MCQ Questions for Class 10 Maths Areas Related To Cricles Quiz 2

Which of the following statement is false?

0%
If we join any two points on a circle we get a diameter of the circle
0%
A diameter of a circle contains the centre of the circle
0%
A semicircle is an arc
0%
The length of a circle is called its circumference

Explanation

$\textbf{Step - 1: Verify, If we join any points on a circle we get a diameter of the circle}$

$\text{A Chord is a line segment that joins any two points of the circle.}$

$\text{The endpoints of this line segments lie on the circumference of the circle.}$

$\therefore \text{Option (A) is false}$

$\textbf{Step - 2: Verify, A diameter of a circle contains the center of the circle}$

$\text{Any interval joining two points on the circle and passing through the center is called a}$

$\text{diameter of the circle.}$

$\therefore \text{Option (B) is True}$

$\textbf{Step - 3: Verify, A semicircle is an arc}$

$\text{The arc of a circle consists of two points on the circle and all of the points on the circle that lie}$

$\text{between those two points.}$

$\text{It's like a segment that was wrapped partway around a circle.}$

$\text{An arc whose measure equals 180 degrees is called a semicircle since it divides the circle in two}$

$\therefore \text{Option (C) is True}$

$\textbf{Step - 4: Verify, the length of a circle is called its circumference}$

$\text{A Circle is a round closed figure where all its boundary points are equidistant from a fixed point}$

$\text{called the center.}$

$\text{The two important metrics of a circle is the area of a circle and the circumference of a circle.}$

$\therefore \text{Option (D) is True}$

$\textbf{Hence, option A is correct as it is false}$

A chord of a circle of radius 10 cm subtends a right angle at its centre.The length of the chord (in cm) is.

0%
$5 \sqrt 2$
0%
$10 \sqrt 2$
0%
$\frac {5}{\sqrt 2}$
0%
$10 \sqrt 3$

Explanation

Consider a circle with centre at O, radius 10 cm. AB is the chord which subtends

$90^{\circ}$ at the centre.
Now, In

$\triangle OAB$ ,

$OA^2 + OB^2 = AB^2$ (Using Pythagoras Theorem)

$10^2 + 10^2 = AB^2$

$AB^2 = 200$

$AB = 10\sqrt{2}$

The perimeter of a square whose area is equal to that of a circle with perimeter

$\displaystyle 2\pi x$ is

0%
$\displaystyle 2\pi x$
0%
$\displaystyle \sqrt\pi x$
0%
$\displaystyle 4\sqrt{\pi x}$
0%
$\displaystyle 4x\sqrt\pi$

If the circumference of a circle is

$\displaystyle \frac{30}{\pi }$ then the diameter of the circle is

0%
$\displaystyle 60\pi$
0%
$\displaystyle \frac{15}{\pi }$
0%
$\displaystyle \frac{30}{\pi^{2} }$
0%
$30$

Explanation

$\displaystyle 2\pi r=\frac{30}{\pi }$

$\displaystyle 2 r=\frac{30}{\pi^{2} }$

The radius of a circle is

$14$ m, then the circumference of a circle is

0%
$616$ m
0%
$88$ m
0%
$154$ m
0%
None of the above

Explanation

$r = 14$ m
Circumference

$= 2\displaystyle \pi r$

$\displaystyle =2\times \frac{22}{7}\times 14= 88m$

The area of the sector of a circle whose radius is 6 m when the angle at the centre is

$\displaystyle 42^{\circ}$ is

0%
$\displaystyle 13.2\:m^{2}$
0%
$\displaystyle 14.2\:m^{2}$
0%
$\displaystyle 13.4\:m^{2}$
0%
$\displaystyle 14.4\:m^{2}$

Explanation

Area of sector

$\displaystyle =\frac{42}{360}\times \pi r^{2}$

$\displaystyle =\frac{42}{360}\times \frac{22}{7}\times6\times6=13.2m^{2}$

The area of the shaded region

0%
$29\ cm^2$
0%
$30.5\ cm^2$
0%
$25\ cm^2$
0%
$32.5\ cm^2$

Explanation

Given,

$AB=8$ cm and

$BC=6$ cm
The diagonal of the inscribed rectangle will be the diameter of the circle.
Thus, by Pythagoras Theorem, we get

$(AC)^2=(AB)^2+(AC)^2$

$=8^2+6^2$

$=64+36$

$=100$

$=>AC=\sqrt{100}$

$=>AC=10$
Thus, radius

$= \frac{10}{2}$

$=5$ cm
Area of the rectangle ABCD

$=8\times 6$

$=48cm^2$
Area of the circle

$=\pi (5)^2$

$=3.14\times 25$

$=78.5 cm^2$
Thus, the area of the shaded portion is

$=(78.5-48)cm^2$

$=30.5 cm^2$

The area of a segment of a circle of radius

$21$ cm if the arc of the segment has a measure of

$60^{\circ}$ is
(Take

$\sqrt 3\, =\, 1.73$ )

0%
$45.27$ sq. cm
0%
$40.27$ sq. cm
0%
$40.8$ sq. cm
0%
none of these

Explanation

Area of sector

$OAB$

$=\, \displaystyle \frac {x}{360}\, \times\, \pi r^2$

$= \displaystyle \frac {60}{360}\, \times\, \displaystyle \frac {22}{7}\, \times\, 21\, \times\, 21$

$=231$ sq. cm
Here, the triangle ABC is an equilateral triangle, so the area of

$\triangle\, OAB\, =\, \displaystyle \frac {\sqrt 3}{4}\, r^2$

$= \displaystyle \frac {\sqrt 3}{4}\,\times\, 21\, \times\, 21$

$=190.73$ sq. cm
Therefore, area of shaded region

$=\, 231\, -\, 190.73$

$=\, 40.27$ sq. cm

The sides of a triangle are

$5$ ,

$12$ and

$13$ units. A rectangle of width

$10$ units is constructed equal in area to the area of the triangle. Then the perimeter of the rectangle is

0%
30 units
0%
26 units
0%
13 units
0%
15 units

Explanation

By Pythagoras theorem, we find that the given triangle is a right angled triangle with

$12$ as height and

$5$ as base.

$\displaystyle \therefore$ Area of the triangle

$\displaystyle =\frac{1}{2}\times12\times5sq. units$

$=30$ sq. units

$\displaystyle \therefore$ Area of the rectangle

$= length \times breadth = 30$

$\displaystyle \Rightarrow Length =\frac{30}{breadth}=\frac{30}{10}=3\, units$

$\displaystyle \therefore$ Perimeter of the rectangle

$= 2 \times ( 10 + 3 )$ Units

$= 26$ units.

A square sheet of paper is converted into a cylinder by rolling it along its length. What is the ratio of the base radius to side of the square ?

0%
$\displaystyle \frac{1}{2\pi}$
0%
$\displaystyle \frac{\sqrt{2}}{\pi}$
0%
$\displaystyle \frac{1}{\sqrt{2\pi }}$
0%
$\displaystyle \frac{1}{\pi}$

Explanation

Let length of each side of the square

$=a$

Base radius of cylinder

$=r$

Surface area of sheet

$={ a }^{ 2 }$

Surface area of cylinder

$=2\pi rh$

But height

$h=a$ , because the square sheet is rolled it along its length.

Thus, surface area of cylinder

$=2\pi ra$

Therefore,

${ a }^{ 2 }=2\pi ra\Rightarrow a=2\pi r\Rightarrow \dfrac { r }{ a } =\dfrac { 1 }{ 2\pi }$

Find the area of the figure shown below

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

Explanation

Area of given figure= sum of the area of 3 rectangle and 2 suares

Area =

$(3\displaystyle \times 1) + (1\displaystyle \times 1) + (3\displaystyle \times 1) + (1\displaystyle \times 1) + (3\displaystyle \times 1)$

$= 3 + 1 + 3 + 3 + 1$

$= 11$ sq cm

1068948_416128_ans_266853ac12e847f8844817fdc44d333a.png

In the given figure , AC is diameter of a circle with radius 5cm. If AB = BC and CD = 8 cm, the area of the shaded region to the nearest whole number is

0%
$\displaystyle 28 \ cm^{2}$
0%
$\displaystyle 29\ cm^{2}$
0%
$\displaystyle 30\ cm^{2}$
0%
$\displaystyle 45\ cm^{2}$

Explanation

Given that, AC is the diameter of a circle with radius

$5 \ cm$
and

$AB=BC, CD=8\ cm$

$\therefore \ AC=10\ cm$
Area of the circle

$=\pi (5)^2$

$=25\pi \ cm^2$

$[\because \ \text{Area of a circle}=\pi r^2 \text{ square units}]$

We know that, angle subtended by the diameter of a circle at any point on the circle is a right angle.

Hence,

$\triangle ABC$ is a right angled triangle with

$\angle B=90^o$

$\therefore AC^2=AB^2+BC^2$ [Pythagoras theorem]

$\Rightarrow100=2AB^2$

$\Rightarrow AB^2=50$

$\Rightarrow AB=\sqrt{50}$

$\Rightarrow AB=5\sqrt2$

$\therefore AB=BC=5\sqrt2$
Area of

$\triangle ABC$

$=\cfrac{1}{2}\times AB\times BC$

$\Rightarrow \cfrac{1}{2}\times 5 \sqrt2 \times 5 \sqrt2$

$=25 \ cm^2$
Similarly,

$\triangle ADC$ is a right angled triangle with

$\angle D=90^o$

$\therefore AC^2=AD^2+DC^2$

$\Rightarrow 10^2=AD^2+8^2$

$\Rightarrow 100=AD^2+64$

$\Rightarrow AD^2=36$

$\Rightarrow AD=6$
Area of

$\triangle ADC$

$=\cfrac{1}{2}\times AD\times DC$

$\Rightarrow \cfrac{1}{2}\times 6\times 8$

$=24 \ cm^2$
Area of shaded region

$=25\pi-25-24$

$\Rightarrow 25\pi -49=29.5 \ cm^2$

The nearest whole number of

$29.5$ is

$30$ .

Hence, the area of the shaded region to the nearest whole number is

$30 \ cm^2$

A right angled isosceles triangle is inscribed in a circle of radius

$r$ . What is the area of the remaining portion of the circle?

0%
$\displaystyle \frac{\pi r^{2}}{2}$
0%
$\displaystyle \left ( \pi -\frac{1}{2} \right )r^{2}$
0%
$\left ( \pi -1 \right )r^{2}$
0%
$\left ( \pi -2 \right )r^{2}$

Explanation

Let the sides containing the right angle be

$a$ units each and

$r$ be the radius of the circle.

The hypotenuse of the right-angled triangle is the diameter of the circle. So, by applying Pythagoras theorem,

$a^{2}+a^{2}=\left ( 2r \right )^{2}$

$\displaystyle \Rightarrow 2a^{2}=4r^{2}$

$\Rightarrow a=\sqrt{2} r$

$\therefore$ Remaining area

$=$ Area of circle

$-$ Area of triangle

$A=\pi r^{2}-\dfrac{1}{2}\left ( r\sqrt{2} \right )\left ( r\sqrt{2} \right )$

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

$=r^{2}\left ( \pi - 1 \right )$

Hence, area of the remaining region is equal to

$(\pi-1)r^2.$

In the diagram shown,

$ABCD$ is a parallelogram and

$CDEF$ is a rectangle. The total area of the whole figure is

0%
$35\, cm^2$
0%
$30\, cm^2$
0%
$25\, cm^2$
0%
$20\, cm^2$

Explanation

Total area of the whole diagram

$=$ Area of parallelogram

$ABCD +$ Area of rectangle

$CDEF$

Area of parallelogram

$ABCD=$ base

$\times$ height

$=5\times 3\ cm^2$

$=15\ cm^2$

Area of rectangle

$CDEF=2\times 5\ cm^2$

$=10\ cm^2$

Let

$A$ be the total area of figure,

$A=(15+10)\ cm^2$

$=25\ cm^2$

Out of the following shapes having equal perimeters, which is the one having the largest area :

0%
circle
0%
equilateral triangle
0%
square
0%
regular pentagon

Explanation

$Let\quad the\quad given\quad perimeter\quad be\quad P.\\ We\quad calculate\quad the\quad areas\quad of\quad the\quad given\quad figures\quad \\ and\quad compare.\\ Option\quad A\longrightarrow circle.\\ The\quad perimeter(circumference)=P\\ \Longrightarrow 2\pi r=P\quad (when\quad r=radius\quad of\quad the\quad circle.)\\ \Longrightarrow r=\frac { P }{ 2\pi } \\ \therefore \quad ar.circle=\pi { r }^{ 2 }=\pi \times { \left( \frac { P }{ 2\pi } \right) }^{ 2 }=\frac { { P }^{ 2 } }{ 4\times 3.14 } =\frac { { P }^{ 2 } }{ 12.56 } .\\ Option\quad B\longrightarrow Equilateral\quad triangle.\\ One\quad side=\frac { P }{ 3 } .\\ \therefore \quad area=\frac { \sqrt { 3 } }{ 4 } { side }^{ 2 }=\frac { \sqrt { 3 } }{ 4 } \times { \left( \frac { P }{ 3 } \right) }^{ 2 }=\frac { { P }^{ 2 } }{ 20.77 } .\\ Option\quad C\longrightarrow Square\\ side=\frac { P }{ 4 } \\ \therefore \quad area=\frac { P }{ 4 } \times \frac { P }{ 4 } =\frac { { P }^{ 2 } }{ 16 } .\\ Option\quad D\longrightarrow Regular\quad pentagon.\\ side=\frac { P }{ 5 } .\\ \therefore \quad ar.pentagon=\frac { 1 }{ 4 } \sqrt { 5\left( 5+2\sqrt { 5 } \right) } \times { side }^{ 2 }\\ =\frac { 1 }{ 4 } \sqrt { 5\left( 5+2\sqrt { 5 } \right) } \times { \left( \frac { P }{ 5 } \right) }^{ 2 }=\frac { { P }^{ 2 } }{ 19.95 } .\\ \therefore \quad comparing\quad the\quad areas\quad we\quad get\\ the\quad ar.circle\quad is\quad the\quad greatest.\\ Ans-\quad Option\quad A\\$

A chord of a circle of radius

$7$ cm. subtends an angle of

$90^o$ at its centre. The ratio of areas of smaller and larger segment is

0%
$2:7$
0%
$1:10$
0%
$1:11$
0%
none of these

Explanation

Area of minor segment

$=\dfrac {\pi \times R^2}{4}-\dfrac {1}{2}\times R^2$

$=\dfrac {(\pi -2)}{4}R^2$ ..........(1)

Area of major segment

$=\pi R^2-\dfrac {(\pi-2)}{4}R^2$

$=\dfrac {[4\pi-(\pi-2)]}{4}R^2$ ............(2)

$\therefore$ required ratio

$=\dfrac {\pi -2}{3\pi+ 2}=\dfrac {8}{80}=\dfrac {1}{10}$

A sector of

$120^{\circ}$ cut out from a circle has an area of

$9\displaystyle \frac {3}{7}$ sq cm. The radius of the circle is

0%
$3$ cm
0%
$2.5$ cm
0%
$3.5$ cm
0%
$3.6$ cm

Explanation

Given, area of an sector

$=9\dfrac {3}{7}$ sq. cm ,

$\theta=120^0$

We know Area of sector

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

$\Rightarrow \displaystyle \frac {\theta}{360}\, \times\, \pi r^2\, =\, 9\, \displaystyle \frac {3}{7}$

$\Rightarrow \displaystyle \frac {120}{360}\, \times\, \displaystyle \frac {22}{7}\, \times\, r^2\, =\, \displaystyle \frac {66}{7}$

$\Rightarrow r^2\, =\, \displaystyle \frac {66}{7}\, \times\, \displaystyle \frac {360}{120}\, \times\, \displaystyle \frac {7}{22}\, =\, 9$

$\Rightarrow r\, =\, \sqrt 9\, =\, 3$ cm

A borsce is tethered to one corner of a rectangular gracy field 40 m by 24 m with a rope 14 m long . Over how much area of the filed can it gerce

0%
154 ${ m }^{ 2 }$
0%
308 ${ m }^{ 2 }$
0%
150 ${ m }^{ 2 }$
0%
77 ${ m }^{ 2 }$

A figure shows a trapezium

$ABCD$ with two circles of equal diameter drawn in it. If

$ON=56\;cm$ and

$PM\;\bot\;BC\,,\ \,PM=42\;cm,\;AD=30\;cm$ and

$BC=64\;cm$ . Find the area of the shaded region.

0%
$722\;cm^2$
0%
$744\;cm^2$
0%
$742\;cm^2$
0%
$732\;cm^2$

Explanation

Let the radius of each circle be

$r.$

Then,

$\begin{aligned}{}4r &= 56\\r &= 14\ cm\end{aligned}$

Area of shaded region

$=$ Area of trapezium

$-2$

$\times$ Area of circle

$=$

$\dfrac { 1 }{ 2 } \times \left( AD+BC \right) \times PM-2\times \pi { r }^{ 2 }$

$=$

$\dfrac { 1 }{ 2 } \times \left( 30+64 \right) \times 42-2\times \dfrac { 22 }{ 7 } \times 14\times 14$

$=1974 - 1232$

$=742\ { cm }^{ 2 }$

A chord of a circle of radius

$12$ cm subtends an angle of

$60^o$ at the centre. Find the area of the corresponding segment of the circle. (Use

$\pi=3.14$ and

$\sqrt 3=1.73$ ).

0%
$17.43cm^2$
0%
$13.08cm^2$
0%
$11.66cm^2$
0%
$9.64cm^2$

Explanation

Here

$r=12cm$ ,

$\theta=60^o$
Area of the corresponding segment

$ABXA$

$=\displaystyle\left[\frac{\pi r^2\theta}{360}-\frac{1}{2}r^2\sin\theta\right]$

$=\displaystyle\left[\frac{3.14\times 12\times 12\times 60^o}{360^o}-\frac{1}{2}\times 12\times \sin 60^o\right]$

$=\displaystyle\left[\frac{314}{100}\times 12\times 2-6\times 12\times \displaystyle\frac{\sqrt 3}{2}\right]cm^2$ .

$\displaystyle\left[\because \sin 60^o=\frac{\sqrt 3}{2}\right]$

$=\displaystyle\left[\frac{7536}{100}-36\times 1.73\right]cm^2$

$=[75.36-62.28]cm^2=13.08cm^2$

A rectangular sheet of acrylic is 50 cm by 25 cm . From it 60 circular buttons, each of diameter 2.8 cm have been cut out. The area of the remaining sheet is

0%
1260.82 $\displaystyle cm^{2}$
0%
880.4 $\displaystyle cm^{2}$
0%
630.4 $\displaystyle cm^{2}$
0%
None of these

Explanation

Required area
= Area of sheet - 60

$\displaystyle \times$ area of 1 button

$\displaystyle=\left ( 50\times 25-60\times \dfrac{22}{7}\times 1.4\times 1.4 \right )cm^{2}$

$\displaystyle =\left ( 1250-369.6 \right )cm^{2}$

$\displaystyle =880.4cm^{2}$

A chord AB of circle of radius

$14 cm$ makes an angle of

$\displaystyle 60^{\circ}$ at the centre of the circle .The area of the minor segment of the circle is
(Use

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

0%
$\displaystyle \dfrac{308}{3}cm^{2}$
0%
$\displaystyle \dfrac{208}{3}cm^{2}$
0%
$\displaystyle \dfrac{108}{3}cm^{2}$
0%
$\displaystyle \dfrac{408}{3}cm^{2}$

ABCD is a flower bed. If

$OA=21cm$ and

$OC=14m$ , find the area of the bed. [Use

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

0%
$161.4m^2$
0%
$192.5m^2$
0%
$212.6m^2$
0%
$257.2.5m^2$

Explanation

We have,

$OA=R=21m$ and

$OC=r=14m$

$\therefore$ Area of the flower bed
=Area of a quadrant of a circle of radius R - Area of a quadrant of a circle of radius r

$=\displaystyle\frac{1}{4}\pi R^2-\frac{1}{4}\pi r^2$

$=\displaystyle\frac{\pi}{4}(R^2-r^2)$

$=\displaystyle\frac{1}{4}\times \frac{22}{7}(21^2-14^2)cm^2$

$[\therefore R=12m$ and

$r=14m]$

$=\displaystyle\left(\frac{1}{4}\times \frac{22}{7}\times(21+14)(21-14)\right)m^2$

$=\displaystyle\left(\frac{1}{4}\times \frac{22}{7}\times 35\times 7\right)m^2=192.5m^2$

It is proposed to add to a square lawn with each side 58 m two circular ends the centre of each circle being the point of intersection of the diagonals of the square the area of the whole lawn is

0%
$415$ $\displaystyle m^{2}$
0%
$4200$ $\displaystyle m^{2}$
0%
$4319.5$ $\displaystyle m^{2}$
0%
$4215.5$ $\displaystyle m^{2}$

Explanation

Let ABCD be a square with side $58$ metres
Let its diagonals intersect at O

Then $OA=OB=OC=OD$ and

$\displaystyle \angle AOD=\angle BOC=90^{\circ}$

With radius $= \displaystyle \dfrac{1}{2}AC\text{ arcs have been drawn namely } AED \ and \ CBF$

$\displaystyle \therefore \text{Radius of each arc}=\dfrac{1}{2}\times diagonal$

$\displaystyle =\dfrac{1}{2}\sqrt{2}\times 58$

$=1.41 \times 29 =40.89 metres.$

$\displaystyle 58\times 58+2\times \left [ \dfrac{22}{7}\times \left ( 40.89 \right )^{2}\times \dfrac{90}{360}-\dfrac{1}{2}\times \left ( 40.89 \right )^{2}90^{\circ} \right ]$

$\displaystyle 33.64+\dfrac{11\times \left ( 40.89 \right )^{2}}{7}-\left ( 40.89 \right )^{2}$

$=(3364+2627.4-1671.9)$ sq metres

$=4319.5$ sq metres

The area of an equilateral triangle is

$49\sqrt 3cm^2$ . Taking each angular point as centre, a circle is described with radius equal to half the length of the side of the triangle as shown in figure. Find the area of the triangle not included in the circle. [Take

$\sqrt 3=1.73$ ]

0%
$5.67cm^2$
0%
$6.12cm^2$
0%
$7.77cm^2$
0%
$7.98cm^2$

Explanation

Area of equilateral

$\Delta ABC=49\sqrt { 3 } { cm }^{ 2 }$

$\dfrac { \sqrt { 3 } }{ 4 } \times { AB }^{ 2 }=49\sqrt { 3 } { cm }^{ 2 }\Rightarrow { AB }^{ 2 }=196\Rightarrow AB=\sqrt { 196 } =14cm$

Now,

$\dfrac { 1 }{ 2 } AB=AF=FB=BD=DC=CE=EA$

$\therefore$

$AF=FB=BD=DC=CE=EA=7cm$

Now, area of triangle not included

$=$

$49\sqrt { 3 } -3\times \dfrac { \theta }{ { 360 }^{ 0 } } \times \pi { r }^{ 2 }$

$=$

$49\sqrt { 3 } -3\times \dfrac { { 60 }^{ 0 } }{ { 360 }^{ 0 } } \times \pi { r }^{ 2 }$

$= 49$

$\times$

$1.73 - 77$

$= 84.77 - 77$

$= 7.77$

${ cm }^{ 2 }$

A square park has each side of

$100m$ . At each corner of the park, there is a flower bed in the form of a quadrant of radius

$14m$ as shown in figure. Find the area of the remaining part of the park. [Use

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

0%
$12843m^2$
0%
$11284m^2$
0%
$9384m^2$
0%
$7382m^2$

Explanation

Area of remaining park

$=$ area of park

$- 4$

$\times$ area of

$1$ quadrant

$={ \left( 100 \right) }^{ 2 }-4\times \dfrac { 1 }{ 4 } \times \pi { r }^{ 2 }$

$= 10000 -$

$\dfrac { 22 }{ 7 } \times 14\times 14$

$= 10000 - 616$

$= 9384$

${ m }^{ 2 }$

Find the area of the sector of a circle when the angle of the sector is

$\displaystyle 63^{\circ}$ and the diameter of the circle is

$20\ cm$ .

0%
$35 \displaystyle \ cm^{2}$
0%
$45 \displaystyle \ cm^{2}$
0%
$55 \displaystyle \ cm^{2}$
0%
$65 \displaystyle \ cm^{2}$

Explanation

Given that, diameter of the circle

$\displaystyle =20 \ cm$

$\displaystyle \therefore \$ Radius of the circle

$\displaystyle \left ( r \right )=10 \ cm$

Also, angle of the sector

$\displaystyle \left ( \theta \right )=63^{\circ}$

We know that, the area of a sector that subtends an angle

$\theta$ at the centre of the circle is

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

where,

$\theta$ is in degrees.

$\therefore \$ Area

$=\dfrac{22}{7}\times \left ( 10 \right )^{2}\times \dfrac{63}{360}\quad \quad \left[\because \ \pi=\dfrac{22}{7}\right]$

$\Rightarrow \displaystyle \dfrac{22}{7}\times 100\times \dfrac{63}{360}$

$\Rightarrow \displaystyle \dfrac{2200}{40}=55 \ cm^{2}$

Hence, the area of the sector is

$55\ cm^2$ .

1141086_345280_ans_8d8540ad3f3649d7b936eb430e7ebcf0.png

A sector is cut off from a circle of radius

$21$ cm The angle of the sector is

$\displaystyle 120^{\circ}$ The length of its arc is [Take

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

0%
$40 cm$
0%
$44 cm$
0%
$35 cm$
0%
$28 cm$

Explanation

Given radius

$(r)=21cm$ and the angle

$(\theta)=120^\circ$

length of arc

$=r\times \theta$

here

$r=21cm,$

$\theta=120^\circ=\dfrac{120}{360}\times 2\pi$

length of arc =

$\dfrac { \theta }{ { 360 }^{ 0 } } \times 2\pi r=\dfrac { { 120 }^{ 0 } }{ { 360 }^{ 0 } } \times 2\times \dfrac { 22 }{ 7 } \times 21=44cm$

If one side of a square is 2.4 m. Then what will be the area of the circle inscribed in the square?

0%
$1.44 \displaystyle\, m^{2}$
0%
$\displaystyle 1\frac{11}{25}\pi$ $\displaystyle m^{2}$
0%
$\displaystyle \frac{11}{25}\pi$ $\displaystyle m^{2}$
0%
None of these

Explanation

The radius of the circle inscribed in the square
of side 2.4m

$\displaystyle r=\dfrac{2.4}{4}m=1.2m$

$\displaystyle \therefore$ Area of the circle

$\displaystyle =\pi r^{2}$ square units

$\displaystyle =\pi \times 1.2 m\times 1.2 m$

$\displaystyle =1.44 \pi m^{2}$

$\displaystyle =1\dfrac{11}{25}\pi m^{2}$

$\displaystyle \therefore$ The required area

$\displaystyle =1\frac{11}{25}\pi m^{2}$

$ABCD$ is a square with side

$a$ . With centers,

$A, B, C$ and

$D$ four circles are drawn such that each circle touches externally the two of the remaining three circles. Let

$\delta$ be the area of the region in the interior of the square that is not covered by the circles. Then the value of

$\delta$ is:

0%
$a^2(1 -\pi )$
0%
$a^2\left ( \dfrac{4-\pi}{4} \right )$
0%
$a^2(\pi -1)$
0%
$\dfrac {\pi a^2}{4}$

Explanation

Area of the shaded region

$=$ Area of square

$-$ Area of 1 Circle (4 quarter = 1 circle)

Area of Square ABCD

$= a^2$

radius of circle

$=\dfrac {a}{2}$

Area of circle

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

$\therefore$ Area of shaded region

$=a^2-\pi \dfrac{a^2}{4}$

$=a^2\bigg(\dfrac{4-\pi}{4}\bigg)$

Hence, option

$B$ is correct.

CBSE Questions for Class 10 Maths Areas Related To Cricles Quiz 2 - 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 2 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

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