MCQ Questions for Class 7 Maths Perimeter And Area Quiz 2

The radius of a circle whose area is equal to the sum of the areas of two circles of radii 5 cm and 12 cm is

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13 cm
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14 cm
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15 cm
0%
17 cm

Explanation

We know that area of a circle of radius

$r = πr ^2$

By given condition,

${ A }_{ 3 }={ A }_{ 1 }+{ A }_{ 2 }$ .....(1)

Since,

$r_1 = 5$ &

$r_2= 12$

So, by (1), we have

${ \pi r }^{ 2 }={ \pi 5 }^{ 2 }+{ \pi 12 }^{ 2 }\\\\ { r }^{ 2 }={ 5 }^{ 2 }+{ 12 }^{ 2 }\\\\ r=\sqrt { 25+144 } =\sqrt { 169 } =13 \text{ cm}$

The radius of a circle is increased by 1 cm. Then the ratio of new circumference to the new diameter is

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$\pi :3$
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$\pi :2$
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$\pi :1$
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$\pi :\frac { 1 }{ 2 }$

Explanation

New radius =

$(r+1)\ \mathrm{cm}$

Ratio =

$2\pi (r+1):2(r+1)=\pi :1$

If the perimeter and area of a circle are numerically equal then the radius of the circle is

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$6$ units
0%
$\displaystyle \pi$ units
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$4$ units
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$2$ units

Explanation

$\textbf{Step 1: Substitute area of circle and perimeter of circle formula in the given condition}$

$\text{Area of circle}=\pi r^2$

$\text{Perimeter of circle}=2\pi r$

$\text{Given that area and perimeter are numerically equal}$

$\implies \pi r^2=2\pi r$

$\implies r=2$

$\textbf{Thus, the required radius of the circle is 2 units}$

From a square metal sheet of side

$28\;cm$ , a circular sheet is cut off. Find the radius of the largest possible circular sheet that can be cut. Also find the area of the remaining sheet.

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$14\;cm,\;148\;cm^2$ .
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$14\;cm,\;168\;cm^2$ .
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$12\;cm,\;168\;cm^2$ .
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$14\;cm,\;164\;cm^2$ .

Explanation

Area of square sheet

$=$

${ (28) }^{ 2 }=784{ cm }^{ 2 }$

The largest circle of diameter equals to the side of square can be cut off from the square sheet.

$\therefore$ Radius of circular sheet

$=$

$\dfrac { 28 }{ 2 } =14cm$

Area of remaining sheet

$=$

$784-\pi { r }^{ 2 }=784-\dfrac { 22 }{ 7 } \times 14\times 14$

$=$

$784 - 616$ =

$168$

${ cm }^{ 2 }$

The difference in the area of a square of perimeter

$88$ m and a circle with same circumference is

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166 $\displaystyle cm^{2}$
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122 $\displaystyle cm^{2}$
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133 $\displaystyle cm^{2}$
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132 $\displaystyle cm^{2}$

Explanation

Perimeter of

$sq=88 cm$

$\displaystyle \therefore$ side of

$sq = 22 cm$

$\displaystyle \Rightarrow$ area of sq

$\displaystyle =484 cm^{2}$

$\displaystyle C=2\pi r=88\Rightarrow r=14$

$\displaystyle \therefore Area=\dfrac{22}{7}\times 14\times 14=616cm^{2}$
Difference in the areas

$\displaystyle =616-484=132 cm^{2}$

The ratio of the outer and inner circumferences of a circular path is

$23:22$ , If the path is

$5\ m$ wide, the radius of the inner circle is:

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$55\ m$
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$110\ m$
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$220\ m$
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$230\ m$

Explanation

Let R and r be the outer and inner radii of the circular path.

Given that,

$\frac{2\pi R}{2\pi r}=\frac{23}{22}$

$=>\frac{R}{r}=\frac{23}{22}$

Let

$R=23x$ and

$r=22x$

It is given that the width of the path is

$5$ m wide

$\therefore R-r=5$ m

$=>23x-22x=5$

$=>x=5$

$\therefore$ the inner radius of the circle is

$=22\times 5$

$=110$ m

The radius of a circle whose area is equal to the sum of the areas of two circles of radii 3 cm and 4 cm is

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5 cm
0%
5.5 cm
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5.8 cm
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6 cm

Explanation

The area of circle

$C_1$ whose radius is

$4\ cm =\pi (4)^{2}$

$=16\pi$ sq cm

And the area of circle

$C_2$ whose radius is

$3\ cm = \pi (3)^{2}$

$=9\pi$ sq cm

Given: Area of the new circle is equal to the sum of areas of circles

$C_1$ and

$C_2$

Let the radius of the new circle be

$R$ cm

Area of big circle

$=16\pi +9\pi =25\pi$ sq cm

$\Rightarrow$

$\pi R^{2}=25\pi$

$\Rightarrow$

$R^{2}=25$

$\Rightarrow$

$R=5$ cm

So the radius of the new circle

$=5$ cm

Area of a square 625 sq m. Then the measure of its side is

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$15$ m
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$25$ m
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$20$ m
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$24$ m

Explanation

We know that,

Area

$=side\times side$

$s \displaystyle \times s \displaystyle= 625 \displaystyle m^{2}$

$s \displaystyle \times s \displaystyle = 25 \displaystyle \times 25$

$\displaystyle=$ 625

$\displaystyle m^{2}$

$s \displaystyle = 25\ m$

The radius of a circle whose area is equal to the sum of the areas of two circles of radii are 5 cm and 12 cm is

0%
13 cm
0%
14 cm
0%
15 cm
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none

Explanation

Given two circle radii are

$5 cm$ and

$12 cm$

Then,

Area of circle of radius 5 cm

$=$

$\pi r^{2}=\pi (5)^{2}=25\pi$ sq cm

Area of circle of radius 12 cm

$=$

$\pi r^{2}=\pi (12)^{2}=144\pi$ sq cm

So area of circle whose area is equal to sum of areas of two circles

$=$

$25\pi +144\pi =169\pi$ sq cm

Let the radius of the bigger circle be

$=$ R cm

$\therefore \pi R^{2}=169\pi$

$\Rightarrow R^{2}=169$

$\Rightarrow R=13 cm$

Find the perimeter of a circle whose radius is 7 cm (in cm)

0%
$\displaystyle 12\pi$
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$\displaystyle 14\pi$
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$\displaystyle 16\pi$
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$\displaystyle 10\pi$

Explanation

Perimeter of a circle

$= 2\pi r$ , where r is the radius of the circle

So, perimeter of the given circle

$= 2 \times \pi \times 7 = 14 \pi$

If the area of the circle be

$\displaystyle 154 \text{ cm}^{2},$ then its radius is equal to:

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$12\text{ cm}$
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$8\text{ cm}$
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$7\text{ cm}$
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None of these

Explanation

Area of the circle

$=154\displaystyle \text{ cm}^{2}$

$\because$ Area

$=\displaystyle \pi r^{2}$

$\therefore \pi r^2=154$

$\Rightarrow$

$\displaystyle r^{2}=\frac{154\times 7}{22}$

$\Rightarrow r^2 = \displaystyle 7\times 7$

$\displaystyle \Rightarrow r=7\text{ cm}$

If the circumference of a circle be

$8.8 \text{ m}$ then its radius is equal to -

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$1.60 \text{ m}$ (approx)
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$1.61 \text{ m}$ (approx)
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$1.40 \text{ m}$ (approx)
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None of these

Explanation

Circumference of a circle,

$\displaystyle C=2\pi r$
or

$\displaystyle r=\frac{C}{2\pi }$

$=\displaystyle \frac{8.8\times 7}{2\times 22}$

$= 1.4 \text{ m}$

If the radius of a circle be

$r$

$cm$ , then its area will be equal to-

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$\displaystyle 2\pi r^{2}cm^{2}$
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$\displaystyle \pi r^{2}cm^2$
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$\displaystyle 2\pi rcm^{2}$
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None of these

Explanation

If radius is

$r$

$cm$ , then

$area=\pi r^2\,\, cm^2$ .

Option B is correct.

What is the circumference of a circle whose radius is 8 cm?

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$\displaystyle 8 \pi$
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$\displaystyle 16\pi$
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$\displaystyle 61 \pi$
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$\displaystyle 24 \pi$

Explanation

Circumference=

$\displaystyle 2\pi r$
=

$\displaystyle 2\times \pi \times 8$
=16

$\displaystyle \pi$ cm

The radius of a circle is increased by

$5$ units. What is ratio of the new circumference and the new diameter?

0%
$x-5$
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$\pi +5$
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$\pi$
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$\pi -1$

Explanation

Let the radius of the circle be

$r$ .

$\therefore$ New circumference

$=2\pi(r+5)$
and new diameter

$=2(r+5)$

$\therefore$ Ratio

$=\cfrac{2\pi(r+5)}{2(r+5)}$

$=\pi$

If circumference of a circle is

$\displaystyle 3\pi$ , then its area is

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$\displaystyle \frac{7\pi }{2}$
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$\displaystyle 9\pi ^{2}$
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$\displaystyle 4\pi ^{2}$
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$\displaystyle \frac{9\pi }{4}$

Explanation

$\Rightarrow$ Circumference of a circle

$=3\pi$

$\Rightarrow$

$2\pi r=3\pi$

$\Rightarrow$

$2r=3$

$\Rightarrow$

$r=\dfrac{3}{2}$

$\Rightarrow$ Area of a circle

$=\pi r^2$

$=\pi\times \left(\dfrac{3}{2}\right)^2$

$=\dfrac{9}{4}\pi$

$\displaystyle \frac{1}{5}\, of\,10 km=$ _____m

0%
2
0%
200
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20
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2000

Explanation

$\displaystyle \frac{1}{5}$ of 10 km

$\displaystyle \frac{1}{5}\times10 \ km$ = 2 km=2000 m

A wall is made up of square bricks of unit length. Then its area is _____

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1 sq unit
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3 sq units
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20 sq units
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24 sq units

Explanation

As 24 square bricks are required for the wall
So Area = 24

$\times$ area of one brick =

$24 \times 1$ = 24 sq units

A figure is formed by putting two squares one on the other as shown below. If the length of each side of the two squares is

$8$ cm, then the perimeter of the formed figure is

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$56$ cm
0%
$64$ cm
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$32$ cm
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$48$ cm

Explanation

New dimensions are

$b = 16$ &

$l=8$

$\displaystyle \therefore$ Perimeter of the figure formed

$\displaystyle = 16+8+16+8= \displaystyle 48$ cm

The total boundary length of circle is called

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area
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volume
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circumference
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diameter

Area of a rectangle is

$120$

$m^2$ and the breadth is

$5 m$ . Then its length is

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$204 m$
0%
$24 m$
0%
$28 m$
0%
$26 m$

Explanation

We know, area of rectangle =

$l \times b$

Since, area

$= 120$

$\therefore$

$l \times b = 120$

$\Rightarrow$ length =

$l =\displaystyle \frac{120}{5}=24 m$

Circumference of a circle is equal to

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$\displaystyle \pi r$
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$2\displaystyle \pi r$
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$\displaystyle \frac{\pi r}{2}$
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$\displaystyle 2+\frac{\pi r}{2}r$

Explanation

The circumference of a circle whose radius is equal to

$r$ is given by

$2\pi r.$

The circumference of a circle is called

0%
chord
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segment
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boundary
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None

Explanation

Perimeter of circle = Circumference ---eq.1

Perimeter = Boundary ---eq.2

from eq.1 & eq.2

Circumference = Boundary

$\therefore$ The circumference of a circle is called boundary.

If the radius of a circle is

$\dfrac{7}{\sqrt{\pi}}$ , what is the area of the circle (in

$cm^2$ )?

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$154$
0%
$\dfrac{49}{\pi}$
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$22$
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$49$

Explanation

Given: Radius of circle

$(r)=\dfrac{7}{\sqrt{\pi}}$

Area of the circle

$=\pi (r)^2$

$=\pi\times (\dfrac{7}{\sqrt{\pi}})^2$

$=\pi\times \dfrac{49}{\pi}$

$=49 cm^2$

Find the area of the triangle shown in figure.

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$20$
0%
$30$
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$35$
0%
$40$

Explanation

We know, area of a triangle

$A=\dfrac{1}{2}\times b\times h$ ,

where

$b$ is the length of the base and

$h$ is the height.

In the figure, the base of the right-angled triangle is 8 and the height is 5.

$\therefore$ Area

$=\dfrac{1}{2}\times 8\times 5$

$=20$

The ______ of a circle is called the circumference.

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area
0%
volume
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perimeter
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radius

The distance, once around the circle is called

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diameter
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center
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circumference
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chord

The circumference of the circle is the boundary of t he

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perimeter
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radius
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diameter
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circle

The circumference of the circle is calculated by the formula

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$4\pi r$
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$2\pi r^2$
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$3\pi r$
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none of these

Explanation

The circumference of the circle is calculated by the formula

$2\pi r$ , where

$r$ is radius of the circle.

The dotted line represents the

0%
center
0%
diameter
0%
circumference
0%
chord

CBSE Questions for Class 7 Maths Perimeter And Area Quiz 2 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths Perimeter And Area Quiz 2 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

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