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.

2

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