[2 Mark Question Answer]

Question 12 Marks

In the figure given below, find the area of shaded region: (All measurements are in cm)

Answer

Area of shaded portion
$= 2 \times 20 + 2 \times 8 + 2 \times (12 + 2)$
$= 40 + 16 + 28 cm^2$
$= 84 cm^2$

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Question 22 Marks

In the figure given below, find the area of shaded region: (All measurements are in cm)

Answer

Area of shaded region = Area of outer region − Area of inner region
$ = 40 \times 40 − 15 \times 15$
$= 1600 − 225$
$ = 1375 cm^2$

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Question 32 Marks

In the figure given below, find the area of shaded region: (All measurements are in cm)

Answer

Area of shaded portion = Area of outer region − Area of unshaded region
$= 40 \times 40 − 32 \times 15$
$= 1600 − 480 cm^2$
$ = 1120 cm^2$

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Question 42 Marks

Find the area of the square whose perimeter is 56 cm.

Answer

$\begin{aligned} & \text { Perimeter of square }=56 \mathrm{~cm} \\ & \Rightarrow 4 \times \text { Side }=56 \mathrm{~cm} \\ & \Rightarrow \text { Side }=\frac{56}{4} \mathrm{~cm} \\ & \Rightarrow \text { Side }=14 \mathrm{~cm} \\ & \therefore \text { Area of square }=(\text { Side })^2=(14)^2 \\ & =14 \times 14 \mathrm{~cm}^2=196 \mathrm{~cm}^2\end{aligned}$

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Question 52 Marks

The diameter of every wheel of a car is 63 cm. How much distance will the car move during the 2000 revolutions of its wheel.

Answer

Diameter of car wheel $(\mathrm{d})=63 \mathrm{~cm}$
$ \therefore \text { Circumference }=\pi d=\frac{22}{7} \times 63=198 \mathrm{~cm} $
Distance covered in 2000 revolutions
$ \begin{aligned} & =2000 \times 198 \mathrm{~cm} \\ & =\frac{2000 \times 198}{100}=3960 \mathrm{~m}=3.96 \mathrm{~km} \end{aligned} $

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Question 62 Marks

A circular racing track has inner circumference 528 m and outer circumference 616 m. Find the width of the track.

Answer

Outer circumference $=616 \mathrm{~m}$
Radius $(R)=\frac{C}{2 \pi}=\frac{616 \times 7}{2 \times 22} \mathrm{~m}=98 \mathrm{~m}$
Inner circumference $=528 \mathrm{~m}$
$\therefore$ Inner radius $(r)=\frac{528 \times 7}{2 \times 22} \mathrm{~m}=84 \mathrm{~m}$
$\therefore$ Width of track $=R-r=98-84=14 m$

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Question 72 Marks

The diameter of a circle is 20 cm. Taking π = 3.14, find the circumference and its area.

Answer

Diameter of circle $(\mathrm{d})=20 \mathrm{~cm}$
$\therefore$ Circumference $=\mathrm{d} \pi=20 \times 3.14=62.8 \mathrm{~cm}$
Radius $(r)=\frac{d}{2}=10 \mathrm{~cm}$
$\therefore$ Area of a circle $=\pi r^2$
$=3.14 \times 10 \times 10=314 \mathrm{~cm}^2$

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Question 82 Marks

Find the base of a triangle whose area is $360 cm^2$ and height is 24 cm .

Answer

Area of triangle $=360 \mathrm{~cm}^2$
and height (h) $=24 \mathrm{~cm}$
$ \begin{aligned} & \therefore \text { Base }=\frac{\text { Area } \times 2}{\text { Height }} \\ & =\frac{360 \times 2}{24}=\frac{720}{24}=30 \mathrm{~cm} \end{aligned} $

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Question 92 Marks

Find the height of a triangle whose base is 18 cm and the area is $270 cm^2$.

Answer

Base of triangle $=18 \mathrm{~cm}$
Area of triangle $=270 \mathrm{~cm}^2$
$ \begin{aligned} & \therefore \text { Height }=\frac{\text { Area } \times 2}{\text { Base }} \\ & =\frac{270 \times 2}{18}=\frac{540}{18}=30 \mathrm{~cm} \end{aligned} $

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Question 102 Marks

Find the area of a triangle whose base is 30 cm and the height is 18 cm.

Answer

Base of triangle $=30 \mathrm{~cm}$
Height of triangle $=18 \mathrm{~cm}$
$\therefore$ Area $=\frac{1}{2}$ base $\times$ height
$=\frac{1}{2} \times 30 \times 18=270 \mathrm{~cm}^2$

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Question 112 Marks

The area of a rhombus is $84 cm^2$ and its perimeter is $56 \ cm$ . Find its height.

Answer

Area of rhombus $=84 \mathrm{~cm}^2$
Perimeter $=56 \mathrm{~cm}$
$\therefore$ Its side $=\frac{56}{4}=14 \mathrm{~cm}$
$\therefore$ Height $=\frac{\text { Area of rhombus }}{\text { Base }}=\frac{84}{14}=6 \mathrm{~cm}$

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Question 122 Marks

If P = perimeter of a rectangle, l= its length and b = its breadth find P, if l = 38 cm and b = 27 cm

Answer

Length (l) = 38 cm
Breadth (b) = 27 cm
Perimeter of a triangle = 2(l + b)
= 2(38 + 27)
= 2(65)
= 130 cm

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Question 132 Marks

Find the perimeter of a rectangle whose length = 22.5 m and breadth = 16 dm.

Answer

Length = 22.5 m
Breadth = 16 dm = 1.6 m
Perimeter of rectangle = 2(l + b)
= 2(22.5 + 1.6)
= 2(24.1) = 48.2 m

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Question 142 Marks

The diagonal of a square is $12 \sqrt{2}$ cm. Find its perimeter.

Answer

$\begin{aligned} & \text { Diagonal of square }=\text { Its side } \times \sqrt{2} \\ & \text { Side } \sqrt{2}=\sqrt{2} \sqrt{2} \\ & \text { i.e. side }=12 \mathrm{~cm} \\ & \text { Perimeter of a square }=4 \times \text { side } \\ & =4 \times 12=48 \mathrm{~cm}\end{aligned}$

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Question 152 Marks

The circumference of a circle is $440 \mathrm{~cm}$. Find its radius and diameter. (Take $\pi=\frac{22}{7}$ )

Answer

(i) Circumference of the circle $=440 \mathrm{~cm}$
$ \begin{aligned} & \text { Radius }=\frac{C}{2 \pi}=\frac{440 \times 7}{2 \times 22} \mathrm{~cm} \\ & =\frac{3088}{44} \mathrm{~cm} \end{aligned} $
(ii) Diameter $=2 \times$ radius
$ =2 \times 70=140 \mathrm{~cm} $

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