WBBSE Solutions For Class 9 Maths Solid Geometry Chapter 5 Construction Drawing Of Parallelograms Equal To The Area Of A Given Triangle

Solid Geometry Chapter 5 Construction Drawing Of Parallelograms Equal To The Area Of A Given Triangle

Chapter 5 Construction Drawing Of Parallelograms Equal To The Area Of A Given Triangle Drawing of parallelograms equal to the area of a triangle: You have already studied in the geometric theorems section that

1. The area of a triangle is half of the area of a parallelogram having equal bases and within the same parallels.
2. The area of all triangles is equal having equal bases and within the same parallels.

By applying these two theorems we can construct a parallelogram equal to the area of a given triangle. In the following construction, it is discussed thoroughly.

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Construction-1

Construct a parallelogram one whose angle is equal to a given angle and whose area is equal to the area of a certain given triangle.

1. Let ABC be a given certain triangle and∠X be a given certain angle.
2. We have to construct a parallelogram one of whose angles is equal to∠X and whose area is equal to the area of the ΔABC.

 

 

Rule of construction:

1. Let us first draw a certain triangle and a certain angle.
2. Let us draw PAQ through A parallel to BC.
3. Let us bisect BC of the ΔABC at D.
4. Let us draw ∠CDE at D of BC equal to ∠X, the side DE of which intersects PQ at E.
5. Let us cut the part EF from EQ equal to DC.
6. Let us then join points C and F.

Thus, DEFC is the required parallelogram.

Proof: DC || EF and DC = EF, DEFC is a parallelogram.

Now, join A, D.

∴ AD is a median of the ΔABC.

We know that the median of any triangle divides it into two triangles of equal areas. AACD=AABC……. …. (1)

∴ \(\Delta \mathrm{ACD}=\frac{1}{2} \Delta \mathrm{ABC}\)……..(1)

Again, ΔACD and DEFC have the same base DC and lie within the same parallels BC and PQ. 

∴ or, ΔABC = \(\frac{1}{2}\) DEFC [ by (1)] 

or, \(\frac{1}{2}\) ΔABC = \(\frac{1}{2}\) DEFC.

or, ΔABC = DEFC.

∴ The areas of the ΔABC and the parallelogram DEFC are equal and by construction

∴ DEFC is the required parallelogram. (Proved)

In the following examples application of the construction-10 is discussed

 

Question 1. Construct a triangle of sides 5 cm, 8 cm, and 11 cm and then construct a parallelogram the area of which is equal to the area of that triangle and one of whose angles is 60° [ Rule of construction and proof must be given. ]

Solution:

Given 

Triangle of sides 5 cm, 8 cm, and 11 cm.

One of whose angles is 60°

Let in triangle ABC, AB = 5 cm; BC= 11 cm and CA = 8 cm, and X be a given certain angle of 60°.

We have to construct a parallelogram, the area of which is equal to the area of ΔABC and one of whose angles is 

 

 

Rule of construction:

1. First, let us draw a straight line BC of length 11 cm with the use of a ruler.

2. Let us now draw an arc of radius 5 cm at B of BC and an arc of radius 8 cm at C. Let the latter one intersect the previous arc at A. A, B, and A, C are joined.
Then a triangle ABC is constructed, the sides of which are 5 cm, 11 cm, and 8 cm.

3. Let us now draw a certain angle X = 60°. 

4. Let us determine D, the mid-point of BC.

5. Let us draw PQ through A parallel to BC.

6. Let us draw 4 CDE at D equal to ∠X = 60°, the side DE of which intersects PQ at E.

7. Let us cut the part EF from EQ equal to DC.

8. C and F. are joined.

Then DEFC is the required parallelogram.

 

 

Proof: A and D are joined.

According to the construction, in ΔABC, AB = 5 cm, BC= 11 cm, and CA = 8 cm.

∴ ΔABC is the required triangle.

Since DC = EF and DC || EF,

∴ DEFC is a parallelogram.

By construction, 4 CDE = X = 60° of  DEFC.

Now, D is the mid-point of BC, 

∴ AD is a median of ΔABC.

∴ ΔACD= \(\frac{1}{2}\) ΔABC……...(1)

Again, ΔACD and DEFC have the same base DC and they lie within the same parallels BC and PQ.

∴ ΔACD = \(\frac{1}{2}\)  DEFC [ by previous theorem ]

ΔABC =  DEFC [ by (1)]

or, ΔABC = \(\frac{1}{2}\)  DEFC.

Hence, DEFC is the required parallelogram.

 

Question 2. Construct a triangle ABC of sides AB = 6 cm, BC = 9 cm, and 2 ABC = 55°. Then construct a parallelogram, one of whose angles is 60°, and the length of one side is equal to half of the length of AC.

Solution:

Given

A triangle ABC of sides AB = 6 cm, BC = 9 cm, and 2 ABC = 55°.

Let in Δ ABC, AB = 6 cm; BC= 9 cm, E = 55°.

We have to construct a parallelogram, the area of which is equal to the area of the Δ ABC and one of whose angles is 60°.

Also, one of which sides is half of AC.

 

 

Method of construction :

1. At first, let us draw by a scale AB = 6 cm, BC = 9 cm, and ∠E = 55°. 

2. Let us draw ∠CBD at B of BC equal to ∠E = 55°.

3. Let us cut the part BA = 6 cm from BD of CBD and let us join A, C.

Then, the required triangle ABC is constructed.

4. Let us draw XY through B of ΔABC parallel to AC.

5. Let us determine P, the mid-point of AC.

6. Let us draw ∠CPQ = 60° at P along with PC.

Let the side PQ of ∠CPQ intersect XY at Q.

7. Let us cut the part QR from QY equal to PC.

8. C and R are joined.

Then, PQRC is the required parallelogram.

Proof: B and P are joined. Now, in ΔABC, AB = 6 cm, BC = 9 cm,∠ABC = < E = 55°

∴ ΔABC is the required triangle. 

Again, in PQRC, PC = QR and PC || QR [ AC || XY]

∴ PQRC is a parallelogram of which ∠CPQ = 60° [ by construction]

Also, P is the mid-point of AC,

∴ PC = \(\frac{1}{2}\) AC and BP is a median of ΔABC.

∴ ΔBPC = \(\frac{1}{2}\) AABC…………(1)

But, Δ BPC and PQRC have the same base PC and they lie within the same parallels AC and XY.

ΔBPC = \(\frac{1}{2}\)  PQRC

or, \(\frac{1}{2}\) ΔABC = \(\frac{1}{2}\) PQRC [by (1)]

or, ΔABC = PQRC

∴ in PQRC, ∠CPQ = 60°, PC =  \(\frac{1}{2}\)  AC and ΔABC = PQRC.

Hence PQRC is the required parallelogram. (Proved)

 

Question 3. In Δ PQR,∠ PQR = 30°, ∠PRQ,= 75″ and QR = 8 cm. Construct a rectangle, the area of which is equal to the area of the ΔPQR. 

Solution:

Given 

Δ PQR,∠ PQR = 30°, ∠PRQ,= 75″ and QR = 8 cm.

Let in ΔPQR, ∠PQR = 30°, ∠PRQ = 75°, and QR = 8 cm.

We have to construct a rectangle, the area of which is equal to the area of the ΔPQR.

Method of construction :

1. At first, let us draw QR = 8 cm by a scale.

2. Let us draw a line segment QT and then the part QR = 8 cm is cut from QT.

3. Let us now draw RQX equal to 30° at Q of QR and QRY equal to 75° at R of QR. 

Let QX and RY intersect at P. 

Then A PQR is the required triangle.

4. Let us draw AB through P parallel to QR.

5. Let us draw the perpendicular-bisector DE of QR, which intersects AB at E.

6. Let us cut EF from EB equal to DR.

7. R and F are joined.

8. ∴ DEFR is the required rectangle.

 

 

Proof: By construction, in ΔPQR, PQR = 30°, PRQ= 75°, and QR = 8 cm. 

∴ ΔPQR is the required triangle.

Again, in the quadrilateral DEFR, EF = DR, EF || DR and EDR = 90°, [ by construction]

∴ DEFR is a rectangle. 

Now, P and D are joined.

Since D is the mid-point of QR, 

∴ PD is a median of Δ PQR.

∴ ΔPDR = \(\frac{1}{2}\) ΔPQR…………..(1)

Again, ΔPDR and    DEFR have the same base DR and they lie within the same parallels DR and AB.

∴ ΔPDR = \(\frac{1}{2}\)   DEFR, 

or, ΔPQR = \(\frac{1}{2}\)   DEFR [by (1)]

or, ΔPQR = \(\frac{1}{2}\)   DEFR, i.e., area of ΔPQR = area of  DEFR.

 Hence DEFR is the required rectangle. (Proved)

 

Question 4. Construct an equilateral triangle of sides 6-5 cm and construct a parallelogram, the area of which is equal to the area of that triangle and one of whose angles is 45°.

Solution:

Given

An equilateral triangle of sides 6-5 cm.

One of whose angles is 45°.

Let ΔABC be an equilateral triangle of sides 6.5 cm. 

We have to construct it. 

Then we have also to construct a parallelogram, the area of which is equal to the area of the ΔABC, one of whose angles is 45°.

Method of construction:

1. Let us first draw a straight line of length 6.5 cm.

2. Now, let us draw a horizontal line BX and let us cut the part BC equal to a from BX. 

3. Let us draw an arc with center B of BC and a radius equal to 6-5 cm.

Taking C of BC as the center and with the same radius, let us draw another arc on the same side as the previous arc. 

Let the two arcs intersect each other at A.

4. Let us now join A, B, and A, C. 

Then ΔABC is the required triangle.

5. Now, let us draw a line GH through A of ΔABC parallel to BC.

6. Let us draw the perpendicular bisector AD of BC, which intersects BC at D.

7. Let us draw CDE equal to 45° at D, the side DE of which intersects GH at E.

8. Let us now cut the part EF equal to DC from EH.

9. C and F are joined.

10. Then DEFC is the required parallelogram.

 

 

Proof: By construction, in ΔABC, AB = BC = CA = 6.5 cm, 

∴ ΔABC is the required triangle.

Also, in the quadrilateral DEFC, DC = EF, DC || EF and CDE = 45°,

∴ DEFC is a parallelogram, one of whose angles is 45°.

Again, AD is the perpendicular bisector of BC [ by construction]

Since Δ ABC is equilateral, 

∴ AD is a median of ΔABC.

∴ ΔACD = \(\frac{1}{2}\) ΔABC…………..(1)

 Now, ΔACD and DEFC has the same base DC and they lie within the same parallels DC and GH.

∴ ΔACD = \(\frac{1}{2}\)  DEFC 

or, \(\frac{1}{2}\) ΔABC = \(\frac{1}{2}\) DEFC. [ by (1)]

or, ΔABC = \(\frac{1}{2}\) DEFC

∴ DEFC is the required parallelogram. (Proved)

 

Question 5. Construct an isosceles triangle, the length of each of whose equal sides is 8 cm, and the length of the base is 5 cm. Then construct a parallelogram, the area of which is equal to the area of the triangle, one angle of whose is equal to one of the two equal angles of that triangle and one side of whose is half of the two equal sides of the very triangle. [Give construction signs only]

Construction:

Given 

An isosceles triangle, the length of each of whose equal sides is 8 cm, and the length of the base is 5 cm.

 

Here, BCDE is the required parallelogram, the area of which is equal to the area of the ΔABC, one of its angles is equal to ABC and one side is half of AB.

 

Question 6. Construct an isosceles triangle, each of whose equal sides is of length 8 cm and the angle between the equal sides is 30°. Then construct a rectangle, the area of which is equal to the area of that triangle. [Give construction signs only]

Construction:

Given

An isosceles triangle, each of whose equal sides is of length 8 cm and the angle between the equal sides is 30°.

 

Here, in ΔABC, AB = BC = 8 cm and < ABC = 30°; 

∴ ΔABC is the required triangle.

Again in, DEFC, CDE = 90°, DC = EF and DC || EF,

∴ DEFC is the required rectangle

 

Question 7. Construct a triangle of sides 3 cm, 4 cm and 6 cm and then construct a parallelogram, the area of which is equal to the area of that triangle and one of whose angles is 30°.

[Give construction signs only]

Solution:

Given 

Triangle of sides 3 cm, 4 cm and 6 cm.

One of whose angles is 30°.

Let the lengths of AB, AC, and BC of ΔABC be 3 cm, 4 cm, and 6 cm respectively. 

We have to construct the ΔABC. 

Then we have also to construct a parallelogram, the area of which is equal to the area of ΔABC.

Construction: 

 

 

Here, in ΔABC, AB = 3 cm, AC = 4 cm and BC 6 cm. 

∴ ΔABC is the required triangle. and ΔABC = DEFC

Also in DEFC, ∠CDE = 30°, DC = EF and DC || EF and ΔABC =   DEFC

∴ DEFC is the required parallelogram.