MCQ Questions for Class 10 Maths Areas Related To Cricles Quiz 3

Size of a tile is

$9$ inches by

$9$ inches. The number of tiles needed to cover a floor of

$12$ feet by

$18$ feet is

0%
$384$
0%
$32$
0%
$24$
0%
$216$

Explanation

Number of tiles

$\displaystyle =\dfrac{Area\, of\, floor}{Area\, of\, 1 tile}$

$\displaystyle =\dfrac{12\times 12\times 18\times 12}{9\times 9}=384$

The portion of a circle between two radii and an arc is called

0%
Sector
0%
Segment
0%
Chord
0%
Secant

Find the area of portion between the 2 semicircles.

0%
$\displaystyle \frac { \pi }{ 2 } { \left( 28 \right) }^{ 2 }$
0%
$\displaystyle \frac { \pi }{ 2 } { \left( \frac { 28 }{ 2 } \right) }^{ 2 }$
0%
$\displaystyle \frac { \pi }{ 2 } { \left( 28 \right) }^{ 2 }-\dfrac{\pi}{2}{ \left( \frac { 28 }{ 2 } \right) }^{ 2 }$
0%
None

Explanation

Area of small semicircle

$\displaystyle \frac { \pi { r }^{ 2 } }{ 2 } =\frac { \pi }{ 2 } \left( \frac { 28 }{ 2 } \times \frac { 28 }{ 2 } \right)$

$\displaystyle =\frac { \pi }{ 8 } \times { \left( 28 \right) }^{ 2 }$
Area of large semicircle

$\displaystyle =\frac { { \pi r }_{ 1 }^{ 2 } }{ 2 } =\frac { \pi }{ 2 } \left( 28\times 28 \right)$
Area of shaded portion = Area (Large - Small) semicircle

$\displaystyle =\frac { \pi }{ 2 } { \left( 28 \right) }^{ 2 }-\frac { \pi }{ 2 } { \left( \frac { 28 }{ 2 } \right) }^{ 2 }$

$\displaystyle =\frac { \pi }{ 2 } \left( { 28 }^{ 2 }-\frac { { 28 }^{ 2 } }{ 4 } \right)$

Arc of a sector is equal to-

0%
Length of arc $\times$ radius
0%
$\displaystyle \frac{sector angle}{360^{\circ}}\times circumference of circle$
0%
$\displaystyle \frac{sector angle}{360^{\circ}}\times (area of circle)$
0%
None of these

Explanation

The yellow section in the above figure is called sector.

Formula for the Arc of a Sector is given by

$\dfrac{\text{Sector of angle}}{360^{\circ}}\times \text{Area of circle}$

So,

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

1906358_404423_ans_8db594c2fe914b08b6c5fc0c969ceccd.png

In given figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB then CD is equal to

0%
2 cm
0%
3 cm
0%
4 cm
0%
5 cm

Explanation

OC is perpendicular on chord AB

$\therefore OC$ bisects the chord AB

$\Rightarrow AC=CB$

Now,

$AC+CB=AB$

$\Rightarrow AC+CB=8$

$\Rightarrow AC=\frac{8}{2}$

$\Rightarrow 4$ cm

$\triangle OCA$ is a right angled triangle

$\therefore AO^2=AC^2+OC^2$

$\Rightarrow 5^2=4^2+OC^2$

$\Rightarrow 5^2-4^2=OC^2$

$\Rightarrow OC^2=9$

$\Rightarrow OC=3$

Since, OD is the radius of the circle

$\therefore OA=OD=5$ cm

$CD=OD-OC$

$\Rightarrow 5-3=2$ cm

Area of shaded figure is

0%
2400 sq m
0%
48 sq m
0%
50 sq m
0%
98 sq m

Tick the correct answer in the following:
Area of a sector of angle

$\theta$ (in degrees) of a circle with radius R is

0%
$\dfrac {\theta}{180}\times 2\pi R$
0%
$\dfrac {\theta}{180}\times \pi R^{2}$
0%
$\dfrac {\theta}{3600}\times 2\pi R$
0%
$\dfrac {\theta}{720}\times 2\pi R^{2}$

Explanation

Area of a sector with angle

$p$

$=\dfrac{\theta}{360}\times\pi R^2$

$=\dfrac{\theta}{360\times2}\times\pi R^2\times2$

$=\dfrac{\theta}{720}\times2\pi R^2$

Hence, Option

$D$ is correct

What is the area of a sector with a central angle of

$100$ degrees and a radius of

$5$ ? (Use

$\pi = 3.14$ )

0%
$21.80$
0%
$11.56$
0%
$12.46$
0%
$15.75$

Explanation

Area =

$\dfrac{n}{360}\pi r^2$
=

$\dfrac{100}{360}\pi 5^2$
=

$6.944\pi$
= 21.80

Find the area of a sector in radians whose central angle is

$45^o$ and radius is

$2$ .

0%
$\dfrac{\pi}{3}$
0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{2}$
0%
$\dfrac{\pi}{6}$

Explanation

Given:

$\theta = 45^o$ r =2

Area of sector

$=$

$\dfrac{\theta}{360}{\pi}r^2$

$\dfrac{45^o}{360}{\pi}2^2$

$\dfrac{\pi}{2}$

Find the area of the shaded segment(in sq. units)

0%
$88.5$
0%
$90.5$
0%
$86.5$
0%
$89$

Explanation

Area of shaded part = area sector OAPB - ar

$\Delta OAB$

Area sector OAPB =

$\dfrac { \theta }{ { 360 }^{ 0 } } \times \pi { r }^{ 2 }$

$\dfrac { { 120 }^{ 0 } }{ { 360 }^{ 0 } } \times \dfrac { 22 }{ 7 } \times 12\times 12$

$= 150.86$ sq. units

Now, area

$\Delta OAB$

Draw OM

$\bot$ AB, which results in two

${ 30 }^{ 0 }-{ 60 }^{ 0 }-{ 90 }^{ 0 }$ triangles, whose sides are in ratio

$\dfrac { 1 }{ 2 } :\dfrac { \sqrt { 3 } }{ 2 } :1$

$\therefore$ In

$\Delta OMA\Rightarrow \quad OM=\dfrac { 12 }{ 2 } =6,\quad AM=\dfrac { 12\sqrt { 3 } }{ 2 } =6\sqrt { 3 } ,\quad OA=12$

$\therefore$ area

$\Delta OMA=\dfrac { 1 }{ 2 } \times AM\times OM=\dfrac { 1 }{ 2 } \times 6\sqrt { 3 } \times 6=18\sqrt { 3 }$

And area

$\Delta OAB=2\times ar\Delta OMA=2\times 18\sqrt { 3 } =36\sqrt { 3 } =62.28$ sq. units

$\therefore$ Area of shaded region

$= 150.86 - 62.28 = 88.56$ sq.units

What is the length of the arc?

0%
$20.445$ ft
0%
$40.445$ ft
0%
$60.5$ ft
0%
$80.445$ ft

Explanation

Solution:

Length of the arc

$=r\theta$

$=11ft.\times\cfrac{315^\circ}{360^\circ}\times2\pi$

$=60.5ft.$

Hence, C is the correct option.

Length of the arc

$DE$ = ?

0%
$22.32 m$
0%
$12.32 m$
0%
$24.12 m$
0%
$10.45 m$

Explanation

Arc length =

$\dfrac{\theta}{360}2 \pi r$
=

$\dfrac{160}{360}\times 2 \times \pi \times 8$
=

$22.32 m$

What is the length of the given arc?

0%
$18.26$ km
0%
$28.26$ km
0%
$38.26$ km
0%
$48.26$ km

Explanation

Given, radius

$=12$ km

Arc length

$=$

$\dfrac{\theta}{360}2 \pi r$

$=$

$\dfrac{135}{360}\times 2 \times \pi \times 12$

$=$

$28.26$ km

Therefore, length of given arc is

$28.26$ km.

The area of a sector is

$120\pi$ and the arc measure is

$160^o$ . What is the radius of the circle?

0%
$16.43$
0%
$11.43$
0%
$12.23$
0%
$10.43$

Explanation

$A_{sector}= \dfrac{n}{360}\pi r^2$

$120\pi= \dfrac{160}{360}\pi r^2$

$270=r^2$

$r = 16.43$

What is the area of the sector?

0%
170 $cm^2$
0%
100 $cm^2$
0%
110 $cm^2$
0%
130 $cm^2$

Explanation

$A_{sector}= \dfrac{n}{360}\pi r^2$

$= \dfrac{135}{360}\pi 12^2$

$169.714$

$\approx 170 cm^2$

Determine the length of the arc

$ADC$ .

0%
$65.2 ft$
0%
$75.4 ft$
0%
$86.1 ft$
0%
$90.2 ft$

Explanation

Radius, $r=16\ ft$

Central Angle, $\theta = 270^o$

Arc length $=\dfrac{\theta}{360^o}\times 2 \pi r$

$=$ $\dfrac{270^o}{360^o}\times 2 \times \pi \times 16$

$=$ $75.4\ ft$

Therefore, the length of the given arc is $75.4\ ft.$

Find the arc length of the sector.

0%
$3\pi$
0%
$2\pi$
0%
$\pi$
0%
$\frac{\pi}{2}$

Explanation

Arc length =

$\frac{\theta}{360}2 \pi r$
=

$\frac{90}{360}2 \pi 2$
=

$\pi$

Determine the area of the shaded segment.

0%
$10$
0%
$11$
0%
$12$
0%
$13$

Explanation

$A_{segment}=A_{sector}-A_{triangle}$

$A_{sector}= \dfrac{n}{360}\pi r^2$

$= \dfrac{60}{360}\pi 6^2$
= 18.84

$cm^2$

$A_{triangle}=\dfrac{1}{2}bh$

$b=sin60^0*6=$

$3\sqrt{3}$

$h=cos60^0*6=$

$3$

$A_{triangle}=\dfrac{1}{2}bh$
=

$\dfrac{1}{2}3\sqrt{3}\times 3$
=

$7.79$

$A_{segment}=18.84-7.79 \approx 11$

Calculate the area of a segment of a circle with a central angle of

$165$ degrees and a radius of

$4$ . Express answer to nearest integer.

0%
$10$
0%
$20$
0%
$30$
0%
$40$

Explanation

$A_{segment}=A_{sector}-A_{triangle}$

$A_{sector}= \dfrac{n}{360}\pi r^2$

$= \dfrac{165}{360}\pi 4^2$
= 23.02

$cm^2$

$A_{triangle}=\dfrac{1}{2}bh$
b =

$2\sqrt{3}$
h = 2

$A_{triangle}=\dfrac{1}{2}bh$
=

$\dfrac{1}{2}2\sqrt{3}\times 2$
=

$3.46$

$A_{segment}=23.02-3.46 \approx 20$

Calculate the arc length for the given diagram.

0%
$1.23\text{ in}$
0%
$2.23\text{ in}$
0%
$4.23\text{ in}$
0%
$5.23\text{ in}$

Explanation

Arc length of a sector having radius

$r$ and central angle

$\theta$ is given by,

$l = \dfrac{\theta }{{360}} \times 2\pi r$

So, by using the above formula,

$\begin{aligned}{}l &= \frac{\theta }{{360}} \times 2\pi r\\& = \frac{{50}}{{360}} \times 2 \times \pi \times 6\\& = 5.23\text{ in}\end{aligned}$

Points

$A$ and

$B$ lie on circle with centre

$O$ (not shown).

$AO=3$ and

$\angle AOB ={120}^{\circ}$ . Find the area of sector

$AOB.$

0%
$\dfrac{\pi}{3}$
0%
$\pi$
0%
$3 \pi$
0%
$9 \pi$

Explanation

Area of a sector is given by

$\cfrac{\theta}{360} \times \pi \times r^2$ where

$\theta$ is the angle made by the sector,

$r$ is the radius of the circle.

Here,

$\theta = 120^\circ$ and

$r = 3$

$\therefore$ Area of sector

$= \cfrac{120}{360} \times \pi \times 9$

$= 3\pi$

In the figure,

$ABCD$ is a rectangular and

$\angle F = 90^{\circ}$ . If

$\overline{FD}=\overline{DC}, \overline{AB}=6$ , and the perimeter of rectangular

$ABCD$ is

$30$ , Calculate

$\overline{FA}$

0%
$6$
0%
$3\sqrt{5}$
0%
$8$
0%
$5\sqrt{3}$

Explanation

Rectangle

$ABCD$ has its side

$AB$ as

$6$ and its perimeter as

$30$ .

So, side

$AB = CD = 6$ and

$BC = DA = 9$

Since

$FD = DC, FD$ also becomes equal to

$6$ .

Now in

$\Delta AFD, \angle F = 90^o$

So,

$(AF)^2 + (FD)^2 = (AD)^2$

$\Rightarrow (AF)^2 + 36 = 81$

$\Rightarrow (FA)^2 = 45$

$\Rightarrow FA = 3\sqrt{5}$

The length of the arc

$OP$ is

0%
$16.28 cm$
0%
$12.28 cm$
0%
$15.28 cm$
0%
$19.28 cm$

Explanation

Arc length =

$\dfrac{\theta}{360}2 \pi r$
=

$\dfrac{65}{360}\times 2 \times \pi \times 17$
=

$19.28 cm$

The trapezoid is divided into a rectangle and two triangles as shown in the above figure. Find the combined area of the two shaded triangles. ( All lengths are in inches)

0%
$4$
0%
$6$
0%
$9$
0%
$12$
0%
$14$

Explanation

The base of the triangles is

$=\dfrac{1}{2}\times (12-6)=3$ .

Hence the two triangles are right angles isosceles as base and height are both equal to

$3$ units.

Area of one triangle

$=\dfrac{1}{2}\times \mathrm{Base}\times \mathrm{Height} =\dfrac{1}{2}\times 3\times 3=\dfrac{9}{2} \ \mathrm{sq. units}$
Therefore, the required area is

$2\times ($ Area of the triangle

$)$

$=2\times \dfrac{9}{2}\ \text{sq. units}$

$=9\ \mathrm{sq. units}$

What is the area of the shaded segment?

0%
$1$ $m^2$
0%
$2$ $m^2$
0%
$3$ $m^2$
0%
$4$ $m^2$

Explanation

From figure, we have

$A_{sector}= \dfrac{n}{360}\pi r^2$

$= \dfrac{30}{360}\pi 4^2$

$= 4.18$

$cm^2$

$A_{triangle}=\dfrac{1}{2}bh$

$b=sin60^0*4=$

$2\sqrt{3}$

$h=cos60^0*4=$

$2$

$A_{triangle}=\dfrac{1}{2}bh$

$=$

$\dfrac{1}{2}2\sqrt{3}\times 2$

$=$

$3.46$

$A_{segment}=4.18-3.46 \approx 1\space\ m^2$

Find the area of the shaded segment.

0%
$11$
0%
$12$
0%
$13$
0%
$14$

Explanation

$A_{sector}= \dfrac{n}{360}\pi r^2$

$= \dfrac{60}{360}\pi 6^2$
=

$18.84\ cm^2$

$A_{triangle}=\frac{1}{2}bh$

$b=sin60^0*6=$

$3\sqrt{3}$

$h=cos60^0*6=$

$3$

$A_{triangle}=\dfrac{1}{2}bh$
=

$\dfrac{1}{2}3\sqrt{3}\times 3$
=

$7.79$

$A_{segment}=18.84-7.79 \approx 11 \space\ cm^2$

In figure, if the area of square

$ABCD$ is

$5$ , calculate the area of square

$BEFD$ .

0%
$7.07$
0%
$8.25$
0%
$10.00$
0%
$12.50$
0%
$25.00$

Explanation

Given, the area of square

$ABCD$ is

$5$

Then side of

$ABCD$ square

$=$

$\sqrt{5}$

The side of square

$BDEF$ is the hyptenuse of triangle

$BCD$

Then side

$BD =$

$\sqrt{(\sqrt{5})^{2}+(\sqrt{5})^{2}}=\sqrt{5+5}=\sqrt{10}$

Then area of square

$BDEF=$

$($

$\sqrt{10}$

$)^{2}= 10$

In Figure 5, rectangle

$ABCD$ is inscribed in a circle. If the radius of the circle is

$2$ and

$\overline{CD} = 2$ , find the area of the shaded region.

0%
$0.362$
0%
$0.471$
0%
$0.577$
0%
$0.707$
0%
$0.866$

Explanation

Given,

$CD=2=AB$ and

radius is

$2$

Let

$O$ be the center of circle.

Area of shaded region

$=$ Area of sector made by arc

$AB$ with center - area of triangle

$AOB$ .

Area of shaded region

$=\dfrac{\text{Length of arc CD}}{2\pi r}\times \pi r-\dfrac{\sqrt 3}{4} a^2$

Area of shaded region

$=$

$\dfrac { 2 }{ 2\pi \times 2 } \times \dfrac { 22 }{ 7 } \times 2\times 2 - \dfrac { 1 }{ 2 } \times 2\times 2\times \dfrac { \sqrt { 3 } }{ 2 }$

$=\dfrac{88}{42}-\sqrt { 3 }$

$=2.095-1.732$

$= 0.362$ sq. units

The geometric figure consists of a right triangle and

$2$ semicircles. Find the area of the triangle if the diameter of the two circles are

$25 \ m$ and

$12 \ m$ respectively.

0%
$150 \ m^2$
0%
$140 \ m^2$
0%
$142 m^2$
0%
$117 m^2$
0%
$15 m ^2$

Explanation

If the the diameters of the circles are

$25$ m and

$12$ m,

then sides of the triangle are

$25$ m and

$12$ m

Hence area of the triangle

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

$\times$ base

$=\quad \dfrac{1}{2}\times 25\times 12=150$ m

$^2$

$ABCD$ is a square of side

$1$ unit and

$B, D$ are centers of two circles of radius

$1\ unit$ . Find the area of the portion which is common to both circles

0%
$\dfrac {\pi}{2}$
0%
$\dfrac {1}{2}$
0%
$\dfrac {\pi}{4} - \dfrac {1}{2}$
0%
$\dfrac {\pi}{2} - 1$

Explanation

We know that ABCD is a square, hence triangle ABC is right-angled.

Area of shaded region

$= 2$

$\times$ area of mirror segment

Now, Area of minor segment

$=$ area of sector

$-$ area of

$\triangle ABC$

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

$=\dfrac { 1 }{ 2 } \left( \dfrac { \pi }{ 2 } -1 \right)$ ....{Taking

$\dfrac{1}{2}$ Common from both the brackets}

$\therefore$ Area of shaded region =

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

Hence, the area of the shaded region is

$=0.570 \,\,units^2$

CBSE Questions for Class 10 Maths Areas Related To Cricles Quiz 3 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Areas Related To Cricles Quiz 3 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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