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

**Exercise 15.2**

1.Â If Fig. 15.26, ABCD is a parallelogram, AE âŠ¥ DC and CF âŠ¥ 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 =Â Height Ã—Â Base

AE is base on CD

AEÃ—CD = 8cmÃ—16cm

Also, area is

CFÃ—AD = 10cmÃ—AD

So,

Equating Both

8cmÃ—16cm = 10cmÃ—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(â–³ABD) = ar(â–³BCD) = ar(â–³ABC) = 1/2 ar(âˆ¥gm ABCD)

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

CD = AB = 16cm

Since, area of parallelogram =Â Height Ã—Â Base

AE is base on CD

AEÃ—CD = 8cmÃ—16cm

Also, area is

CFÃ—AD = 10cmÃ—AD

So,

Equating Both

8cmÃ—16cm = 10cmÃ—AD

AD = 128cm/10cm = 12.8cm

**Solution**
We know that,

Area of a parallelogram ABCD = ADÃ—CF --- (1)

Again, area of parallelogram ABCD = CDÃ—AE --- (2)

Comparing equation (1) and (2),

ADÃ—CF = CDÃ—AE

â‡’ 6Ã—10 = DÃ—8

â‡’ D = 60/8 = 7.5 cm

âˆ´ AB = DC = 7.5 cm

[âˆµ Opposite side of a parallelogram are equal.]

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

^{2Â }. If E and F are mid points of sides AB andÂ CD respectively, then find the area of parallelogram AEFD.**Solution**4. If ABCD is a parallelogram, then prove that ar(â–³ABD) = ar(â–³BCD) = ar(â–³ABC) = 1/2 ar(âˆ¥gm ABCD)

**Solution**
Given,

ABCD is a parallelogram.

To prove:

ar(â–³ABD) = ar(â–³BCD) = ar(â–³ABC) = ar(â–³ACD) = 1/2 ar(âˆ¥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(â–³ABC) = ar(â–³ACD) = 1/2 ar(âˆ¥gm ABCD) --- (1)

Since, BD is the diagonal.

Then,

ar(â–³ABD) = ar(â–³BCD) = 1/2 ar(âˆ¥gm ABCD) --- (2)

Compare equation (1) and (2)

âˆ´ arâ–³ABC) = ar(â–³ACD) = ar(â–³ABD)= ar(â–³BCD) = 1/2 ar(âˆ¥gm ABCD).