# R.D. Sharma Solutions Class 9th: Ch 15 Areas of Parallelogram and Triangles Exercise 15.2

#### Chapter 15 Areas of Parallelogram and Triangles R.D. Sharma Solutions for Class 9th Exercise 15.2

**Exercise 15.2**

1. If Fig. 15.26, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm. AE = 8 cm and CF = 10 cm, find AD.

Opposite side of parallelogram are equal, therefore,

CD = AB = 16cm

Since, area of parallelogram = Height × Base

AE is base on CD

AE×CD = 8cm×16cm

Also, area is

CF×AD = 10cm×AD

So,

Equating Both

8cm×16cm = 10cm×AD

AD = 128cm/10cm = 12.8cm

2. In Q.No. 1, if AD = 6 cm, CF = 10 cm, and AE = 8 cm, find AB.

3. Let ABCD be a parallelogram of area is 124 cm

4. If ABCD is a parallelogram, then prove that ar(△ABD) = ar(△BCD) = ar(△ABC) = 1/2 ar(∥gm ABCD)

**Solution**Opposite side of parallelogram are equal, therefore,

CD = AB = 16cm

Since, area of parallelogram = Height × Base

AE is base on CD

AE×CD = 8cm×16cm

Also, area is

CF×AD = 10cm×AD

So,

Equating Both

8cm×16cm = 10cm×AD

AD = 128cm/10cm = 12.8cm

**Solution**
We know that,

Area of a parallelogram ABCD = AD×CF --- (1)

Again, area of parallelogram ABCD = CD×AE --- (2)

Comparing equation (1) and (2),

AD×CF = CD×AE

⇒ 6×10 = D×8

⇒ D = 60/8 = 7.5 cm

∴ AB = DC = 7.5 cm

[∵ Opposite side of a parallelogram are equal.]

3. Let ABCD be a parallelogram of area is 124 cm

^{2 }. If E and F are mid points of sides AB and CD respectively, then find the area of parallelogram AEFD.**Solution**4. If ABCD is a parallelogram, then prove that ar(△ABD) = ar(△BCD) = ar(△ABC) = 1/2 ar(∥gm ABCD)

**Solution**
Given,

ABCD is a parallelogram.

To prove:

ar(△ABD) = ar(△BCD) = ar(△ABC) = ar(△ACD) = 1/2 ar(∥gm ABCD)

Proof:

We know that diagonal of a parallelogram divides it into two equilaterals.

Since,

AC is the diagonal.

AC is the diagonal.

Then,

ar(△ABC) = ar(△ACD) = 1/2 ar(∥gm ABCD) --- (1)

Since, BD is the diagonal.

Then,

ar(△ABD) = ar(△BCD) = 1/2 ar(∥gm ABCD) --- (2)

Compare equation (1) and (2)

∴ ar△ABC) = ar(△ACD) = ar(△ABD)= ar(△BCD) = 1/2 ar(∥gm ABCD).