MCQ Questions for Class 7 Maths Perimeter And Area Quiz 10

If the perimeter of a circle is

$132$ cm, find its area.

0%
$1356$ sq. cm
0%
$1386$ sq. cm
0%
$1340$ sq. cm
0%
$1436$ sq. cm

Explanation

Let the radius of the circle is

$r$ cm

Perimeter of a circle

$=2\pi r=132$

$\Rightarrow 2\times \cfrac{22}{7}\times r=132$

$\Rightarrow r=\cfrac{132\times 7}{22\times 2}$

$\Rightarrow r=21$ cm

Area of the circle

$=\pi r^{2}$

$\Rightarrow =\cfrac{22}{7}\times 21\times 21$

$\Rightarrow =1386$

$cm^{2}$

Find the area of a circle whose radius is

$6$

$cm$ .

0%
$26$ $c{m^2}$
0%
$46$ $c{m^2}$
0%
$56$ $c{m^2}$
0%
$36$ $c{m^2}$
0%
none

Explanation

The radius of circle is

$6$

$cm$

The area of circle is given as

$πr^2$

$\implies π(6)^2$

$cm^2$

$\implies 36π$

$cm^2$

One diagonal of a parallelogram is

$7cm$ and the perpendicular distance of this diagonal from either of the outlying vertices is

$27cm$ . The area of the parallelogram (in sq.cm) is:

0%
$180$
0%
$183$
0%
$189$
0%
$198$

Explanation

Area of parallelogram

$=2 \times$ area of

$\triangle ACD$

$=2\times \dfrac{1}{2}\times 27 \times 7$

$189\ cm^2$

1374833_1161681_ans_fd0ed655a7f54c09946a74ee23ef9913.png

A race track is in the form of ring whose inner and outer circumstances are

$352\ meters$ and

$396\ meters$ respectively. Find the width of the track.

0%
$7\ meters$
0%
$14\ meters$
0%
$14\pi\ meters$
0%
$7\pi\ meters$

Explanation

Let,

$R, r$ be the outer and inner radius of the track then,

$2\pi R = 396 \ m$

$2\pi r = 352 \ m$

Now ,

$2\pi R - 2\pi r = 396 - 352$

$\implies 2\pi (R - r) = 44$

$\implies 2\times \dfrac{22}{7} \times (R - r) = 44$

$\implies R - r = 44 \times \dfrac{7}{44}$

$\therefore R - r = 7 \ m$

i.e., The width of the track

$= 7 \ m$

Option A is correct.

1116380_1183082_ans_4ecb27bedb5a46bf984990ded7e6e2d7.png

The radius of a circular wheel is

$1.75\ m$ . The number of revolutions that it will make in covering

$11\ kms$ is:

0%
$1000$
0%
$100000$
0%
$100$
0%
$10$

Explanation

Let It take 'n' revolution

Then

$n\times 2\pi r = 11,000$

$\displaystyle \Rightarrow n\times2\times \frac{22}{7}\times 1.75 = 11000$

$\displaystyle \therefore n = 1000$

If the area of a parallelogram is

$144 \operatorname { cm } ^ { 2 }$ and its base is

$9 cm$ . then its height is

0%
$8 cm$
0%
$12 cm$
0%
$24 cm$
0%
$16 cm$

Explanation

Area of parallelogram

$=$ base

$\times$ height

Given : base

$=9\ cm,$

area

$=144\ cm^2$

Let height be

$h$

$\Rightarrow$

$144\ {cm}^{2}=9cm\times h$

$\Rightarrow$

$h$

$=\cfrac{144\ {cm}^{2}}{9\ cm}$

$\therefore$

$h=16\ cm$

If the circumference of a circle is 88 cm, then the area of a circle (in sq. cm) is:

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

Explanation

Given,

Circumference of the circle,

$2\pi r=88$

$\therefore r=\dfrac{44}{\pi}=14$

Area of circle,

$A=\pi r^2$

$=\pi \times 14^2$

$=196 \pi$

The area of a circle whose diameter is 1.4 cm is

0%
$1.54\ cm^2$
0%
$0.77\ cm^2$
0%
$6.16\ cm^2$
0%
None of these

Explanation

Area of cirlce

$=\pi r^2$

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

$=\dfrac{22}{7} \left ( \dfrac{1.4}{2} \right )^2$

$=1.54m^2$

Find the circumference of the circles with the radius 14cm :

$(Take \ \pi = \frac{22}{7})$

0%
88 cm
0%
28 mm
0%
21 cm
0%
90 cm

Explanation

Circumference

$= 2\pi r$

$r = 14 \,cm$

$= 2\pi r = 2\times \frac{22}{7}\times 14$

$= 2\times 22\times 2$

$= 4\times 22$

$= 88 \,cm$

An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression

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

0%
0%
0%
0%
$None\ of\ these$

Explanation

$(a)$ Diameter

$=2.8$ cm

$\Rightarrow$ Radius

$=\dfrac{2.8}{2}=1.4$ cm

Perimeter of a circle

$=2\pi r\Rightarrow$ Perimeter of a semicircle

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

$\therefore$ Perimeter of the figure

$=\pi r+$ diameter

$=\dfrac{22}{7}\times\dfrac{14}{10}+2.8$ cm

$=4.4+2.8=7.2$ cm

$(b)$ Perimeter of the semi-circular part

$=\pi r$

$=\dfrac{22}{7}\times 1.4$ cm

$=4.4$ cm

Perimeter of the remaining part

$=1.5+2.8+1.5=5.8$ cm

Total perimeter

$=4.4+5.8=10.2$

$(c)$ Perimeter of the semi-circular part

$=\dfrac{\pi d}{2}=\dfrac{22}{7}\times \dfrac{2.8}{2}=4.4$ cm

$\therefore$ Perimeter of the figure

$=4.4+2+2=8.4$ cm

Since,

$7.2$ cm

$<8.4$ cm

$<10.2$ cm

$\therefore$ perimeter of figure

$b$ has the longest round.

Find the area of circle whose circumference is

$220cm$

0%
3850
0%
3500
0%
3700
0%
3900

Explanation

Circumference of circle

$=220\,cm$ [ Given ]

Circumference of circle

$=2\pi r$

$\Rightarrow$

$220=2\times\dfrac{22}{7}\times r$

$\Rightarrow$

$r=220\times \dfrac{7}{22}\times\dfrac{1}{2}$

$\Rightarrow$

$r=5\times 7$

$\therefore$

$r=35\,cm^2$

$\Rightarrow$ Area of circle

$=\pi r^2$

$=\dfrac{22}{7}\times 35\times 35$

$=22\times 5\times 35$

$=3850\,cm^2$

The area of the concentric circles are

$962.5\ cm^{2}$ and

$1368\ cm^{2}$ . Find the width of the ring formed by them.

0%
$2.1\ cm$
0%
$1 cm$
0%
$5.5\ cm$
0%
$3.5\ cm$

Explanation

Let the radii of the bigger and smaller circles be

$R\space\mathrm{cm}$ and

$r \space\mathrm{cm}$ , respectively.

Now, Area of the bigger circle

$=\pi R^2$

$\Rightarrow 1386=\frac{22}{7}\times R^2$

$\Rightarrow R^2=\frac{1386\times7}{22}$

$\Rightarrow R^2=441$

$\Rightarrow R=21$

$\therefore$ Radius of bigger circle

$(R)=21\space\mathrm{cm}$

Similarly, Area of the smaller circle

$=\pi r^2$

$\Rightarrow 962.5=\frac{22}{7}\times r^2$

$\Rightarrow r^2=\frac{962.5\times7}{22}$

$\Rightarrow r^2=306.25$

$\Rightarrow r=17.5$

$\therefore$ Radius of smaller circle

$(r)=17.5\space\mathrm{cm}$

Hence, Width of Ring

$= R-r=21-17.5=3.5\space\mathrm{cm}$

Thus,

$\text{D}$ is the correct option.

1939108_1393572_ans_50b4bead6ed2457e8a7d54f92751b13d.png

The area of parallelogram if the base is

$36cm$ and height is

$45cm$

0%
1620
0%
1800
0%
1250
0%
1640

Explanation

The base of parallelogram is

$36cm$

The height is

$45cm$

The area of parallelogram is

$36\times 45=1620cm^2$

The area of triangle with base

$4\ cm$ and height is

$3\ cm$

0%
$6\ cm^2$
0%
$12\ cm^2$
0%
$24\ cm^2$
0%
$3\ cm^2$

Explanation

The area of triangle is

$=\dfrac 12\times height \times base \\=\dfrac 12\times 3\times 4=3\times 2=6\ cm^2$

If the area of a circle is

$154{ cm }^{ 2 }$ , then the perimeter is

0%
11 cm
0%
22 cm
0%
44 cm
0%
55 cm

If The side and altitude of a parallelogram are 9 cm and 6 cm respectively, then its area is

0%
$27 cm ^{2}$
0%
$236cm ^{2}$
0%
$54 cm ^{2}$
0%
$45 cm ^{2}$

Explanation

the area of a parallelogram is the product of a side and height or altitude.

here Side =

$9 cm$ , Altitude=

$6cm$

so, area=

$9\times6$ sq. cm=

$54$ sq. cm

The radius of two concentric circles is

$4\ cm$ and

$3\ cm.$ The area of the bounded region will be:-

0%
$22\ cm^{2}$
0%
$12\ cm^{2}$
0%
$32\ cm^{2}$
0%
$18\ cm^{2}$

Explanation

Given:-

$R=4\ cm$

$r=3\ cm$

Hence, the area of the bounded region will be,

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

$=\dfrac {22}7(16-9)$

$=\dfrac {22}7\times 7\ cm^2$

1342227_1574660_ans_9b72a5ed4563424a8e865aa483799467.png

By converting the 80.2km into the hectare, the answer will be

0%
0.08020ha
0%
8020ha
0%
802ha
0%
0.802ha

By converting the 4.8mm into the cm, the answer will be

0%
0.048cm
0%
0.48cm
0%
48cm
0%
480cm

Ram has a plate in circular form which has design in

$7$ meter radius .Find the area of design in

$cm^2$

0%
$154cm^2$
0%
$15400cm^2$
0%
$1540000cm^2$
0%
$.154cm^2$

Explanation

Radius of circular plate

$(r)=7\space\mathrm{m}$

$=7\times100\space\mathrm{cm}\quad[\because 1\space\mathrm{m}=100\space\mathrm{cm}]$

$=700\space\mathrm{cm}$

Now, Area of design

$=\pi r^2$

$=\dfrac{22}{7}\times700^2$

$=1540000\space\mathrm{cm^2}$

So,

$\text{C}$ is the correct option.

By converting the

$0.0287m^2$ into the

$cm^2$ , the answer will be

0%
$28.7 \, cm^2$
0%
$2870 \, cm^2$
0%
$287 \, cm^2$
0%
None of these

Area of triangle PQR is

$100\, cm^2$ (Fig ). If altitude QT is

$10\, cm$ , then its base PR is.

0%
$20 \,cm$
0%
$15 \,cm$
0%
$10 \,cm$
0%
$5 \,cm$

Explanation

Option (a) is correct.
Give, area of triangle

$PQR =100\, cm^2$ .

We know drat Area of triangle

$PQR \times \dfrac {1}{2} PR \times QT$

$\Rightarrow 100 = \dfrac {1}{2} \times PR \times 10$

$\Rightarrow PR =\dfrac {100\times 2}{10}$

$\Rightarrow PR = 20\, cm$

Area of parallelogram ABCD is not equal to.

0%
$DE \times DC$
0%
$BE \times AD$
0%
$BF \times DC$
0%
$BE \times DC$

Explanation

Option (a) is correct.
Area of parallelogram

$=$ Base

$\times$ corresponding height
Hence, area of parallelogram

$ABCD$

$AD \times BE =BC \times BE$

$\because [AD//BC]$

Also,

$BF\times CD =BF\times AB$

$\because[AB //CD]$

Ratio of areas of

$\Delta MNO, \Delta MOP$ and

$\Delta MPQ$ in Fig. is

0%
$2: 1: 3$
0%
$1: 3:2$
0%
$2: 3: 1$
0%
$1:2:3$

Explanation

Option (a) is correct
Area of triangle

$MNO = \frac {1}{2} \times NO \times MO$

$=10\, cm^2$
Area of triangle

$MOP= \frac{1}{2} \times PQ \times MQ$

$\frac {1}{2} \times 2 \times 5$

$= 5\, cm^2$
Therefore,

$=105$
Area of triangle

$MOP= \frac{1}{2} \times PQ \times MQ$

$\frac {1}{2} \times 6 \times 5$

$= 15\, cm^2$
Therefore, required ratio

$=10:5:15 =2:1:3$

In Fig, if

$PR = 12\, cm, QR = 6\, cm$ and

$PL= 8\, cm$ , then QM is

0%
$6\, cm$
0%
$9\, cm$
0%
$4\, cm$
0%
$5\, cm$

Explanation

Given that

$PL =8\, cm$

$PR =12\,cm$

$QR = 6\, cm$
Now, in right-angled triangle PLR
Using Pythagoras theorem,

$\Rightarrow PR^2 = PL^2 + LR^2$

$\Rightarrow LR^2 = 144 - 64$

$=\sqrt {80} = 4\sqrt {5}\ cm$

$LR = LQ +QR$

$\Rightarrow LQ = LR - QR = (4\sqrt{5} - 6)\ cm$

Now area of triangle PLR

$A =\dfrac {1}{2}\times \left ( 4\sqrt {5} \right )\times 8$

$= 16\sqrt{5}\ cm^2$

Now, area of triangle PLQ

$A =\dfrac {1}{2}\times \left ( 4\sqrt {5} - 6\right )\times 8$

$= (16\sqrt{5}-24) \,cm^2$

Hence, area of triangle PLR

$=$ Area of triangle PLQ

$+$ Area of triangle PQR

So, area of triangle PLR

$-$ Area of triangle PLQ

$=$ Area of triangle PQR

$\Rightarrow 16 \sqrt {5} - (16 \sqrt {5} -24)$ ,

$\Rightarrow$ Area of triangle PQR

$= 24 \, cm^2$

$\Rightarrow \dfrac {1}{2}\times PR\times QM = 24$

$\Rightarrow \dfrac {1}{2} \times 12 \times QM =24$

$\therefore QM = 4\,cm$

A wire is bent to form a square of side

$22\, cm$ If the wire is rebent to form a circle, its radius is.

0%
$22\, cm$
0%
$14\, cm$
0%
$11\, cm$
0%
$7\, cm$

Explanation

Option (b) is correct.
Given, side of a square is

$22\, cm$ The wire has same length.
Hence, perimeter of square arid circumference of circle are equal.
ACQ,
Circumference of circle

$=$ Perimeter of square

$= 2 \times \pi \times r = 4\times (side)$

$\Rightarrow 2\times \dfrac {22}{7}\times r = 4\times 22$

$\Rightarrow r =\dfrac {4\times 22 \times 7}{2\times 22}$

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

$14\, cm$ is the radius of the circle.

Area of circular garden with diameter

$8\, m$ is.

0%
$12.56\, m^2$
0%
$25.12 \, m^2$
0%
$50.28 \, m^2$
0%
$200.96 \, m^2$

Explanation

Given, diameter is

$8\;\text{m}$
Hence, radius

$=\dfrac{8}{2}\;\text{m}= 4\;\text{m}$
Therefore, area of the circular garden

$=\pi r^2$

$=\dfrac {22}{7}\times 4\times 4$

$= 50.28\;\text{m}^2$

In Fig., EFGH is a parallelogram, altitudes FK and Fl are 8 cm and ,10 cm respectively. If

$EF =10\, cm$ , then area of EFGH is

0%
$20\, cm^2$
0%
$32\, cm^2$
0%
$40\, cm^2$
0%
$80\, cm^2$

Explanation

Option (c) is correct
Area of parallelogram (EFGH)

$=$ Base

$\times$ corresponding height

$= 10\times 4$

$= 40\, cm^2$

Area of a rectangle and the area of a circle are equal. If the dimensions of the rectangle are

$10\,cm \times 11\, cm$ , then radios of the circle is

0%
$21\, cm$
0%
$10.5\, cm$
0%
$14\, cm$
0%
$7\, cm$

Explanation

Option (d) is correct.
Given, Length of rectangle

$=l=14 \,cm$
Breadth of rectangle

$= b =11\,cm$
Area of circle

$=$ Area of rectangle

$\Rightarrow \pi r^2 = l\times b$

$\Rightarrow \dfrac {22}{7}\times r^2 = 14\times 11$

$\Rightarrow r^2 =\dfrac {14\times 11\times 7}{22}$

$\Rightarrow r^2 =\sqrt {49}$

$\Rightarrow r= 7\, cm$
Therefore,

$7\, cm$ is the radius of the circle.

Ratio of circumference of a circle to its radius is always

$2 \pi : 1$

0%
True
0%
False

Explanation

We know that, for a circle with radius

$r$ , the circumference

$=2\pi r$

$\therefore \ \text{Circumference : Radius}$

$= 2 \pi r : r =2\pi :1$ .

Hence, the given statement is true.

CBSE Questions for Class 7 Maths Perimeter And Area Quiz 10 - MCQExams.com

0:0:1

Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths Perimeter And Area Quiz 10 - MCQExams.com

0:0:1

Practice Class 7 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.