# RD Sharma Solutions Class 9 Areas Of Parallelograms And Triangles Exercise 15.2

## RD Sharma Solutions Class 9 Chapter 15 Exercise 15.2

### RD Sharma Class 9 Solutions Chapter 15 Ex 15.2 Download

Q 1. If figure, ABCD is a parallelogram, AE $\perp$ DC and CF $\perp$ AD. If AB = 16 cm, AE = 8 cm, and CF = 10 cm, Find AD.

Solution:

Given that,

In parallelogram ABCD, CD = AB = 16 cm                                $\left [ ∵ Opposite\;side \;of \;a\; parallelogram\;are\;equal\right ]$

We know that,

Area of parallelogram = Base $\times$ Corresponding altitude

Area of parallelogram ABCD = CD $\times$ AE = AD $\times$ CF

16 cm $\times$ cm = AD $\times$ 10 cm

AD = $\frac{16\times 8}{10}$ cm = 12.8 cm

Thus, The length of AD is 12.8 cm.

Q 2. In Q 1, if AD = 6 cm, CF = 10 cm, and AE = 8 cm, Find AB.

Solution:

We know that,

Area of a parallelogram ABCD = AD $\times$ CF $\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 1 \right )$

Again area of parallelogram ABCD = CD $\times$ AE$\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 2 \right )$

Compare equation(1) and equation(2)

AD $\times$ CF  = CD $\times$ AE

$\Rightarrow 6\times 10=D\times 8$

$\Rightarrow D=\frac{60}{8}$ = 7.5 cm

$∴ AB=DC=7.5\;cm$                                     $\left [ ∵Opposite\;side \;of \;a\; parallelogram\;are\;equal\right ]$

Q 3. Let ABCD be a parallelogram of area 124 $cm^{2}$ . If E and F are the mid-points of sides AB and CD respectively, then find the area of parallelogram AEFD.

Solution:

Given,

Area of a parallelogram ABCD = 124 $cm^{2}$

Construction: Draw AP$\perp$DC

Proof:-

Area of a parallelogram AFED = DF $\times$ AP  $\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 1 \right )$

And  area of parallelogram EBCF = FC $\times$ AP$\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 2 \right )$

And DF = FC  $\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 3 \right )$                                               $\left [ F \;is\;the\;midpoint\;of\;DC \right ]$

Compare equation (1), (2)  and (3)

Area of parallelogram AEFD = Area of parallelogram EBCF

$∴ Area \;of\; parallelogram\; AEFD=\frac{Area \;of\; parallelogram\;ABCD}{2}$ = $\frac{124}{2}$ = $62 \;cm^{2}$

Q 4. If ABCD is a parallelogram, then prove that

Ar $\left ( \Delta ABD \right )$ = Ar $\left ( \Delta BCD \right )$ = Ar $\left ( \Delta ABC \right )$ = Ar $\left ( \Delta ACD \right )$ = $\frac{1}{2}Ar\left ( //^{gm} ABCD \right )$.

Solution:

Given:-

ABCD is a parallelogram,

To prove : – Ar $\left ( \Delta ABD \right )$ = Ar $\left ( \Delta BCD \right )$ = Ar $\left ( \Delta ABC \right )$ = Ar $\left ( \Delta ACD \right )$ = $\frac{1}{2}Ar\left ( //^{gm} ABCD \right )$.

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

Since, AC is the diagonal.

Then,  Ar $\left ( \Delta ABC \right )$ = Ar $\left ( \Delta ACD \right )$ = $\frac{1}{2}Ar\left ( //^{gm} ABCD \right )$ $\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 1 \right )$

Since, BD is the diagonal.

Then ,  Ar $\left ( \Delta ABD \right )$ = Ar $\left ( \Delta BCD \right )$ = $\frac{1}{2}Ar\left ( //^{gm} ABCD \right )$ $\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 2 \right )$

Compare equation (1) and (2)

$∴$ Ar $\left ( \Delta ABC \right )$ = Ar $\left ( \Delta ACD \right )$ = Ar $\left ( \Delta ABD \right )$ = Ar $\left ( \Delta BCD \right )$ = $\frac{1}{2}Ar\left ( //^{gm} ABCD \right )$..