## 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.

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** BC^{2} = AB^{2} + AC^{2}.

**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 AB^{2 }= BC.BD …….(1)

Again, ΔABC ~ ΔACD

∴ \(\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{DC}}{\mathrm{AC}}\) or AB^{2} = BC.BD …….(2)

Now adding (1) and (2) we get,

AB^{2 }+ AC^{2} = BC.BD + BC.CD = BC (BD + CD) = BC.BC = BC^{2}.

∴ BC^{2 }= AB^{2} + AC^{2}. (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** c^{2 }= a^{2} + b^{2}.

**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 = c^{2}.

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 = a^{2} + b^{2} + 2ab – 2ab – a^{2 }+ b^{2}

∴ c^{2 }= a^{2 }+ b^{2}. (Proved)

**Method 2.** Let ABC be a right-angled triangle in which ∠A is a right angle.

**To prove** BC^{2} = AB^{2 }+ AC^{2}.

**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.

∴ BC^{2} = AB^{2 }+ AC^{2}. (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., AB^{2} = BC^{2}+ AC^{2}.

**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 AB^{2} = BC^{2} + AC^{2}.

= EF^{2} + DF^{2} [∵ by construction BC = EF and AC = DF.]

= DE^{2 }[by Pythagoras theorem]

∴ AB^{2} = DE^{2 }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 AB ^{2} + AC^{2} = 2 (BD^{2} + AD^{2})**

**Given:** Let in ΔABC, D is a mid-point of BC, i.e., BD = CD.

**To prove** AB^{2} + AC^{2} = 2 (BD^{2 }+ AD^{2})

**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,

AB^{2} = AE^{2} + BE^{2 }[by Pythagoras’ theorem]

= AE^{2} + (BD – DE)^{2}

= AE^{2} + BD^{2} – 2 BD.DE + DE^{2}

= AE^{2} + DE^{2}+ BD^{2} – 2BD.DE

= AD^{2} + BD^{2} – BC.DE [∵ in ΔAED, ∠E = right angle

∴ by Pythagoras’ therorem, AE^{2} + DE^{2} = AD^{2} and ∵ 2BD = BC],

∴ AB^{2} – AD^{2} + BD^{2} – BC.DE…….(1)

Again, from the right-angled triangle AEC we get,

AC^{2 }= AE^{2} + CE^{2}

= AE^{2}+ (CD + DE)^{2}

= AE^{2} + CD^{2} + 2CD.DE + DE^{2}

= AE^{2} + DE^{2} + CD^{2} + 2CD.DE

= AD^{2 }+ CD^{2} + 2CD.DE [∵ AE^{2}+ DE^{2} = AD^{2}]

= AD^{2} + CD^{2} + BC.DE [∵ 2CD = BC]

∴ AC^{2} = AD^{2} + CD^{2} + BC.DE ……(2)

Now, adding (1) and (2) we get,

AB^{2} + AC^{2} = AD^{2} + BD^{2} – BC.DE + AD^{2} + CD^{2} + BC.DE

= 2AD^{2} + BD^{2} + CD^{2}

= 2AD^{2} + BD^{2} + BD^{2 }[∵ CD = BD]

= 2 AD^{2} + 2BD^{2}

= 2 (AD^{2 }+ BD^{2})

∴ AB^{2} + AC^{2} = 2 (BD^{2 }+ AD^{2}) (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

**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**

- 34 m
- 17 m
- 26 m
- 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, OB^{2} = OA^{2} + AB^{2}.

or, OB^{2 }= (24^{2} + 10^{2}) 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 AD ^{2} =**

- \(\frac{3}{2}\) DC
^{2} - 2DC
^{2} - 3DC
^{2} - 4DC
^{2}

**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, AB^{2} = AD^{2} + BD^{2}……(1)

or, BC^{2 }= AD^{2} + DC^{2} [∵ ΔABC is equilateral and AD ⊥ BC, D is the mid-point of BC.]

or, (2DC)^{2} = AD^{2 }+ DC^{2} or, 4DC^{2}= AD^{2}+ DC^{2}

or, AD = 3DC^{2}

∴ 3. 3DC^{2} is correct

**Example 3. ABC is an isosceles triangle of which AC = BC and AB ^{2} = 2AC^{2}. Then value of ∠C will be**

- 30°
- 90°
- 45°
- 60°

**Solution:**

Given

ABC is an isosceles triangle of which AC = BC and AB^{2} = 2AC^{2}.

∵ AB^{2} = 2AC^{2}, ∴ AB^{2} = AC^{2 }+ AC^{2} = AC^{2} + BC^{2} [∵ 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**

- 9 m
- 10 m
- 11 m
- 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,

BD^{2} = BE^{2} + DE^{2}

= (AB – AE)^{2} + DE^{2}

= (13 – DC)^{2} + DE^{2 }[∵ AE = DC]

= (13 – 7)^{2} + AC^{2} [∵ DC = 7m and DE = AC]

= 6^{2 }+ AC^{2 }= 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**

- 13 cm
- 26 cm
- 52 cm
- 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,

AB^{2} = AO^{2 }+ BO^{2}

or, AB^{2} = {(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

**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, 25x^{2} = 9x^{2 }+ 16x^{2}

∴ 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, AB^{2} = AO^{2} + BO^{2} = (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

**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,

BC^{2} = AB + AC^{2 }= (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,

AC^{2} = AB^{2} + BC^{2}

or, (13)^{2} = (12)^{2} + BC^{2 }[∵ AB = 12 cm]

or, 169 = 144 + BC^{2 }

or, BC^{2 } = 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

**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, BC^{2 } = (2a + 1)^{2 } sq-cm = 4a- + 4a + 1 sq-cm.

AB^{2 }= (2a – 1)^{2 }sq-cm = 4a^{2 } – 4a + 1 sq-cm

AC^{2 }= (2 √2a)^{2 }sq-cm = 8 a sq-cm.

Now, AB^{2 } + AC^{2 } = (4a^{2 } – 4a + 1 + 8a) sq-cm = (4a^{2 } + 4a + 1) sq-cm = BC^{2 }

∴ BC^{2 } = AB^{2 } + AC^{2 }.

∴ 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,

PQ^{2 }= PO^{2 }+ OQ^{2}

= {(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,

RQ^{2 }= PR^{2} + PQ^{2}

= {(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 OA^{2} + OC^{2 }= OB^{2 }+ OD^{2}

or, 5^{2} + OC^{2} = 6^{2} + 8^{2} [∵OA = 5 cm, OB = 6 cm, OD = 8 cm]

or, 25 + OC^{2} = 36 + 64 or, OC^{2} = 100 – 25

or, OC^{2} = 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, AC^{2 }= AD^{2 }+ CD^{2}

= (4^{2} + 2^{2}) sq-cm [∵ AD = 4 cm, DC = 2 cm] = 20 sq-cm.

Now, from the right-angled ΔABD we get,

AB^{2} = AD^{2} + BD^{2} = (4^{2} + 8^{2}) sq-cm [∵ AD = 4 cm, BD = 8 cm] = 80 sq-cm.

Again, BC = BD + CD = 8 cm + 2 cm = 10 cm

∴ BC^{2} = (10)^{2} sq-cm =100 sq-cm

∴ BC^{2 }= 100 sq-cm = 20 sq-cm + 80 sq-cm

= AC^{2} + AB^{2}, BC^{2} = AC^{2} + AB^{2}

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,

AC^{2} = AB^{2} + BC^{2} = (3)^{2} + (4)^{2} sq-cm = 25 sq-cm.

∴ AC = √25 cm = 5 cm.

Now, from the right-angled triangle ABD we get,

AB^{2} = BD^{2} + AD^{2}…… (1)

Again, from the right-angled triangle BCD we get,

BC^{2} = BD^{2} + CD^{2}……(2)

From (1) we get, BD^{2} = AB^{2} – AD^{2 }= (3)^{2} – AD^{2} [∵ AB = 3 cm] = 9 – AD^{2} …..(3)

From (2) we get, BD^{2} = (BC)^{2} – (CD)^{2}

= (4)^{2} – CD^{2} [∵ BC = 4 cm]

= 16 – CD^{2 }………(4)

Comparing (3) and (4) we get,

9 – AD^{2 }= 16 – CD^{2}

or, CD^{2} – AD^{2} =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, BD^{2} = 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

**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 :**

- 8 cm, 15 cm, 17 cm
- 9cm, 11 cm, 6 cm

**Solution:**

1. ∵ 8^{2} + 15^{2} = 64 + 225 = 289 = 172, i.e., 8^{2} + 15^{2} = 17^{2},

∴ The triangle constructed by the given three sides will be a right-angled triangle.

2. ∵ 9^{2} + 6^{2} = 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, OB^{2} = BD^{2} + OD^{2}……(1)

Again, from the right-angled triangle AOC we get, OA^{2} = AC^{2} + OC^{2}

or, OB^{2} = AC^{2} + OC^{2}….. (2) [∵ OA = OB = length of the ladder]

Then from (1) and (2) we get BD^{2} + OD^{2} = AC^{2} + OC^{2}

or, 9^{2} + OD^{2 }= (12)^{2} + OC^{2 }or, OD^{2} – OC^{2} = 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 OB^{2} = 9^{2 }+ OD^{2} [∵ BD = 9 m]

or, (15)^{2} = 9^{2} + OD^{2} [∵ length of the ladder OB = 15 m]

or, 225 = 81 + OD^{2} or, OD^{2} = 225 – 81

or, OD^{2} = 144 or, OD = √144 = 12.

From (2) we get, OB^{2} = AC^{2} + OC^{2} or, (15)^{2} = (12)^{2} + OC^{2}

or, 225 = 144 + OC^{2} or, OC^{2} = 225 – 144 or, OC^{2} = 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,

AB^{2} = AO^{2} + BO^{2} or, (10)^{2} = 6^{2} + BO^{2}

or, 100 = 36 + BO^{2} or, BO^{2} = 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 PS ^{2} + QR^{2} = PR^{2 }+ QS^{2}.**

**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, PS^{2} = PQ^{2} + QS^{2}….. (1)

Again, PR^{2} = PQ^{2} + QR^{2}

or, QR^{2} = PR^{2} – PQ^{2}……..(2)

Now adding (1) and (2) we get,

PS^{2} + QR^{2 }= PQ^{2 }+ QS^{2}+ PR^{2} – PQ^{2}

or, PS^{2} + QR^{2} = PR^{2} + QS^{2}

Hence PS^{2} + QR^{2} = PR^{2 }+ QS^{2}. (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** 4AB^{2} = AC^{2} + BD^{2}

**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, AB^{2} = OA^{2} + OB^{2}

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, 4AB^{2} = AC^{2} + BD^{2}

Hence 4AB^{2} = AC^{2} + BD^{2}. (Proved)

**Example 6. ABC is an equilateral triangle. AD is perpendicular to BC. Prove that AB ^{2 }+ BC^{2}+ CA^{2} = 4AD^{2}.**

**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, AB^{2} = AD^{2} + BD^{2}…….(1)

Again, from the right-angled triangle ACD we get,

AC^{2} = AD^{2}+ CD^{2}…… (2)

Then adding (1) and (2) we get,

AB^{2} + AC^{2} = 2AD^{2} + BD^{2} + CD^{2}

or, AB^{2} + AC^{2} = 2AD^{2}+ \(\left(\frac{1}{2} B C\right)^2+\left(\frac{1}{2} B C\right)^2\) [∵ D is the midpoint of BC.]

or, AB^{2} +AC^{2} = 2AD^{2} + \(\frac{1}{2}\) BC^{2}

or, AB^{2} +AC^{2} = \(\frac{1}{2}\)(4AD^{2} + BC^{2})

or, 2AB^{2} + 2AC^{2 }= 4AD^{2} + BC^{2}

or, 2AB^{2} + AC^{2} + AC^{2} = 4AD^{2} + AB^{2} [∵ BC^{2} = AB^{2}]

or, 2AB^{2} – AB^{2} + BC^{2 }+ AC^{2} = 4AD^{2} [∵ AC = BC]

or, AB^{2} + BC^{2 }+ CA^{2} = 4AD^{2}

Hence AB^{2} + BC^{2} + CA^{2} – 4AD^{2} (Proved)

**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 BQ ^{2} + PC^{2 }= BC^{2}+ PQ^{2}**

**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.

∴ AP^{2} + AQ^{2} = PQ^{2}……(1)

Again, from the right-angled triangle APC

we get, AP^{2} + AC^{2} = PC^{2}….. (2)

Again, from the right-angled triangle ABC we get, AB^{2} + AC^{2} = BC^{2}…….(3)

Also from right-angled triangle ABQ we get, AB^{2} + AQ^{2}= BQ^{2}….. (4)

Now, adding (2) and (4) we get,

PC^{2} + BQ^{2} = AP^{2} + AC^{2} + AB^{2} + AQ^{2}

= (AB^{2} + AC^{2}) + (AP^{2} + AQ^{2})

= BC^{2} + PQ^{2} [from (3) and (1)]

Hence BQ^{2} + PC^{2 }= BC^{2} + PQ^{2} (Proved)

**Example 8. If the diagonals of the quadrilateral ABCD intersect each other orthogonally, then prove that AB ^{2} + CD^{2 }= BC^{2} + DA^{2}.**

**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,

AB^{2} = AO^{2} + BO^{2} …….(1)

From the right-angled triangle COD, we get, CD^{2} = CO^{2} + DO^{2} ……(2)

Similarly, BC^{2} = BO^{2} + CO^{2} ……(3) [from the right-angled triangle BOC]

and DA^{2} = AO^{2} + DO^{2} …….(4) [from the right-angled triangle AOD]

Now, adding (1) and (2) we get,

AB^{2} + CD^{2} = AO^{2} + BO^{2} + CO^{2} + DO^{2}

= (AO^{2} + DO^{2}) + (BO^{2} + CO^{2})

= DA^{2} + BC^{2} [from (4) and (3)].

Hence AB^{2} + CD^{2} = BC^{2} + DA^{2}. (Proved)

**Example 9. AD is the height of the triangle ABC. If AB > AC, then prove that AB ^{2}– AC^{2} = BD^{2}– CD^{2}.**

**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, AB^{2} = AD^{2} + BD^{2}

Again, from the right-angled triangle ACD we get, AC^{2} = AD^{2} + CD^{2}……(1)

Now, subtracting (2) from (1) we get,

AB^{2} – AC^{2} = AD^{2} + BD^{2 }– AD^{2} – CD^{2}

or, AB^{2} – AC^{2} = BD^{2} – CD^{2}

Hence AB^{2} – AC^{2} = BD^{2}– CD^{2}. (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 AC ^{2} + BP^{2} = AB^{2} + CP^{2}.**

**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** AC^{2} + BP^{2} = AB^{2} + CP^{2}.

**Construction:** Let us join A, P.

**Proof:** AC^{2} + BP^{2} = AE^{2} + CE^{2} + BE^{2 }+ PE^{2}

[From right-angled triangles AEC and BEP respectively by Pythagoras’ theorem.]

∴ AC^{2} + BP^{2} = (AE^{2 }+ PE^{2}) + (CE^{2} + BE^{2})

= AP^{2} + BC^{2} [From the right-angled triangles ΔAEP and ΔBEC by Pythagoras’ theorem]

= (AD^{2} + PD^{2}) + (BD^{2} + CD^{2}) [From the right-angled triangles ΔAPD and ΔBCD by Pythagoras’ theorem]

= (AD^{2} + BD^{2}) + (PD^{2} + CD^{2})

= AB^{2 }+ CP^{2} [From right-angled triangles ΔABD and ΔCPD by Pythagoras theorem]

Hence AC^{2} + BP^{2} = AB^{2} + CP^{2} (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 AD ^{2} + BD^{2} = 2CD^{2}**

**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** AD^{2} + BD^{2} = 2CD^{2}.

**Construction:** Let us draw CE ⊥ AB, where CE intersects AB at E.

**Proof:** From the right-angled triangle ABC we get,

AB^{2} = AC^{2} + BC^{2} = AC^{2} + AC^{2} [∵ BC = AC] = 2 AC∵

or, (AD + BD)^{2 }– 2 AC^{2} or, AD^{2} + BD^{2} + 2AD.BD = 2 AC^{2}

or, AD^{2} + BD^{2 }= 2AC^{2} – 2AD.BD

= 2AC^{2} – 2(AE – DE) (BE + DE)

= 2 AC^{2}– 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.]

= 2AC^{2} – 2 (AE^{2}– DE^{2})

= 2AC^{2} – 2 AE^{2} + 2 DE^{2}

= 2 (AC^{2} – AE^{2}) + 2 DE^{2}

= CE^{2} + 2DE^{2} [AC^{2} = AE^{2} + CE^{2}]

= 2 (CE^{2} + DE^{2}) = 2CD^{2} [from the right-angled triangle CED by Pythagoras’ theorem]

Hence AD^{2} + BD^{2} = 2 CD^{2} (Proved)

**Example 12. In ΔABC, ∠A = right angle. If CD is a median, then prove that BC ^{2} = CD^{2} + 3AD^{2}.**

**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** BC^{2} = CD^{2} + 3AD^{2}.

**Proof:** From the right-angled triangle ABC by Pythagoras’ theorem we get

BC^{2} = AC^{2} + AB^{2 }…..(1)

= AC^{2} + (2AD)^{2} [∵ AB = 2AD]

= AC^{2} + 4AD^{2}

= (AC^{2} + AD^{2}) + 3AD^{2}

= CD^{2} + 3AD^{2} [Since from the right-angled triangle ΔACD we get, AC^{2} + AD^{2} = CD^{2}]

Hence BC^{2} = CD^{2} + 3AD^{2} (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 AZ ^{2} + BX^{2} + CY^{2} = AY^{2} + CX^{2} + BZ^{2}**

**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** AZ^{2} + BX^{2} + CY^{2} = AY^{2} + CX^{2} + BZ^{2}.

**Proof:** From the right-angled triangle AOZ by Pythagoras’ theorem

we get, OA^{2} = AZ^{2} + OZ^{2}.

or, AZ^{2 }= OA^{2}– OZ^{2 }….(1)

Similarly, BX^{2}– OB^{2}– OX^{2}….(2)

and CY^{2} = OC^{2} – OY^{2}…. (3)

Now adding (1), (2) and (3) we get,

AZ^{2} + BX^{2} + CY^{2} = (OA^{2} + OB^{2} + OC^{2}) – (OZ^{2} + OX^{2} + OY^{2})

= (OA^{2} – OY^{2}) + (OC^{2} – OX^{2}) + (OB^{2} – OZ^{2})

= AY^{2 }+ CX^{2} + BZ^{2}

[From the right-angled triangles ΔAOY, ΔCOX and ΔBOZ by Pythagoras’ theorem.]

∴ AZ^{2} + BX^{2} + CY^{2} = AY^{2} + CX^{2} + BZ^{2}. (Proved)

**Example 14. In the ΔRST, ∠S = right angle. X and Y are the midpoints of RS and ST respectively. Prove that RY ^{2} + XT^{2} = 5 XY^{2}.**

**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** RY^{2} + XT^{2} = 5 XY^{2}.

**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, XY^{2} = SX^{2} + SY^{2}…..(2)

Similarly, from the right-angled triangle ΔRSY by Pythagoras’ theorem we that RY2 = RS2 + SY2…….(2)

Now, RY^{2} + XT^{2} = RS^{2} + SY^{2} + SX^{2} + ST^{2} [from (2) and XT^{2} = SX^{2} + ST^{2}]

= (2SX)^{2 }+ SY^{2} + SX^{2} + (2SY)^{2} [∵ RS = 2SX and ST = 2SY]

= 4SX^{2} + SY^{2} + SX^{2} + 4SY^{2}

= 5 SX^{2} + 5SY^{2} = 5 (SX^{2} + SY^{2})

= 5 XY^{2} [from (1)]

Hence RY^{2} + XT^{2} = 5XY^{2}. (Proved)

**Example 15. If in ΔABC, AD ⊥ BC, then prove that AB ^{2} + CD^{2} = AC^{2} + BD^{2}.**

**Solution:**

**Given:**

If in ΔABC, AD ⊥ BC,

In ΔABC, AD ⊥ BC,

∴ from the right-angled triangle ΔABD by Pythagoras’ theorem we get, AB^{2} = AD^{2 }+ BD^{2} = AC^{2}= CD^{2} + BD^{2}

[∵ in right-angled triangle ACD, AC^{2} = AD^{2} + CD^{2}]

or, AB^{2} + CD^{2 }= AC^{2} + BD^{2}

Hence AB2 + CD^{2 }= AC^{2} + BD^{2} (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** AC^{2} = 2AB^{2}.

**Proof:** ABC is a right-angled triangle of which ∠B = right angle.

∴ by Pythagoras’ theorem we get, AC^{2} = AB^{2} + BC^{2}

= AB^{2} + AB^{2} [∵ in the square ABCD, AB = BC = CD = DA]

∴ AC^{2} = 2AB^{2} (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 OA ^{2} + OC^{2} = OB^{2} + OD^{2}**

**Solution:**

**Given:**

O is any point inside a rectangle ABCD.

Let O be any point inside the rectangle ABCD.

**To prove** OA^{2} + OC^{2} = OB^{2} + OD^{2}

**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, OA^{2} = OP^{2} + AP^{2}……(1)

Similarly, OC^{2} = OQ^{2} + CQ^{2}…..(2) [in ΔCOQ]

Then adding (1) and (2) we get,

OA^{2} + OC^{2} = OP^{2} + AP^{2} + OQ^{2} + CQ^{2} = AP^{2} + OQ^{2} + CQ^{2} + OP^{2}

= BQ^{2} + OQ^{2} + PD^{2 }+ OP^{2} [∵ AB || PQ || DC, ∴ AP = BQ and PD = CQ.]

= OB^{2} + OD^{2}

[∵ in right-angled triangle ΔBOQ, BQ^{2} + OQ^{2} = OB^{2} and in right-angled ΔDOP, PD^{2} + OP^{2} = OD^{2}]

Hence OA^{2} + OC^{2} = OB^{2} + OD^{2} (Proved)

**Example 18. ABCD is a rhombus. Prove that AB ^{2} + BC^{2} + CD^{2}+ DA^{2} = AC^{2}+ BD^{2}.**

**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** AB^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2} + BD^{2}.

**Proof:** From the right-angled triangle AOB by Pythagoras’ theorem, we get,

AB^{2} = OA^{2} + OB^{2}….(1)

Similarly, BC^{2} = OB^{2} + OC^{2}…..(2)

CD^{2} = OC^{2} + OD^{2}……(3)

DA^{2} = OD^{2} + OA^{2}……..(4)(

Then adding (1), (2), (3) and (4) we get,

AB^{2} + BC^{2} + CD^{2} + DA^{2} = 2 (OA^{2} + OB^{2} + OC^{2} + OD^{2})

= 2 (OA^{2} + OB^{2} + OA^{2} + OB^{2})

[∵ O is the midpoint, ∴ OC = OA and OD = OB.]

= 2 (2OA^{2} + 2OB^{2})

= 4OA^{2} + 4OB^{2} = (2OA)^{2} + (2OB)^{2}

= AC^{2} + BD^{2} [∵ 2OA = AC and 2OB = BD]

Hence AB^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2} + BD^{2} (Proved)

**Example 19. In ΔABC, AD ⊥ BC. Prove that AB ^{2} – BD^{2} = AC^{2} – CD^{2}.**

**Solution:**

**Given:**

In ΔABC, AD ⊥ BC.

Given that AD ⊥ BC.

∴ from the right-angled triangle ABD by Pythagoras’ theorem we get, AB^{2} = AD^{2 }+ BD^{2}…..(1)

Again, in right-angled triangle ACD, we get, AC^{2} = AD^{2} + CD^{2}….(2)

Now, subtracting (2) from (1) we get,

AB^{2} – AC^{2} = BD^{2} – CD^{2}

Hence AB^{2} – BD^{2} = AC^{2} – CD^{2}. (Proved)

**Example 20. In ΔABC, AD ⊥ BC which intersects BC at D. If BD = 3 CD, then prove that 2AB ^{2 }= 2AC^{2}**

**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,

AB^{2} = AD^{2} + BD^{2}…..(1)

Again, in right-angled triangle ACD we get,

AC^{2} = AD^{2} + CD^{2} ……..(2)

Now, subtracting (2) from (1) we get, AB^{2} – AC^{2} = BD^{2} – CD^{2}

= (3CD)^{2 }– CD^{2} = 9CD^{2} – CD^{2} = 8CD^{2}

or, 2AB^{2} – 2AC^{2} = 16 CD^{2}

or, 2AB^{2} – 2AC^{2} = BC^{2} [∵ BC^{2} = (BD + CD)^{2 }= (3CD + CD)^{2} = (4CD)^{2} = 16 CD^{2}.]

or, 2AB^{2} = 2AC^{2} + BC^{2}.

Hence 2AB^{2} = 2AC^{2} + BC^{2}. (Proved)

**Example 21. In the isosceles triangle ABC, AB = AC and BE is perpendicular to AC from B. Prove that BC ^{2} = 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, AB^{2} = BE^{2 }+ AE^{2}

= BE^{2} + (AC – CE)^{2}

= BE^{2}+ AC^{2} – 2AC x CE + CE^{2}

= BE^{2} + CE^{2} + AC^{2} – 2AC x CE

= BC^{2} + AB^{2} – 2AC x CE

or, 0 = BC^{2} – 2AC x CE

or, BC^{2} = 2AC x CE

Hence BC^{2} = 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 CD ^{2 }= 2BD^{2}.**

**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** CD^{2} = 2BD^{2}.

**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,

CD^{2} = DE^{2} + CE^{2} = DE^{2} + DE^{2} [∵ CE = DE]

= 2DE^{2} = 2BD^{2} [∵ DE = BD.]

Hence CD^{2} = 2BD^{2}. (Proved)

**Example 23. P is an external point of the square ABCD. If PA > PB, then prove that PA ^{2 }– PB^{2} = PD^{2} – PC^{2}.**

**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 PA^{2} – PB^{2} = PD^{2} – PC^{2}.

**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,

PA^{2} = AE^{2} + PE^{2}…..(1)

Similarly, PB^{2} = BF^{2} + PF^{2} …..(2)

Now, subtracting (2) from (l) we get,

PA^{2} – PB^{2} = AE^{2} + PE^{2} – BF^{2} – PF^{2}

= AE^{2 }+ (PD – DE)^{2 }– BF^{2} – (CP – CF)^{2}

= AE^{2} + PD^{2} + DE^{2} – 2PD.DE – BF^{2} – CP^{2} – CF^{2}+ 2CP.CF.

= AE^{2} + DE^{2} + PD^{2 }– CP^{2} – (BF^{2} + CF^{2}) – 2PD.DE + 2CP.CF

= AD^{2} + PD^{2} – CP^{2} – BC^{2} – 2PD.DE + 2CP.CF

= BC^{2 }+ PD^{2} – CP^{2} – BC^{2 }– 2PD.DE + 2CP.CF

= AD^{2} – 2PD.DE + 2CP.CF – BC^{2} + PD^{2} – PC^{2}

= AD^{2} – AD^{2} + BC^{2} – BC^{2} + PD^{2} – PC^{2}

[∵ 2PD.DE = AD^{2} and 2CP.CF = BC^{2} ]

= PD^{2 }– PC^{2}

Hence PA^{2}– PB^{2} = PD^{2} – PC^{2}. (Proved)

**Example 24. In the right-angled triangle ABC, ∠A = 1 right angle. BE and CF are two medians of ΔABC. Prove that 4 (BE ^{2} + CF^{2}) = 5BC^{2}.**

**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, BE^{2} = AB^{2} + AE^{2}…..(1)

Similarly, CF^{2} = AC^{2 }+ AF^{2} ……(2)

∴ adding (1) and (2) we get, BE^{2} + CF^{2} = AB^{2} + AC^{2} + AE^{2} + AF^{2}

= AB^{2} + AC^{2} + \(\left(\frac{1}{2} \mathrm{AC}\right)^2+\left(\frac{1}{2} \mathrm{AB}\right)^2\)

= AB^{2} + AC^{2} + \(\frac{1}{4}\) AC^{2} + \(\frac{1}{4}\) AB^{2}

= \(\frac{5}{4}\) AB^{2}+ \(\frac{5}{4}\) AC^{2} = \(\frac{5}{4}\)(AB + AC)^{2}

= \(\frac{5}{4}\) BC^{2 }[ ∵ in the right-angled triangle ABC, AB^{2} + AC^{2 }= BC^{2}]

or, 4 (BE^{2} + CF^{2}) = 5 BC^{2}.

Hence 4 (BE^{2} + CF^{2}) = 5 BC^{2}. (Proved)

**Example 25. A perpendicular AD is drawn on BC from the vertex A of the acute ΔABC. Prove that AC ^{2} = AB^{2} + BC^{2} – 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** AC^{2} = AB^{2} + BC^{2} – 2BC.BD.

**Proof:** AD ⊥ BC, ΔACD is a right-angled triangle and hence by Pythagoras’ theorem we get,

AC^{2} = AD^{2} + DC^{2}

= AD^{2 }+ (BC – BD)^{2}

= AD^{2 }+ BC^{2} + BD^{2} – 2BC.BD

= AD^{2} + BD^{2} + BC^{2} – 2BC.BD

= AB^{2} + BC^{2} – 2BC.BD [∵ in right-angled triangle ΔABD, AD^{2 }+ BD^{2} = AB^{2}]

Hence AC^{2 }= AB^{2} + BC^{2 }– 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 = \(\frac{\sqrt{3}}{4}\) a^{2} sq-units…..(1)

∴ Δ QBC = \(\frac{\sqrt{3}}{4}\) x BC^{2} sq- units [side = BC]……(2)

Similarly, Δ PAB = \(\frac{\sqrt{3}}{4}\) x AB^{2} sq- units……(3)

Now, ∵ ΔABC is a right-angled triangle,

∴ by Pythagoras’ theorem, we get, BC^{2} = AB^{2}+ AC^{2}

or, \(\frac{\sqrt{3}}{4}\) BC^{2} = AB^{2} + \(\frac{\sqrt{3}}{4}\) AC^{2} [multiplying by \(\frac{\sqrt{3}}{4}\)]

or, ΔQBC = ΔPAB + ΔCRA [from (1), (2) and (3)]

Hence ΔQBC = ΔPAB + ΔCRA (Proved)