MCQ Questions for Class 7 Maths Perimeter And Area Quiz 3

In which figure circumference of the circle is shown in green?

0%
figure 2
0%
figure 3
0%
figure 1
0%
figure 4

The area of triangle is equal to the area of a rectangle whose length and breadth are

$20\ cm$ and

$15\ cm$ respectively. Calculate the height of the triangle if its base measures

$30\ cm$ .

0%
$20\ cm$ .
0%
$30\ cm$ .
0%
$40\ cm$ .
0%
None of these

Explanation

The length of rectangle

$= 20cm$

The breadth of rectangle

$= 15cm$

The area of rectangle

$= l\times b$

$\Rightarrow$

$= 20cm\times 15cm=300cm^2$

The area of triangle

$= 300cm^2$

The base of triangle is

$= 30cm$

By given,

$\dfrac 12\times 30\times h=300\\\implies h=20cm$

The height of triangle

$=20cm$ .

$80$ students of the same height stand with both hands stretched all along the sides of a rectangular garden, each student covering a length of

$1.75$ m. Then the perimeter of the garden is

0%
$1400$ m
0%
$140$ m
0%
$14$ m
0%
$1400$ km

Explanation

It is given that

$80$ students of same height stand with both hands stretched all along the sides of a rectangular garden, each student covering a length of

$1.7$ m.

Perimeter of the garden

$=$ Length covered by

$80$ students.

$\Rightarrow 80\times$ Length covered by single student.

$\Rightarrow 80\times 1.75=140$

So, option B is correct answer.

Which figure has an area of

$20$ sq. units and a perimeter of

$18$ units?

Explanation

Side of one square

$=1$ unit
Area of one square

$=1\times 1=1$ sq. unit
(A) Area of

$18$ squares

$=18$ sq. units
Perimeter

$=2(6+3)=18$ units
(B) Area of

$18$ squares

$=18$ sq. units
Perimeter

$=2(9+2)=22$ units
(C) Area of

$16$ squares

$=16$ sq. units
Perimeter

$=2(8+2)=20$ units
(D) Area of

$20$ squares

$=20$ sq. units
Perimeter

$=2(5+4)=18$ units.

The given figure is made up of two rectangles. Find its total area.

0%
$144cm^2$
0%
$216cm^2$
0%
$240cm^2$
0%
$300cm^2$

Explanation

Area of the bigger rectangle

$=(18\times 8)cm^2$

$=144cm^2$
Area of the smaller rectangle

$=((30-18)\times (8-6)) = (12\times 6)cm^2$

$=72cm^2$

$\therefore$ Total area = area of smaller rectangle + area of bigger rectangle

$=(72+144)cm^2=216cm^2$ .

If the outer and inner radii of a ring are

$10$ cm and

$8$ cm, then its area is nearly

0%
$113.443$ sq. cm
0%
$113.343$ sq. cm
0%
$113.243$ sq. cm
0%
$113.143$ sq. cm

Explanation

We know that area of a circle is

$πr^2$ , where

$r$ be a radius of the circle.

Let us assume

$R = 100$

$cm$ &

$r = 8$

$cm$

$\therefore$ Required area

$= \pi (R^2 - r^2 ) = \pi (100^{2} - 8^{2}) = 36π = 113.143\ cm^{2}$ .

A circular park, 42 m in diameter, has a path 3.5 m wide running around it on the outside. Find the cost of gravelling the path at Rs. 4 per

$m^2$ .

0%
$Rs. 2002$
0%
$Rs. 1652$
0%
$Rs. 1672$
0%
$Rs. 2048$

Explanation

Internal Radius = 21 m, External Radius

$= 21 + 3.5 = 24.5$ m
Area of path =

$\pi(24.5)^2- \pi(21)^2$

$\Rightarrow\pi[(24.5)^2-(21)^2]$

$= 500.5$
Cost of travelling

$= 500.5 \times 4 = Rs.2002$
1109093_796882_ans_e3d3c330d4604c5d89d5a4eaf373209e.png

Circumference

$C$ of the circular ring is:
where

$C_{outer}$ and

$C_{inner}$ denotes the circumference of outer and inner circles respectively.

0%
$C_{outer}+C_{inner}$
0%
$C_{outer}-C_{inner}$
0%
$C_{outer}C_{inner}$
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$2(C_{outer}+C_{inner})$

Explanation

Total circumference of the edges of the ring is sum of both inner and outer circumferences.

$C=C_{outer}+C_{inner}$

Perimeter of a triangle is the sum of the lengths of all the _____ sides.

0%
$2$
0%
$4$
0%
$5$
0%
none of these

Explanation

$\Rightarrow$ Perimeter of a triangle is the sum of the lengths of all the

$3$ sides.

$\Rightarrow$ In figure, ABC is a triangle.

$\Rightarrow$ Perimeter of

$\triangle ABC=AB+BC+CA$

$\therefore$ Perimeter of a triangle is the sum of the lengths of all the three sides.

The shaded area enclosed between the two rings is called _____.

0%
Circle
0%
Spherical shell
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Circular ring
0%
Sphere

Find the area of the circular park of whose radius is

$4.5 m$ .

0%
$33.85\,{{m}^{2}}$
0%
$53.55\,{{m}^{2}}$
0%
$43.85\,{{m}^{2}}$
0%
$63.585\,{{m}^{2}}$

Explanation

Given that,

Radius $r=4.5\,m.$

Then, find area of circular park

We know that,

$Area\,of\,circle\,=\,\pi {{r}^{2}}=3.14\times {{\left( 4.5 \right)}^{2}}$

$=3.14\times 4.5\times 4.5=63.585\,{{m}^{2}}$

Hence, this is the answer.

A circular ring is 5 cm wide. The difference of outer and inner radius is

0%
$5\:cm$
0%
$2.5\:cm$
0%
$10\:cm$
0%
none of these

Explanation

Since, we are given width of a circular ring which is nothing but a difference of radii the outer and inner circle.

Hence,

$R - r = 5 cm$ .

1237441_1195537_ans_a4ffd66e9e6840a3b2dc8f7f72303f6d.png

If the circumference of a circle is

$352$ metres, then its area (approx)in

$\text{ m }^{ 2 }$ is

0%
$9859$
0%
$8956$
0%
$6589$
0%
$5986$

Find the circumference of the circle with the following radius : 10 cm

0%
$20 \pi \ cm$
0%
$10 \pi \ cm$
0%
$30 \pi \ cm$
0%
None of these

Explanation

Radius, r = 10 cm

We know, Circumference =

$2\pi r = 2 \times \pi \times 10 =20 \pi \ cm$

The area of circle with radius

$24\ cm$ is

0%
$576\pi$
0%
$24\pi$
0%
$12\pi$
0%
$48\pi$

Explanation

The radius of circle =

$24\ cm$

The area of circle of radius

$r = \pi r^2$

$\therefore$ Required area =

$π \times {24}^2\ cm^2 = 576 \pi\ cm^2$

Find the radius of circle if its area is

$125\pi cm^2$

0%
$5\sqrt 5cm$
0%
$3\sqrt 3cm$
0%
$2\sqrt 2cm$
0%
none

Explanation

The area of circle is

$125\pi cm^2$

The area of circle of radius

$r =\pi r^2$

$\therefore 125\pi cm^2= πr^2$

$r^2=125$

$\Rightarrow r=5\sqrt 5cm$

If the radius of a circle is tripled, The area becomes

0%
$9\ times$
0%
$3\ times$
0%
$remains\ the\ same$
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$\sqrt{3}\ times$

Explanation

Let the radius of the circle be

$r$ .

So Area

$=A=\pi r^2$

If radius is tripled, then radius becomes

$3r$ .

Now, area of circle

$A'=\pi(3r)^2 =9\pi r^2=9A$

Therefore. if the radius of a circle is tripled, the area becomes

$9\ times$ .

Circumference of a circle disc is

$88\, cm$ . Its radius is.

0%
$8\, cm$
0%
$11\, cm$
0%
$14\, cm$
0%
$44\, cm$

Explanation

Option (c) is correct.
We know that.
Circumference

$= 2\pi r =88$

$\Rightarrow 88=2\times \dfrac {22}{7} \times r$

$\Rightarrow r=\dfrac {88\times 7}{2\times 22}$

$\Rightarrow r= 14\, cm$
Therefore,

$14\,cm$ is the radius.

$1m^2 = x\, mm^2$ , then the value of x is.

0%
$1000$
0%
$10000$
0%
$100000$
0%
$1000000$

Explanation

We know that

$1m = 1000mm$ , then

$1m^2 = (1000)^2 mm$

Option (d) is correct.
Given,

$1\,m^2= x\, mm^2$

$\Rightarrow (1000\,mm)^2 = x\,mm ^2$

$\Rightarrow x =1000000$

Area of a circle with diameter

$m$ , radius

$n$ and circumference

$p$ is

0%
$2 \pi \, n$
0%
$\pi \, m^2$
0%
$\pi \, p^2$
0%
$\pi \, n^2$

Explanation

Option (d) is correct.
Given,
Diameter

$= m$
Radius

$= n$ ,
Circumference

$= p$ .
Therefore, area of circle

$= \pi r^2=\pi n^2$

Circumference of a circle of diameter

$5 \, cm$ is

0%
$5 \, cm$
0%
$31.4 \, cm$
0%
$15.7 \, cm$
0%
$1.57 \, cm$

Explanation

Option (c) is correct. Given,
Diameter

$= 5 \,cm$

Hence. Radius

$=\dfrac {5}{2}\, cm$

Now. circumference

$= 2\pi r = 2\times \dfrac {22}{7}\times \dfrac {5}{2}$

$=\dfrac {110}{7}= 15.7\, cm$

State true/false:

To find the cost of a frame of a picture, we need to find the perimeter of the picture.

0%
True
0%
False

Explanation

Frame is attached to outer boundary of picture. Thus to know the cost of frame we should know lengths of the outer boundaries.
To find lengths of the outer boundaries, we need to find the perimeter of picture.

$\textbf{So, the given statement is true.}$

Write true or false and justify your answer. The area of

$\Delta ABC$ is

$8\ cm^{2}$ in which

$AB=AC=4\ cm$ and

$\angle A = 90^{\circ}$

0%
True
0%
False

Explanation

Justification:

Area of triangle =

$\dfrac{1}{2} \times Base \times Altitude$

$= \dfrac{1}{2} \times 4 \times 4$

$= 8cm^{2}$

Hence, the statement

$\text{area of }\Delta ABC\text{ is 8 cm}^{2}\text{ in which }AB = AC = 4\ cm\ and\ \angle A = 90^{\circ}\text{ is True}$ .

State true or false.

The area of a triangle with base

$4\ cm$ and height

$6\ cm$ is

$24\ cm^{2}$ .

0%
True
0%
False

Explanation

Area of triangle =

$\dfrac{1}{2} \times Base \times Altitude$

$= \dfrac{1}{2} \times 4 \times 6$

$= 12cm^{2}$

Hence, the statement

$\text{the area of a triangle with base 4 cm and height 6 cm is 24 cm}^{2}\text{ is False}$ .

A circular park is surrounded by a road

$21\ m$ wide. If the radius of the park is

$105\ m$ , find the area of the road.

0%
$13860\ m^{2}$
0%
$30492\ m^{2}$
0%
$7623\ m^{2}$
0%
$15246\ m^{2}$

Explanation

$O$ is the centre of a circular park of radius

$OA=105\ m$ with a concentric surrounding circular road of width

$AB=21\ m$

To find the area of the road, first we need to find area of the park and then subtract the area of park from total area including road.

Let's find area of a park.

Park forms a circle with radius

$(r)=105\ m$

$\therefore$ Area of the park

$=\pi r^2=\pi (105)^2$

The outer circle represents total area which includes park and road with radius

$(R)=105+21=126$

$\therefore$ Total area

$=\pi R^2=\pi(126)^2$

Thus, area of the road

$=$ Total area

$-$ Area of park

$=\pi(126)^2-\pi(105)^2=\pi(126^2-105^2)=4851\pi=15246\ m^2$

Hence, area of road

$=15246\ m^2$

The perimeter of a quadrant of a circle of radius

$r$ is:

0%
$\dfrac{\pi r}{2}$
0%
$2\pi r$
0%
$\dfrac{r}{2}\left [ \pi +4\right ]$
0%
$2\pi r+\dfrac{r}{2}$

Explanation

Perimeter of quadrant of a circle

$=\cfrac { 1 }{ 4 } \times 2\pi \times r+2r$

$=\cfrac { \pi r }{ 2 } +2r\\ =\cfrac { r }{ 2 } \left[ \pi +4 \right]$

State whether true or false.

The radii of two circles are

$19$ cm and

$9$ cm respectively. The radius of the circle which has the circumference equal to the sum of the circumferences of the two circles is

$26$ cm.

0%
True
0%
False

Explanation

Circumference of a circle,

$C$

$=$

$2\pi r$

$\Rightarrow C_1$ =

$2\times \dfrac { 22 }{ 7 } \times 19$

$\Rightarrow C_2$ =

$2\times \dfrac { 22 }{ 7 } \times 9$

$\Rightarrow { C }_{ 1 }+{ C }_{ 2 }=2\times \dfrac { 22 }{ 7 } \left( 19+9 \right)$

$\Rightarrow 2\times \dfrac { 22 }{ 7 } \times r=176\\$

$r=28$ cm

The base and the corresponding altitude of a parallelogram are

$10\: cm$ and

$3.5\: cm$ , respectively. The area of the parallelogram is

0%
$30\: cm^2$
0%
$35\: cm^2$
0%
$70\: cm^2$
0%
$17.5\:cm^2$

Explanation

The area of the parallelogram is base

$\times$ height

$cm^2$

Area of the parallelogram

$=(10)(3.5)=35\:cm^2$ .

If the sum of the areas of two circles with radii

$R_1$ and

$R_2$ is equal to the area of a circle of radius R, then

0%
$R_{1}+R_{2}=R$
0%
$R_{1}^{2}+R_{2}^{2}=R^{2}$
0%
$R_{1}+R_{2} < R$
0%
$R_{1}^{2}+R_{2}^{2} < R^{2}$

Explanation

$ar.circle\quad with\quad radius\quad { R }_{ 1 }=\pi { { R }_{ 1 } }^{ 2 }\quad and\\ ar.circle\quad with\quad radius\quad { R }_{ 2 }=\pi { { R }_{ 2 } }^{ 2 }.\\ \therefore \quad sum\quad of\quad the\quad areas=\pi \left( { { R }_{ 1 } }^{ 2 }+{ { R }_{ 2 } }^{ 2 } \right) .\\ \quad Again\quad ar.circle\quad with\quad radius\quad R=\pi { { R } }^{ 2 }.\\ \therefore \quad By\quad the\quad given\quad condition\quad \pi \left( { { R }_{ 1 } }^{ 2 }+{ { R }_{ 2 } }^{ 2 } \right) =\pi { { R } }^{ 2 }\\ \Longrightarrow \left( { { R }_{ 1 } }^{ 2 }+{ { R }_{ 2 } }^{ 2 } \right) ={ { R } }^{ 2 }.\\ Ans-\quad Option\quad B.\\$

State true or false.

If the circumference of the circle is

$88$ cm, then its radius is

$14$ cm.

0%
True
0%
False

Explanation

Let radius be

$r$

Circumference of the circle

$=2\times \pi \times r$

$\Rightarrow 88=2\times \dfrac { 22 }{ 7 } \times r$

$\Rightarrow r=14$

Hence the radius is

$14\ cm$

CBSE Questions for Class 7 Maths Perimeter And Area Quiz 3 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths Perimeter And Area Quiz 3 - MCQExams.com

0:0:2

Practice Class 7 Maths Quiz Questions and Answers

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