MCQ Questions for Class 7 Maths Perimeter And Area Quiz 5

The area of the cross-section;

0%
133.5 $cm^{3}$
0%
187.5 $cm^{3}$
0%
138.5 $cm^{3}$
0%
183.5 $cm^{3}$

Explanation

The given diagram can be divided in two parts, rectangle ABCD and triangle CEF.

$AB=12$ cm

$BE=16$ cm

$FD=7.5$ cm

$AD=BC=10$ cm

Now,

$AB=CD$ (Opposite sides of a rectangle)

$=>AB=CF+FD$

$=>12=CF+7.5$

$=>CF=4.5$ cm

Again,

$BE=BC+CE$

$=>16=10+CE$

$=>CE=6$ cm

Area of cross section=Area of rectangle ABCD+

$\triangle$ CEF

$=AB\times AD+\dfrac{1}{2}\times CE\times EF$

$=10\times 12+\dfrac{1}{2}\times 6\times 4.5$

$=133.5cm^2$

Find the area of a ring shaped region enclosed between two concentric circles of radii

$20$ cm and

$15$ cm.

0%
550 $cm^{2}$
0%
425 $cm^{2}$
0%
496 $cm^{2}$
0%
810 $cm^{2}$

Explanation

Let us given

$R = 20$

$cm$ &

$r = 15$

$cm$

The area of a ring shaped region

$=\pi(20)^2-\pi(15)^2$

$=(400-225)\times \dfrac{22}{7}$

$=(175)\times \dfrac{22}{7}$

$=550$ sq. cm

The following figure shows a square cardboard

$ABCD$ of side

$28\ cm.$ Four identical circles of the largest possible size are cut from this card as shown below.
Find the area of the remaining cardboard.

0%
$168$ $cm^{2}$
0%
$158$ $cm^{2}$
0%
$148$ $cm^{2}$
0%
$138$ $cm^{2}$

Explanation

Let the radius of each circle be

$r.$

Then,

$4r=28\ cm$

$r=7\ cm$

Therefore,
Area of

$4$ circle

$=4\times \pi (7)^2$

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

$=616\ cm^2$

Area of square

$=28^2$

$=784\ cm^2$

Area of remaining cardboard

$=$ Area of square

$-$ Area of

$4$ circles

$=784-616$

$=168\ cm^2$

If the difference between the circumference and diameter of a circle is

$30 cm$ then what is the radius of the circle?

0%
$7 cm$
0%
$15 cm$
0%
$22 cm$
0%
$2 cm$

Explanation

Let r be the radius of the circle
Then

$\displaystyle 2\pi r-2r=30\Rightarrow 2r\left ( \pi -1 \right )=30$

$\displaystyle \Rightarrow r\left ( \frac{22}{7}-1 \right )=15\Rightarrow r\times \frac{15}{7}=15\Rightarrow r=\frac{15\times 7}{15}=7\:cm$

The area of a circular plot is

$3850$ square meters. What is the circumference of the plot ?

0%
$240 \text{ m}$
0%
$210 \text{ m}$
0%
$220 \text{ m}$
0%
$260 \text{ m}$

Explanation

$r$ is the radius of the plot then area is given as

$\pi r^2$
as per the question,

$\displaystyle\pi r^{2}=3850$

$\Rightarrow r^{2}=\dfrac{3850\times 7}{22}=1225$

$\displaystyle \Rightarrow r= 35\text{ m}$
since circumference

$=2\pi r$

$=2\times \dfrac{22}{7}\times 35\text{ m}$
$=220\text{ m}$

In the given figure, the area enclosed between two concentric circles is

$808.5\ cm^2$ . The circumference of the outer circle is

$242\ cm$ . Calculate the radius of the inner circle

0%
$32\ cm$
0%
$35\ cm$
0%
$23\ cm$
0%
$48\ cm$

Explanation

Let the radius of the inner circle be

$r$ and the radius of the outer circle be

$R$

Area enclosed between the two concentric circles

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

Given

$2\pi R=242$

$\Rightarrow R=\cfrac{242 \times 7}{22\times 2}$

$\Rightarrow R=\cfrac{77}{2}$

Now,

$\pi(R^2-r^2)=808.5$

$\Rightarrow {\left(\cfrac{77}{2}\right)}^2-r^2=\cfrac{808.5 \times 7}{22}$

$\Rightarrow \cfrac{5929}{4}-r^2=257.25$

$\Rightarrow r^2=1225$

$\Rightarrow r=35$

Thus, Radius of the inner circle

$=35\ cm$

The area enclosed between two concentric circles is

$770$

$cm^2$ . If the radius of the outer circle is

$21$

$cm$ , calculate the radius of the inner circle.

0%
$7$ $cm$
0%
$14$ $cm$
0%
$2.1$ $cm$
0%
$35$ $cm$

Explanation

Let radius of the inner circle be

$r$ and radius of the outer circle be

$R=21cm$

Area enclosed between the two concentric circles

$=\pi(R^2-r^2)=770cm^2$

$\Rightarrow 770=\pi(R^2-r^2)$

$\Rightarrow 770=\dfrac{22}{7}(21^2-r^2)$

$\Rightarrow \dfrac{770\times 7}{22}=441-r^2$

$\Rightarrow 245=441-r^2$

$\Rightarrow r^2=196$

$\Rightarrow r=\sqrt{196}$

$\Rightarrow r=14$

Thus, radius of the inner circle

$=14$

$cm$

A circular ground whose diameter is

$140$ meters is to be fenced by wire three times around its circumference. Find the length of wire needed.

$[$ use

$\displaystyle \pi = \frac {22}{7}$

$]$

0%
$440m$
0%
$1320m$
0%
$660m$
0%
None of these

Explanation

Diameter of circular ground

$=140m$ .
Radius of circular ground

$=70m$ .
Circumference of circular ground

$=2\pi r$

$\displaystyle =2\times \frac {22}{7}\times 70$

$=440m$
Length of wire needed to fenced three times

$=440\times 3$

$=1320m$

Find the the perimeter of the figure given correct to one decimal place.

0%
$56.0m$
0%
$56.6m$
0%
$57.7m$
0%
$57.9m$

Explanation

The perimeter of the given figure

$= AB + CD + EF + GH + 4 \times$ ( Perimeter of a quadrant of the circle )

$....(1)$

Now,

$AB = EF = 20 m - (2 m + 2 m) = 16 m$

Similarly,

$CD = GH = 10 m - (2 m + 2 m) = 6 m$

Now, the perimeter of the quadrant having radius

$2 \,\,m\,\,$

$= \dfrac { 2\pi r }{ 4 }$

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

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

$=\pi \,\,$

So, the perimeter of the

$4$ quadrants having radius

$2\,\, m \,\,= \,\,4\pi\,\,$

Put all the respective values in

$(1)$ , we get

Perimeter of the given figure

$= 16 m + 6 m + 16 m + 6 m + 4$

$\pi \,\,\,$

$= 44 m + (4 \times 3.14 )$

$= 44 + 12.56$

$= 56.56 m \sim 56.6 m$

Hence, the perimeter of the given figure correct to one decimal place is 56.6 m.

The inner circumference of a circular track

$14\ m$ wide is

$440\ m$ . The radius of the outer circle is

0%
$70\ m$
0%
$56\ m$
0%
$77\ m$
0%
$84\ m$

Explanation

Let

$R$ and

$r$ be tha radii of the outer and inner circle.

So by given

$R-r = 14m$

Inner circumference

$= \displaystyle 2\pi r=2\times \frac{22}{7}\times r=\frac{44r}{7}$
Given

$\displaystyle \frac{44r}{7}=440\Rightarrow r=\frac{440\times 7}{44}=70\ m$

$\displaystyle \therefore$ Radius of outer circle

$= 70\ m + 14\ m = 84\ m$

What is the perimeter of the given figure correct to one decimal place?

0%
$56.6$ m
0%
$46.5$
0%
$56.7$
0%
$23.7$

Explanation

Perimeter(P)=AB + EF + CD + GH + 4 x Arc AH

$\displaystyle =2\times AB+2\times EF+$ Circumference of circle with radius 2

$\displaystyle =2\times (20-4)m+2\times (10-4)m+2\times \frac{22}{7}\times 2$
= 32 m + 12 m + 12.57 m =56.6m

The circumference of a circle is

$44$ m then the area of the circle is

0%
$\displaystyle 6084.5\:m^{2}$
0%
$\displaystyle 276.5\:m^{2}$
0%
$\displaystyle 154\:m^{2}$
0%
$\displaystyle 44\:m^{2}$

Explanation

Circumference of circle:

$44$ m

Let the radius of circle be

$r$ .

Therefore,

$2\pi r = 44$

$\Rightarrow r = 7$ m

Area of circle:

$\pi r^2 = \dfrac{22}7\times7^2 = 154\ m^2$

The circumference of a circular field is

$528\ m$ . Then its radius is

0%
$84\ m$
0%
$80\ m$
0%
$76\ m$
0%
$78\ m$

Explanation

Let

$r$ be the radius of the circle

Circumference of the circle

$=528$

$\Rightarrow 2\pi r=528$

$\Rightarrow r=\dfrac{528\times 7}{22\times 2}$

$\Rightarrow r=84\ m$

A circular park has a path of uniform width around it. The difference between outer and inner circumference of the circular path is

$132\ m$ . Its width is _____

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

0%
$22\ m$
0%
$20\ m$
0%
$21\ m$
0%
$24\ m$

Explanation

Given

$\displaystyle 2\pi r_{2}-2\pi r_{1}=132$

$\displaystyle \Rightarrow r_{2}-r_{1}=\frac{132}{2\pi }=\frac{132\times 7}{44}=21\ m$

Find the area of the parallelogram whose base is

$17\ cm$ and height

$0.8\ m$ ?

0%
$\displaystyle 13.6\:cm^{2}$
0%
$\displaystyle 1360\:cm^{2}$
0%
$\displaystyle 13.6\:m^{2}$
0%
$\displaystyle 1360\:m^{2}$

Explanation

Base

$= 17\ cm$
Height

$= 0.8\ m =0.8 \times 100 = 80\ cm$
Area of parallelogram

$= b \times h$

$= 17\times 80$

$= 1360\ cm^2$

If the diameter of a circle is

$14$ cm, then its circumference is

0%
$66\;cm$
0%
$44\;cm$
0%
$33\;cm$
0%
$55\;cm$

Explanation

Since, diameter

$= d=14\;cm$

$\displaystyle C=\pi d=\frac{22}{7}\times 14=44\;cm$

The length and breadth of a rectangle are in the ratio

$3:2$ . If the sides of the rectangle are extended on each side by

$1 m$ , the ratio of length to breadth becomes

$10:7$ . Find the area of the original rectangle in square meters.

0%
$2350\ m^{2}$
0%
$1150\ m^{2}$
0%
$54\ m^{2}$
0%
$1000\ m^{2}$

Explanation

Let length of rectangle

$=3x$
breadth of rectangle

$=2x$
Now,

$\cfrac { 3x+1 }{ 2x+1 } =\cfrac { 10 }{ 7 } \\ 21x+7=20x+10\\ x=3$

$\therefore$ length

$=3\times 3=9m,$ breadth

$=2\times 3=6m$
Area of original rectangle

$=(9\times 6){ m }^{ 2 }\\ =54\ { m }^{ 2 }$

The area of a circle is

$\displaystyle 2464\:m^{2}$ , then the diameter is

0%
$56$ m
0%
$154$ m
0%
$176$ m
0%
None of the above

Explanation

Since, Area of circle

$= 2464m^2$

$\therefore \displaystyle \frac{\pi d^{2}}{4}=2464$

$\Rightarrow \displaystyle d^{2}=\dfrac{2464\times 4}{\dfrac{22}{7}}=3136$

$\Rightarrow \displaystyle d=\sqrt{3136}=56m$

In the given figure,

$P$ is the center of the circle. The area and perimeter respectively, of the figure are

0%
$3256 cm^{2}$ , $234.2 cm$
0%
$2492 cm^{2}$ , $229.2 cm$
0%
$2942 cm^{2}$ , $234.2 cm$
0%
$3256 cm^{2}$ , $229.2 cm$

Explanation

Area of figure

$=$ area of circle + area of rectangular - area of quarter circle.

$=\pi { r }^{ 2 }+l\times b-\cfrac { 1 }{ 4 } \times \pi { r }^{ 2 }\\ =\cfrac { 3 }{ 4 } \pi { r }^{ 2 }+l\times b=\cfrac { 3 }{ 4 } \times 3.14\times 20\times 20+50\times 40\\ =942+2000\\ =2942c{ m }^{ 2 }$

&
Perimeter of figure

$=2\pi r-\cfrac { 2\pi r }{ 4 } +2(l+b)-2r\\ =\cfrac { 3\pi r }{ 2 } +2(l+b)-2r\\ =\cfrac { 3\times 3.14\times 20 }{ 2 } +2(50+40)-2\times 20\\ =94.2 +180-40\\ =234.2cm$

A field is in the form of a parallelogram whose base is

$420\ m$ and altitude is

$36 m$ . Find the cost of watering at

$10$ paise per sq. m.

0%
Rs. $15.120$
0%
Rs. $1512$
0%
Rs. $151.20$
0%
None

Explanation

Base

$= 420\ m$
Height

$= 36\ m$
Area

$= b \times h = 420 \times 36$

$= 15,120 \displaystyle \ m^{2}$
The cost of watering per sq m

$= 10$ paise
Cost of watering the field

$= 15120 \times 0.1$
Rs.

$1512$

The degree measure of the circumference of the circle is always

0%
$\displaystyle 0^{\circ}$
0%
$\displaystyle 90^{\circ}$
0%
$\displaystyle 180^{\circ}$
0%
$\displaystyle 360^{\circ}$

Explanation

$\Rightarrow$ The degree measure for the circumference of a circle is

$360^o$ .

$\Rightarrow$ This is because the sum of total angles in a circle is

$360^o$ .

$\Rightarrow$ Whenever we want to get the length of an arc which is part of the circumference of a circle we will always refer to the total angle of a circle which is

$360^o.$

The ratio between the length and breadth of a rectangular garden is

$5:3$ . If the perimeter of the garden is

$160$ meters, what will be the area of

$5\text{ m}$ wide road around its outside?

0%
$600\displaystyle \text{ m}^{2}$
0%
$1200\displaystyle \text{ m}^{2}$
0%
$900\displaystyle \text{ m}^{2}$
0%
$1000\displaystyle \text{ m}^{2}$

Explanation

Let the length and breadth of the rectangular garden be

$5x$ meter and

$3x$ meters.

Given

$2(5x + 3x) = 160$

$\Rightarrow 16x = 160$

$\Rightarrow x = 10$

$\therefore$ Length of garden

$= 50 \text{ m}$ and Breadth of garden

$= 30 \text{ m}$

Area of

$5 \text{ m}$ wide road around its outside

$=$ Area of outer rectangle

$-$ Area of inner rectangle

$[\because$ length of outer rectangle

$= 50+2\times 5$ and breadth of outer rectangle

$=30+2\times 5 ]$

Area

$= (50 + 10) \times (30 + 10) - 50 \times 30$

$= 60 \times 40 - 50 \times 30$

$= (2400 - 1500)$

$= 900 \text{ m}^{2}$

$\therefore$ Area of road is

$900\text{ m}^2$

Circumference of a circle is equal to

0%
$\pi\;r$
0%
$2\pi\;r$
0%
$\displaystyle \frac{\pi r}{2}$
0%
$\displaystyle2+ \frac{\pi r}{2}r$

Explanation

$c=\pi d=2\pi r$

A wire is bent into the shape as shown. It is made up of

$5$ semicircles. What is the length of the wire? (Take

$\pi =3.14$ )

0%
$27\ cm$
0%
$42.39\ cm$
0%
$45\ cm$
0%
$27.92\ cm$

Explanation

Let the length of the wire be

$l.$

The length of the wire is equal to the arc length of the semi-circles drawn on the diameters

$AB,\ BC,\ CD,\ EF$ and

$ED.$

$l= \pi \times 1.5 +\pi \times 3 +\pi \times 4.5 +\pi \times 3 +\pi \times 1.5$

$= \pi \times \left ( 1.5+3+4.5+3+1.5 \right )$

$= 3.14 \times 13.5$

$= 42.39\ cm$

The number of revolutions a wheel of diameter

$40 cm$ makes in travelling a distance of

$176 m$ is:

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

0%
$140$
0%
$150$
0%
$160$
0%
$166$

Explanation

Diameter of wheel

$=40cm,$ Radius

$=20$ cm
Distance covered in

$1$ revolution

$=2\pi r=2\times \cfrac { 22 }{ 7 } \times 20\\ =\cfrac { 880 }{ 7 } cm$
total distance

$=176m=17600cm$

$\therefore$ No. of revolutions

$=\cfrac { 17600 }{ 880/7 } =140$

Find the area of the figure shown above.

0%
$9$ sq cm
0%
$10$ sq cm
0%
$11$ sq cm
0%
$15$ sq cm

Explanation

Area

$=$ (Area of rectangle

$ABDC$ )

$\times 2+$ Area of rectangle

$DGEF$

$=2\times \left( 3\times 1 \right) +5\times 1$

$=6+5$

$=11$ sq cm

Samuel wanted to implant some vertical stones along the boundary of his plot at a distance of

$10$ m each. If length of the plot is

$30$ m and the breadth is

$15$ m, then the number of stones used is:

0%
$450$
0%
$45$
0%
$9$
0%
$10$

Explanation

Perimeter of plot

$=2(l+b)=2(30+15)=90m$

Distance between two stones

$=10m$

Therefore, number of stones

$=\cfrac { 90 }{ 10 } =9$

Area of the shaded figure is:

0%
$2400\ sq. m$
0%
$48\ sq. m$
0%
$50\ sq. m$
0%
$98\ sq. m$

Explanation

We know that area of rectangle of sides

$l$ &

$b$ =

$l \times b$

Area of the shaded region

$=$ Area of

$A+$ Area of

$B$

$=8\times 6+10\times 5$

$=48+50=98\ sq.\ m$

The adjacent sides of a parallelogram are

$8\ m$ and

$5\ m$ . The distance between the longer sides is

$4\ m$ . What is the distance between the shorter sides?

0%
$6.4\ m$
0%
$6\ m$
0%
$8.6\ m$
0%
$None\ of\ these$

Explanation

Area of the parallelogram

$=$ Longer side

$\times$ Distance between them

$= 8 m \times 4 m = 32 \displaystyle\ m^{2}$
Also, area of the parallelogram

$=$ Shorter side

$\times$ Distance between shorter sides

$\displaystyle \Rightarrow 32\ m^{2}=5\ m \times$ Distance between shorter sides

$\displaystyle \Rightarrow$ Distance between shorter sides

$\displaystyle \frac{32\ m^{2}}{5\ m}$

$= \displaystyle 6.4\ m$

The base of a parallelogram is three times its height. If the area of the parallelogram is

$75$ sq cm, then its height is

0%
$5 cm$
0%
$\displaystyle 5\sqrt{2}cm$
0%
$\displaystyle 3\sqrt{2}cm$
0%
$15 cm$

Explanation

let height of ||g be 'b', then base of ||g be 3b.
area of ||g

$=$ base

$\times$ height

$75c{ m }^{ 2 }=3b\times b\\ { b }^{ 2 }=25$

$b=5$ cm, therefore height

$=5$ cm

CBSE Questions for Class 7 Maths Perimeter And Area Quiz 5 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths Perimeter And Area Quiz 5 - MCQExams.com

0:0:2

Practice Class 7 Maths Quiz Questions and Answers

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