction: Let us join the points Q, S; R, T; P, S and S, T.

Proof: PS ⊥ PQ, ∴ ∠QPS = 90°

∴ ∠QPS is a semi-circular angle.

∴ QS is a diametet, of which O is the mid point ∴ OQ = OS = radius

Similarly, PT ⊥ PR, ∴ ∠RPT = 90°

∴ ∠RPT is the semi-circular angle.

∴ RT is a diameter of which O is the mid-point.

∴ OR = OT = Radius.

Now, in Δ’s OQR and ΔOST, OQ = OS [radii of same circle]

∴ OR = OT [for similar reason] and ∠QOR = ∠SOT [opposite angles]

∴ ΔOQR ≅ ΔOST [by the S-A-S condition of congruency]

∴ QR = ST [similar sides of congruent triangles]

Hence RQ = ST. (Proved)

Example 6. ABC is an acute-angled triangle. AP is the diameter of the circumcircle of the triangle ABC; BE and CF are perpendiculars on AC and AB respectively and they intersect each other at the point Q. Prove that BPCQ is a parallelogram. 

Solution:

Given :

ABC is an acute-angled triangle. AP is the diameter of the circumcircle of the triangle ABC; BE and CF are perpendiculars on AC and AB respectively and they intersect each other at the point Q.

ΔABC is an acute angled triangle. BE ⊥ AC and CF ⊥ AB and BE and CF intersect each other at a point Q.

AP is the diameter of the circumcircle of ΔABC.

To prove: BPCQ is a parallelogram.

Construction: Let us join O, C.

Proof: AP is a diameter of the circle. ∴ ∠ABP and ∠ACP are both semi circular angles.

∴ ∠ABP = ∠ACP = 90°……(1)

CF ⊥ AB, ∴ ∠CFB = 90°

∴ ∠FBQ = 90° – ∠FQB…….(2)

Again, BE⊥ AC, ∴ ∠BEC = 90°, ∴ ∠ECQ = 90° – ∠CQE . . . (3)

But ∠FQB = ∠CQE [Opposite angles]

∴ from (2) and (3) we get, ∠FBQ = ∠ECQ

Then ∠PBQ = ∠PBA – ∠FBQ = ∠ACP – ∠ECQ

[∠PBA =∠ACP and ∠FBQ = ∠ECQ] = ∠PCQ

∴  two opposite angles ∠PBQ and ∠PCQ of the quadrilateral BPCQ are equal.

Hence by the property of parallelogram, BPCQ is a parallelogram.

Example 7. The internal and external bisectors of the vertical angle of a triangle intersect the circumcircle of the triangle at the points P and Q. Prove that PQ is a diameter of the circle. 

Solution:

The internal and external bisectors of the vertical angle of a triangle intersect the circumcircle of the triangle at the points P and Q.

Given: AP is the internal bisector of the vertical angle ∠A of the ΔABC and AE is the external bisector of the vertical ∠A of ΔABC and they intersect the circumcircle of the ΔABC at the points P and Q respectively.

To prove: PQ is the diameter of the circle.

Proof: ∠BAC + ∠CAD = 1 straight angle = 180° or, \(\frac{1}{2}\)∠BAC + \(\frac{1}{2}\) ∠CAD = 90° [dividing by 2]

or, ∠PAC + ∠CAE = 90° [AP is the internal bisector of ∠A and AE is the external bisector of ∠A]

or, ∠PAQ = 90°

∴ ∠PAQ is a semicircular angle.

Hence PQ is a diameter of the circle. (Proved)

Example 8. AB and CD are two diameters of a circle. Prove that ADBC is a rectangle.

Solution:

Given :

AB and CD are two diameters of a circle.

AOB and COD are two diameters of the circle with centre at O.

To prove: ADBC is a rectangle.

Construction: Let us join A, D; D, B; B, C and C, A.

Proof: AOB is a diameter. ∴ ∠ACB and ∠ADB are two semi-circular angles.

∠ACB = 1 right angle and ∠ADB = 1 right angle.

∴ Two opposite angles of the quadrilateral ADBC are both right angles.

Similarly, COD is a diameter. ∴ ∠CAD and ∠CBD are both semi-circular angles.

∴ ∠CAD = 1 right angle and ∠CBD = 1 right angle.

∴ Two other opposite angles of the quadrilateral ADBC are also right angles.

Again, in ΔAOD and ΔBOC, OA = OB, OD = OC [radii of same circle] and ∠COD = ∠BOC [opposite angles]

∴ ΔAOD ≅ ΔBOC [by the S-A-S condition of congruency]

∴ AD = BC [similar sides of congruent triangles] .

Similarly, it can be proved that AC = BD.

∴ opposite sides of the quadrilateral are equal and each of its angles is a right angle.

Hence ADBC is a rectangle. (Proved)

Example 9. AB is diameter and AC is a chord of the circle with centre at O. The radius parallel to AC intersects the circle at D. Prove that D is the mid-point of the arc BC.

Solution:

Given :

AB is diameter and AC is a chord of the circle with centre at O. The radius parallel to AC intersects the circle at D.

AB is a diameter of the circle with centre parallel to AC intersect the arc BC at the point D.

We have to prove that D is the mid-point of the arc BC.

Construction: Let us join O, C.

Proof: AC || OD and OC is the transversal of them,

∴ ∠OCA = ∠COD [alternate angle]…….. (1)

Again OA = OC [radii of same circle]

∴ ∠OCA = ∠OAC……..(2)

Now, ∠BOC = ∠BOD + ∠COD

or, internally opposite (∠OCA + ∠OAC) = ∠BOD + ∠COD

or, ∠COD  + ∠OAC = ∠BOD + ∠COD [by (1)]

or, ∠OAC = ∠BOD or, ∠COD = ∠BOD [by (1)]

∴ ∠BOD = ∠COD, i.e. the two central angles produced by BD and CD are equal.

∴ arc BD = arc CD.

Hence D is the mid-point of the arc BC. (Proved)

Example 10. Two circles intersect each other at the points P and Q. A straight line passing through P intersects one circle at A and the other circle at B. A straight line passing through Q intersect the first circle at C and the second circle at D. Prove that AC || BD.

Solution:

Given :

Two circles intersect each other at the points P and Q. A straight line passing through P intersects one circle at A and the other circle at B. A straight line passing through Q intersect the first circle at C and the second circle at D.

Let two circles with centres at O and O’ intersect each other at the points P and Q respectively.

A straight line passing through P intersects the circle with centre O at the point A and the circle with centre O’ at the point B.

Another straight line passing through Q also intersects the first circle at C and the second circle at D.

We have to prove that AC || BD.

Construction: Let us join P, Q.

Proof: ΔPQC is a cyclic quadrilateral.

∴ ∠PAC + ∠PQC = 180° …….(1) [opposite angles of a cyclic quadrilateral are supplementary.]

Again, BPQD is also a cyclic quadrilateral.

∴ ∠PBD + ∠PQD = 180°……. (2) [for similar reason]

Now, adding (1) and (2) we get,

∠PAC + ∠PBD +∠PQC + ∠PQD = 360°

or, ∠PAC + ∠PBD + 180° = 360° [∠PQC + ∠PQD = 1 straight angle =180°] .

or, ∠PAC + ∠PBD = 360° – 180°

or, ∠PAC + ∠PBD = 180° .

i.e, the sum of two adjacent angles on the same side of the transversal AB of the two line segments AC and BD is 180°.

Hence AC || BD (Proved)

Example 11. ABC is a cyclic equilateral triangle. If D be any point on the circular arc BC on the opposite side of the point A, then prove that DA = DB + DC

Solution:

Given :

ABC is a cyclic equilateral triangle. If D be any point on the circular arc BC on the opposite side of the point A

Let ABC be a cyclic equilateral triangle inside the circle with centre at O.

D is any point on the circular arc BC on the opposite side of the point A.

We have to prove that DA = DB + DC.

Construction: Let us cut a part DE from DA equal to DC, i.e., DC = DE and let us join C, E.

Proof: In ΔCDE, DC = DE [as per construction]

∴ ∠DCE = ∠DEC …….(1)

∴ ΔABC is an equilateral triangle,

∠BAC – ∠ACB = ∠CBA = 60° and AB = BC = CA……..(2)

Now, two angles in circle produced by the arc AC are ∠ADC and ∠ABC

∴ ∠ADC = ∠ABC = 60° [by (2)]

∴ ∠CDE = 60°……(3)

∴ in ΔCDE, ∠CDE + ∠DCE + ∠DEC = 180° or, 60° + ∠DCE + ∠DCE = 180° [from (1) and (3)]

or, 2 ∠DCE = 180° – 60° or, ∠DCE = \(\frac{120^{\circ}}{2}\) = 60°.

∴ ∠DEC = 60°, i.e., each and every angle of ΔCDE is 60°, ΔCDE is equilateral.

∴ CD = DE = CE………(4)

Now, ∠ACE + ∠BCE = ∠ACB = 60° = ∠DCE = ∠BCD + ∠BCE

∴ ∠ACE = ∠BCD ……….(5)

Again, in Δ’s ACE and ΔBCD

∠CAE = ∠CBD [both are angles in circle produced by the same chord CD.]

∴ ∠ACE = ∠BCD [by (5)] and AC = BC [ΔABC is equilateral]

∴ ΔACE = ΔBCD [by the A-A-S condition of congruency]

∴ AE = BD [similar sides of congruent triangles]…….. (6)

Therefore, DA = DE + AE = DC + BD [DE = DC and AE = BD]

Hence DA = DB + DC (Proved).