MCQ Questions for Class 7 Maths Perimeter And Area Quiz 7

The perimeter of circle is

$\displaystyle \pi$ cm, then the area

0%
$\displaystyle \frac{\pi^{2}}{4}$
0%
$\displaystyle \frac{\pi^{2}}{2}$
0%
$\displaystyle \frac{\pi}{2}$
0%
$\displaystyle \frac{\pi}{4}$

Explanation

Given, Circumference

of a circle of radius r

$= 2\pi r = \pi$

$r = \dfrac {1}{2}$

Area of a circle of radius r

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

Area of circle of radius $\dfrac {1}{2} = \pi { ( \dfrac {1}{2} )}^{ 2 } = \dfrac { \pi }{ 4 }$

The cost of fencing a circular field at the rate of Rs. 12 per meter is Rs.The field is to be ploughed at Rs. 2 per

$\mathrm {m^{2}}$ , then cost of ploughing is:

$\displaystyle \left ( \text{use }\pi =\frac{22}{7} \right )$

0%
$Rs. 3850$
0%
$Rs.1925$
0%
$Rs. 660$
0%
$Rs. 2585$

Explanation

Given that at a rate of $Rs.\ 12$ per meter, the cost of fencing the circular field is $Rs. 1320$

Hence, circumference of the field $= \dfrac {1320}{12} = 110 \ m$
We know that, circumference of a circle is $2\pi r$

$\therefore \ 2\pi r=110$

$\Rightarrow 2 \times \dfrac {22}{7} \times r = 110$

$\Rightarrow r = \dfrac {35}{2} \ m$

We also know that the area of a circle is $\pi r^2$

So, area of the field $= \dfrac {22}{7} \times \dfrac {35}{2} \times \dfrac {35}{2} = \dfrac {1925}{2} \ {m}^{2}$

Given that the cost of ploughing one $m^2$ area is $Rs. \ 2$ .

Hence, cost of ploughing the field at $Rs. \ 2$ per ${m}^{2} = \dfrac {1925}{2} \times 2 = Rs. \ 1925$ .

Hence, option B is correct.

Four circular cardboard pieces, each of radius 7 cm are placed in such a way that each piece touches the two other pieces. The area of the space enclosed by the four pieces is:

0%
$\displaystyle 21{ cm }^{ 2 }$
0%
$\displaystyle 42{ cm }^{ 2 }$
0%
$\displaystyle 84{ cm }^{ 2 }$
0%
$\displaystyle 168{ cm }^{ 2 }$

Explanation

Area of enclosed space = Area of square - 4 x Area of quadrant

$= \quad side \times \quad side - 4 \times \dfrac {πr^2}{4}$

$\displaystyle ={ \left( 14 \right) }^{ 2 }-4$

$\displaystyle \left( \frac { 1 }{ 4 } \times \frac { 22 }{ 7 } \times { 7 }^{ 2 } \right)$

$\displaystyle =196-154$

$\displaystyle =42{ cm }^{ 2 }$

1170097_352235_ans_846a24fab1de400e8909521c9e989af0.PNG

One side of a parallelogram is

$8 \ cm$ . If the corresponding altitude is

$6\ cm$ , then its area is given by

0%
$24 \ cm^2$
0%
$36 \ cm^2$
0%
$40 \ cm^2$
0%
$48 \ cm^2$

Explanation

As we know,

Area of Parallelogram

$=Base \times Height$

Here,

$Base=8\ cm, Height=6\ cm$

So, area of the Parallelogram

$=8\times 6\ cm^2$

$=48\ cm^2$ .

The perimeter of sheet of paper in the shape of a quadrant of a circle is 25 cm , then area of the paper is

0%
$\displaystyle \frac{77}{2}cm^{2}$
0%
$\displaystyle \frac{77}{2}\left ( \frac{25}{18} \right )^{2}cm^{2}$
0%
$\displaystyle \frac{231}{2}cm^{2}$
0%
None

Explanation

Let the radius of the quadrant be $r$

Given,
Perimeter of the square paper $= \dfrac {\pi r}{2} + 2r =25 cm$

$\Rightarrow \dfrac {22}{7} \times \dfrac {r}{2} + 2r = 25$

$\Rightarrow \dfrac {22r + 28r}{14} = 25$

$\Rightarrow \dfrac {50r}{14} = 25$

$\Rightarrow r = 7 cm$

Area of the square paper quadrant of radius $r$

$= \dfrac {\pi {r}^{2}}{4} \\ = \dfrac {\dfrac {22}{7} \times 7 \times 7}{4} = \dfrac {77}{2} {cm}^{2}$

A race track is in the form of a ring whose inner circumference is

$440 \ \mathrm{m}$ & outer circumference is

$506 \ \mathrm{m}$ The width of the track is:

0%
$7 \ \mathrm{m}$
0%
$21 \ \mathrm{m}$
0%
$10.5 \ \mathrm{m}$
0%
$14 \ \mathrm{m}$

Explanation

Let the radius of the outer circle be R and inner circle be r.

Outer circumference

$= 506 \ \mathrm{m}$

$\implies 2 \times \frac {22}{7} \times R = 506 \ \mathrm{m}$

$R = 80.5\ \mathrm{ m}$

Inner circumference

$= 440 m$

$\implies 2 \times \frac {22}{7} \times r = 440 \ \mathrm{m}$

$r = 70 \ \mathrm{ m}$

Hence, width of the track

$= (80.5 - 70)\ \mathrm{m} = 10.5 \ \mathrm{ m}$

The area ofthe shaded portion in the given figure is

0%
7.5 $\pi$ sq. units
0%
6.5 $\pi$ sq. units
0%
5.5 $\pi$ sq. units
0%
4.5 $\pi$ sq. units

Explanation

Given that:

Radius of bigger semi circle

$R=5$ units

Area of Bigger semi circle

$A=\dfrac{\pi}{2} R^2\Rightarrow 12.5\ {\pi}$ unit

Radius of smaller semi circle

$r=5-1\Rightarrow 4$ units

Area of Bigger semi circle

$a=\dfrac{\pi}{2} r^2\Rightarrow 8\pi$ unit

Hence,

Area of shaded portion

$=A-a\Rightarrow 12.5\ \pi-8\ \pi\Rightarrow 4.5\ \pi$ unit

$^2$

A bicycle wheel has diameter 1m. If the bicycle travels one kilometer, then the number of revolutions the wheel make is.

0%
$\dfrac {1}{\pi }$
0%
$\dfrac {100}{\pi }$
0%
$\dfrac {500}{\pi }$
0%
$\dfrac {1000}{\pi }$

Explanation

Let the number of revolution of the wheel is

$n$ .

Since,

$\text{Radius} = \dfrac{1}{2}$
Then,

$n\ \times$ Circumference of wheel = Distance travelled by bicycle

$n \times 2\pi \times \frac {1}{2}=1\ \text{km}$

$n \times \pi =1000\ \text{m}$

$n=\frac {1000}{\pi }\ \text{m}$

A rope by which a cow is tethered is increased from 16 m to 23 m. How much additional ground can the cow graze now?

0%
800 $m^2$
0%
8580 $m^2$
0%
858 $m^2$
0%
858 $cm^2$

Explanation

Given that, a rope by which a cow is tethered is increased from

$16\ m$ to

$23\ m$

To find out: The additional area that the cow can graze now

We know that area of a circle of radius

$r =πr^2$

$\therefore \$ Additional ground area

$= \pi ((23)^2 - (16)^2)$

$\Rightarrow \dfrac{22}{7}(529-256)\quad \quad \left[\because \ \pi=\dfrac{22}{7}\right]$

$\Rightarrow \dfrac{22}{7}(273)$

$= 858 \ m^2$

Hence, the cow can now graze an additional area of

$858 \ m^2$ .

1971151_373597_ans_13b523704f47402fae273abefd36bcab.png

If the side of square is

$21$ cm, then the circumference of the circle in the figure is:

0%
$33$ cm
0%
$1386$ cm
0%
$66$ cm
0%
$132$ cm

Explanation

From the figure, we can say that,

Side of square

$=$ diameter of the circle

$\therefore$ radius

$r=\dfrac{21}{2}=10.5cm$

Now Circumference of the circle

$=2\pi r$

$=2\times \dfrac{22}{7}\times \dfrac{21}{2}=66cm$

∴radius=212=10.5cm

1292331_375721_ans_db60334d1a404371bf74069f37a21d4a.png

If the difference between the circumference and radius of a circle is 37 cm, then the area of the circle is

0%
$111 cm^2$
0%
$148 cm^2$
0%
$259 cm^2$
0%
$154 cm^2$

Explanation

Let, the radius be $r cm$

Circumference will be,

$2πr=2\times 3.14r=6.28r$

According to question, $6.28r-1.r=37$

$\Rightarrow 5.28r=37$

$\Rightarrow r=\dfrac{37}{5.28}$

$\Rightarrow r=$ approx. $7 cm$

Area $=\pi {{r}^{2}}=\dfrac{22}{7} \times 7\times 7=154c{{m}^{2}}$

Hence, this is the answer.

If the circumference of a circle is

$88$ cm, then the area of the circle (in sq.cm) is:

$\displaystyle \left ( \text{take} \ \pi = \frac{22}{7} \right )$

0%
$\displaystyle 196\pi$
0%
$\displaystyle 92\pi$
0%
$\displaystyle 64\pi$
0%
$\displaystyle 48\pi$

Explanation

$Circumference =\displaystyle 2\pi r=88\Rightarrow r=\frac{44\times 7}{22}=14\ cm$

$\displaystyle \therefore area=\pi r^{2}=\pi \times 14\times 14$

$cm^{2}=196\pi \ cm^{2}$

If the circumference of a circle is reduced by 50%, then the area will be reduced by

0%
50%
0%
25%
0%
75%
0%
12.5%

Explanation

Let the original radius be

$r$ .

So, the area of circle

$=\pi r^2$

$....... (1)$

Since, the circumference of the circle is reduced by

$50\%$ .

It means that the radius of the circle is also reduced by

$50\%$ .

Then,

The new radius

$=0.5r$

Therefore, the new area

$=\pi (0.5r)^2$

$=0.25\pi r^2$

Therefore, the required

$\%$

$=\dfrac{\pi r^2-0.25\pi r^2}{\pi r^2}\times 100$

$=\dfrac{0.75\pi r^2}{\pi r^2}\times 100$

$=75\%$

Hence, this is the answer.

One side of a parallelogram is 8 cm If the corresponding altitude is 6 cm then its area is given by

0%
24 $\displaystyle cm ^{2}$
0%
36 $\displaystyle cm ^{2}$
0%
40 $\displaystyle cm ^{2}$
0%
48 $\displaystyle cm ^{2}$

Explanation

Given One side of parallelogram is 8 cm and altitude is 6 cm

Here base=8 cm and height(altitude)=6 cm

We know that area of parallelogram=

$base \times hieght$

Here base=8 cm and height(altitude)=6 cm

Then area of parallelogram=

$8\times 6=48 cm^{2}$

If the radius of the circle is increased by

$100%$ , then the area is increased by

0%
$100$ %
0%
$200$ %
0%
$300$ %
0%
$400$ %

The area of the shaded region in the following figure is

0%
$\displaystyle 2000$ $\displaystyle m^{2}$
0%
$\displaystyle 90$ $\displaystyle m^{2}$
0%
$\displaystyle 45$ $\displaystyle m^{2}$
0%
$\displaystyle 89$ $\displaystyle m^{2}$

Explanation

$\textbf{Step - 1 : Find the area part by part}$

$\text{Given figure is,}$

$\text{As we know that, the area of a rectangle is = Length}\times\text{Breadth}$

$\text{Area of the shaded rectangle breadth wise}=1\times50=50\text{ m}^2$

$\text{Area of the shaded rectangle length wise}=1\times40=40\text{ m}^2$

$\text{The area of square }{=(\text{Side})^2}$

$\text{Area of the square formed at the extreme center position}=1^2=1\text{ m}^2$

$\textbf{Step - 2 : Find the area of the shaded region,}$

$\text{Area of the shaded region}=50+40-1=89\text{ m}^2$

$\textbf{Hence, option D}\textbf{ is the correct choice.}$

Diya has a photo frame, which is in the shape of square. If she wants to know the space occupied by it, then she needs to find its ____.

0%
perimeter
0%
area
0%
can't say anything
0%
none of these

There is gab between two rooms in Riya'house in a triangular shape of area

$512$

$cm^2$ . Find its area in

$mm^2$

0%
$51.2$ $mm^2$
0%
$5120$ $mm^2$
0%
$51200$ $mm^2$
0%
$512000$ $mm^2$

The side of a square is

$2\ cm$ and semicircles are constructed on each side of the square, then the area of the whole figure is

0%
$(4 + 8\pi)\ cm^2$
0%
$(4 + 4 \pi)\ cm^2$
0%
$4 \pi\ cm^2$
0%
$8 \pi\ cm^2$

Explanation

We know that area of semi circle of radius

$r$ is

$\dfrac{πr^2}{2}$ and area of square od side

$a$ is

$a^2$

Given that:

Side of square

$=2\ cm$

Total area

$=$ Area of square

$+\ 4\times$ Area of semi circle

Total area

$=2^2$

$+\ 4\times\dfrac{\pi\times 2^2}{2}$

$=(4+8\pi)\ cm^2$

The perimeter of the following shaded portion of the figure is

0%
40 m
0%
40.07 m
0%
35.72m
0%
35 m

Explanation

Then length and breadth of rectangle is 17 m and 4 m respectively.

Then perimeter of rectangle=

$2( L+B)=2(17+4)=42$

The radius of one fourth circle given in figure is 1 m

Then perimeter of one fourth circle=

$\dfrac{1}{4}\times (2\pi r)=\dfrac{1}{4}\left ( 2\pi (1) \right )=\dfrac{1}{2}\pi$

Then perimeter of four one fourth circle=

$4\times \dfrac{1}{2}\pi =2\times 3.14=6.28 m$

Then perimeter of shaded figure

$= 42-6.28=35.72 m$

If the radius of a circle is 35 m, then the circumference of a circle is:

0%
$110\ m$
0%
$220\ m$
0%
$70\ m$
0%
None of these

Explanation

Given that, the radius of a circle is,

$r = 35\ m$

To find out: The circumference of the circle.

We know that, the circumference of a circle is

$\displaystyle 2 \pi r$

$\therefore \$ Circumference of the given circle

$=2 \times \dfrac{22}{7} \times 35\quad \quad \left[\because \ \pi=\dfrac{22}{7}\right]$

$\therefore \$ Circumference

$= 220\ m$

Hence, if the radius of a circle is

$35\ m$ , the circumference of the circle is

$220\ m$ .

Find the area of the parallelogram whose base is 16 cm and height 0.4 m ?

0%
440 $cm^2$
0%
740 $cm^2$
0%
640 $cm^2$
0%
None of these

Explanation

Base = 16 cm
Height = 0.4 m
= 0.4

$\times$ 100 = 40 cm
Area of parallelogram = b

$\times$ h
= 16

$\times$ 40
= 640

$cm^2$

What will be area of a parallelogram with base 6 cm and altitude 3.5 cm?

0%
28 square cm
0%
20 square cm
0%
21 square cm
0%
None of these

Explanation

Area of parallelogram = Base

$\times$ Altitude
= 6 cm

$\times$ 3.5 cm
= 21 square cm

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

0%
$\displaystyle 42\pi$
0%
$\displaystyle 49\pi$
0%
$\displaystyle 54\pi$
0%
$\displaystyle 60\pi$

Explanation

Area of a circle

$= \pi {r}^{2}$ , where r is the radius of the circle
Since

$r = 7$
So, area of the given circle

$= \pi \times 7 \times 7 = 49 \pi$

The height of a parallelogram of area

$\displaystyle 350 cm^{2}$ and base 25 cm is

0%
12 cm
0%
13 cm
0%
14 cm
0%
15 cm

Explanation

Given the area of parallelogram is 350 sq cm and base is 25 cm

Then area of parallelogram if height h and base 25

$\therefore 350=base \times height\Rightarrow 25\times h=350\Rightarrow h=14$ cm

A well of diameter

$150\text{ cm}$ has a

$30\text{ cm}$ wide parapet running around it. Find the area of the parapet.

0%
$16971.428$ $\text{cm}^2$
0%
$1697.142$ $\text{cm}^2$
0%
$169.714$ $\text{cm}^2$
0%
$16971.42$ $\text{cm}^2$

Explanation

Diameter of well

$= 150\text{ cm}$
Radius of well =

$\displaystyle \frac{150}{2} = 75 \text{ cm}$
Width of parapet

$= 30 \text{ cm}$
Now, parapet of the well forms two circles.
One is the inner circle and the second is the outer circle.
Radius of inner circle

$(r) = 75 \text{ cm}$
Radius of outer circle

$R =$ inner radius

$+$ width of parapet

$= 75 + 30 = 105 \text{ cm}$
Area of parapet

$=$ Area of outer circle

$-$ Area of inner circle

$=\pi R^2 - \pi r^2 = \pi (R^2 - r^2)$

$=\displaystyle \frac{22}{7} [(105)^2 - (75)^2]$

$=\displaystyle \frac{22}{7} \times 5400$

$=\displaystyle \frac{118800}{7}$

$= 16971.428 \text{ cm}^2$

If a square of area

$3$ square units is inscribed in a circle, then the area of the circle is

0%
$\displaystyle \frac{3}{2}\pi\ sq.cm$
0%
$\displaystyle 9\pi ^{2}\ sq.cm$
0%
$\displaystyle 6\pi\ sq.cm$
0%
$\displaystyle \sqrt{3}\pi \ sq.cm$

Explanation

Given, area of a square

$=3$ sq.cm

Let

$a$ be side of square.

$\Rightarrow$

$a^2=3$

$\Rightarrow$

$a=\sqrt{3}$ cm

Diagonal

$=a\sqrt{2}$

$=\sqrt{3}\times \sqrt{2}$ cm

$=\sqrt{6}$ cm

Diameter of circle

$=$ diagonal of square

$=\sqrt{6}$ cm

Radius

$=\dfrac{\sqrt{6}}{2}\ cm$

$\Rightarrow r=\dfrac{\sqrt{3}}{\sqrt2}\ cm$

$\therefore$ Area of the circle

$=\pi r^2$

$=\pi\times \left(\dfrac{\sqrt{3}}{\sqrt2}\right)^2\ cm^2$

$=\dfrac{3}{2}\pi\ cm^2$

Hence, the area of the circle is

$\dfrac{3}{2}\pi\ cm^2$

1354221_427267_ans_8a50fc1d29874d2baa782fb01da16698.png

If a circle is inscribed in a square of area

$4$ sq units, then the area of the circle is

0%
$\displaystyle \pi$
0%
$\displaystyle \frac{\pi }{2}$
0%
` $\displaystyle \frac{\pi }{4}$
0%
$\displaystyle \frac{3\pi }{4}$

Explanation

$\Rightarrow$ Area of a square

$=4$ square units.

Let

$a$ be side of square.

$\Rightarrow$

$a^2=4$

$\Rightarrow$

$a=2$ units.

Diameter of circle and side of square are equal.

$\therefore$

$r=\dfrac{2}{2}=1$ unit.

$\Rightarrow$ Area of a circle inscribed in a square

$=\pi r^2$

$=\pi\times (1)^2$

$=\pi$

1354208_427266_ans_243f933103404da9acfff70a6357f943.png

The diameter of a wheel is

$98\ cm$ The number of revolutions it will have to make to cover a distance of

$1540\ m$ is

0%
500
0%
600
0%
700
0%
800

Explanation

Given the diameter of wheel is

$98\ cm$

Then radius of wheel =

$\dfrac{98}{2}=49\ cm$

Then circumference of wheel

$=2\pi r$

$=2\times \dfrac{22}{7}\times 49=308\ cm$

the number of revolutions to be made to cover a distance of

$1540\ m=\dfrac{1540\times 100\ cm}{308\ cm}$

$=\dfrac{154000}{308}=500$

A parallelogram has an area of

$125$

$m^{2}$ and a height of

$5\ m.$ Find the base.

0%
$250$ m
0%
$25$ m
0%
$270$ m
0%
$28$ m

Explanation

Area of a parallelogram

$=base\times height$ . Let the base be 'b'.
Hence

$125m^{2}=5b$ or

$b=25m$
Therefore

$base=25m$ .

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

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Practice Class 7 Maths Quiz Questions and Answers

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

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Practice Class 7 Maths Quiz Questions and Answers

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