SChand CLASS 9 Chapter 16 Area of plane figures Exercise 16(C)

 Exercise 16 C

Question-1

Find the area of a parallelogram whose base and height are as given below:

$\begin{array}{|l|l|l|l|l|}\hline & \text { (i) } & \text { (ii) } & \text { (iii) } & \text { (iv) } \\\hline \text { Base } & 8 \mathrm{~cm} & 2.8 \mathrm{~cm} & 12 \mathrm{~mm} & 6.5\mathrm{~m} \\\hline \text { Height } & 3 \mathrm{~cm} & 5 \mathrm{~cm} & 8.7 \mathrm{~mm} & 4.8\mathrm{~m} \\\hline\end{array}$

Answer-1

Area of parallelogram $=$ Base $x$ height

(i) Area $=8 \times 3 \mathrm{~cm}^{2}$

$=24 \mathrm{~cm}^{2}$

(ii) Area $=(2.8 \times 5) \mathrm{cm}^{2}$

$=14 \mathrm{~cm}^{2}$

(iii) Area $=12 \times 8.7 \mathrm{~mm}^{2}$

$=104.4 \mathrm{~mm}^{2}$

(iv) Area $=(6.5 \times 4.8) \mathrm{m}^{2}$

$=31.20 \mathrm{~m}^{2}$

Question-2

The area of a parallelogram is $11 / 2$ ares. Its base is $20 \mathrm{~m}$. Find its height. $\left(1 \operatorname{arc}=100 \mathrm{~m}^{2}\right)$

Answer-2

Area $=(3 / 2 \times 100) \mathrm{m}$

$=(3 \times 50) \mathrm{m}^{2}$

$=150 \mathrm{~m}^{2}$

$\therefore 150=20 \times h$

$\mathrm{h}=7.5 \mathrm{~m}$

Question-3

In a parallelogram $A B C D, A B=8 \mathrm{~cm}, B C=5 \mathrm{~cm}$, perp. From $A$ to $D C=3 \mathrm{~cm} .$ Find the length of the perp. drawn from $B$ to $A D$.

Answer-3

Question-4

A parallelogram has side $34 \mathrm{~cm}$ and $20 \mathrm{~cm} .$ One of its diagonals is $42 \mathrm{~cm} .$ Calculate its area.

Answer-4

At first we have to calculate the area of the triangle having sides $34 \mathrm{~cm}, 20 \mathrm{~cm}$ and $42 \mathrm{~cm}$.

Now,

$\mathrm{s}=(34+20+42) / 2$

$=48 \mathrm{~cm}$

$\therefore$ Area of triangle

$=\sqrt{48(48-34)(48-20)(48-42)}$

$=\sqrt{48 \times 14 \times 28 \times 6}$

$=\sqrt{4 \times 12 \times 2 \times 7 \times 4 \times 7 \times 2 \times 3}$

$=\sqrt{4 \times 4 \times 2 \times 2 \times 7 \times 7 \times 36}$

$=4 \times 2 \times 7 \times 6$

$=336 \mathrm{~cm}^{2}$

$\therefore$ Area of parallelogram

$=2 \times$ Area of triangle if separating boundary of diagonal

$=2 \times 336$

$=672 \mathrm{~cm}^{2} .$

Question-5

$A B C D$ is a parallelogram with side $A B=12 \mathrm{~cm}$. Its diagonals $A C$ and $B D$ are the lengths 20 $\mathrm{cm}$ and $16 \mathrm{~cm}$ respectively. Find the area of $\| \mathrm{gm}$ ABCD

Answer-5

Let $O$ be the intersecting point of $A C$ and $B D$

We know,

diagonals of a parallelogram bisect each other

$\mathrm{OA}=\mathrm{OC}=(1 / 2) \times \mathrm{AC}=10 \mathrm{~cm}$

$O B=O D=(1 / 2) \times B D=8 \mathrm{~cm}$

in $\triangle \mathrm{AOB}$

$\mathrm{OA}=10 \mathrm{~cm}$

$\mathrm{OB}=8 \mathrm{~cm}$

$\mathrm{AB}=12 \mathrm{~cm}$

By using Heron's formula

$\operatorname{ar}(\Delta \mathrm{AOB})=\sqrt{s}(s-a)(s-b)(s-c)$

$s=(a+b+c) / 2$

$=(12+8+10) / 2=$

$30 / 2=15 \mathrm{~cm}$

$\begin{aligned}&\operatorname{ar}(\Delta A O B)=\sqrt{1} 5 \times 3 \times 7 \times 5 \\&=15 \sqrt{7} \mathrm{~cm}^{2}\end{aligned}$

we know that the diagonals of a parallelogram divides it into four equal triangles $=>\operatorname{ar}(\Delta A O B)=\operatorname{ar}(\Delta B O C)=\operatorname{ar}(\Delta C O D)=\operatorname{ar}(\Delta A O D)=15 \sqrt{7} \mathrm{~cm}^{2}$

$\operatorname{ar}(\mathrm{ABCD})=4^{\star} 15 \sqrt{7}=60 \sqrt{7} \mathrm{~cm}^{2}$ (Ans.)

Question-6

What is the area of a rhombus which has diagonals of $8 \mathrm{~cm}$ and $10 \mathrm{~cm}$.

Answer-6

Area of rhombus $=1 / 2 \times 8 \times 10 \mathrm{~cm}^{2}$

$=40 \mathrm{~cm}^{2}$

Question-7

The area of a rhombus is $98 \mathrm{~cm}^{2}$. If one of its diagonals is $14 \mathrm{~cm}$, what is the length of the other diagonal?

Answer-7

Let other diagonal be $=a$

Area of rhombus $=1 / 2 \times$ Product of diagonals

$98=1 / 2 \times 14 \times a$

Or, $a=14 \mathrm{~cm}$

Question-8

PQRS is a rhombus.

(i) If it is given that $\mathrm{PQ}=3 \mathrm{~cm}$, calculate the perimeter of $\mathrm{PQRS}$

(ii) If the height of the rhombus is $2.5 \mathrm{~cm}$, calculate its area,

(iii) The diagonals of a rhombus are $8 \mathrm{~cm}$ and $6 \mathrm{~cm}$ respectively. Find its perimeter.

Answer-8

(i)since all sides of rhombus are same in length

Therefore, perimeter of Rhombus $P Q R S$ is $=4 \times$ length of one side $=4 \times 3=12 \mathrm{c} \cdot \mathrm{m}$

(ii)Area of rhombus $=$ base $x$ height

$\begin{aligned}&=3 \times 2.5 \\&=7.5 \mathrm{~cm}^{2}\end{aligned}$

(iii) PQRS is a rhombus and we know that all four sides of a rhombus are of equal length.

And, In $\triangle$ POQ

$\mathrm{PQ}$ is hypotenuse, $\mathrm{OP}$ is base and $\mathrm{OQ}$ is perpendicular.

Using Pythagoras Theorem -

$\begin{aligned}&\Rightarrow(\mathrm{PQ})^{2}=(\mathrm{OP})^{2}+(\mathrm{OQ})^{2} \\&\Rightarrow(\mathrm{PQ})^{2}=(3)^{2}+(4)^{2} \\&\Rightarrow(\mathrm{PQ})^{2}=9+16 \\&\Rightarrow(\mathrm{PQ})^{2}=25 \\&\Rightarrow \mathrm{PQ}=\sqrt{2} 5 \\&\Rightarrow \mathrm{PQ}=5 \mathrm{~cm}\end{aligned}$

So, length of each side of the given rhombus is $5 \mathrm{~cm}$.

Perimeter of rhombus $=4 \times$ side

$\begin{aligned}&\Rightarrow 4 \times 5 \\&=20 \mathrm{~cm}\end{aligned}$

So, perimeter of the rhombus PQRS is $20 \mathrm{~cm}$.

Question 9

 

The sides of a rhombus are $5 \mathrm{~cm}$ each and one diagonal is $8 \mathrm{~cm}$, calculate,

(i) The length of the other diagonal and

(ii) The area of the rhombus.

Answer-9

(i)Let half the length of second diagonal be $x$

As two diagonals and a side form right angled triangle ;

$\Rightarrow x^{2}+4^{2}=5^{2}$

$\Rightarrow x=3$

$\Rightarrow$ length of other diagonal $=2 x=6 \mathrm{~cm}$

(ii)Area of rhombus $=(1 / 2)$ D1 $\times$ D2

Area of rhombus $=(1 / 2) 8 \times 6$

$=24 \mathrm{~cm}^{2}$

Question-10

In the given figure, $A B C X$ is a rhombus of side $5 \mathrm{~cm}$. angles $B A D$ and $A D C$ are right angles. If $\mathrm{DC}=8 \mathrm{~cm}$, calculate the area of $\mathrm{ABCX}$.

Answer-10

(IMAGE TO BE ADDED)

In fig.

Angle BAD = Angle ADC

But they're co interior angles. Thus, AB \| CD.

Also, AX $\|$ BC, this implies, ABCX is a parallelogram.

Therefore, AB = CX = $5 \mathrm{~cm}$ (opposite sides of parallelogram are equal)

Also, AX $=$ BC $=5 \mathrm{~cm}$

Now, DX $=$ DC-CX $=8-5=3 \mathrm{~cm}$

In $\triangle$ ADX, using Pythagorean theorem, we get,

AD $=4 \mathrm{~cm}$

Now, in parallelogram ABCX,

Base $=5 \mathrm{~cm}$

Height $=4 \mathrm{~cm} .$

Therefore, ar $(A B C X)=$ Base $\times$ Height

$=20 \mathrm{~cm}^{2}$