MCQ Questions for Class 10 Maths Areas Related To Cricles Quiz 1

Identify the center of the circle.

0%
$A$
0%
$O$
0%
$D$
0%
$C$

Explanation

Center of a circle is a point inside the circle and is at an equal distance from all of the points on its circumference.

In the given figure,

$O$ is such point.
So,

$O$ is the center of the given circle.

Hence, option

$B$ is correct.

What is the length of arc AB making angle of

$126^0$ at center of radius

$8$ ?

0%
$2.6\displaystyle \pi$
0%
$5.6\displaystyle \pi$
0%
$7.6\displaystyle \pi$
0%
$\displaystyle \frac{1}{2}\pi$

Explanation

Setting a proportion
AB : OB : : 126 : 360

$\displaystyle \frac{\overline{AB}}{2\pi r}=\frac{126}{360}$

$\displaystyle \overline{AB}=\left ( \frac{126}{360} \right )\times 2\pi r$

$\displaystyle \overline{AB}=\left ( \frac{126}{360} \right )\times 2\pi \times 8$

$\displaystyle \overline{AB}=5.6\pi$

Two concentric circles intersect at _____ number of points.

0%
$2$
0%
$0$
0%
$1$
0%
None of these

The diameter of a circle is

$10 cm$ . Find the length of the arc, when the corresponding central angle is

$270^o$ .

$(\pi =3.14)$

0%
$23.55 cm$
0%
$20.57 cm$
0%
$13.57 cm$
0%
$11.57 cm$

Explanation

Radius of the circle

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

Length

of an arc subtending an angle

$\theta = \dfrac { \theta }{ 360 } \times 2\pi R$ where R is the radius of the circle.

So, length of the arc

$= \dfrac {270} {360} \times 2 \times 3.14\times 5 = 23.55 cm$

The length of an arc of a sector of a circle of radius r units and of centre angle

$\theta$ is

$\displaystyle \frac{\theta}{360^{\circ}} \times \pi r^2$ .

0%
True
0%
False
0%
Neither
0%
Either

Explanation

Length of arc subtending angle

$\theta$ =

$\frac{\theta}{360} \times 2 \pi r$ . Note that area will have the dimension of square of length. So

$\frac{\theta}{360} \times \pi r^{2}$ , cannot be the length.

The area of a sector of angle p (in degrees) of a circle with radius R is

0%
$\displaystyle \frac{p}{360} \times 2 \pi R$
0%
$\displaystyle \frac{p}{180}\times \pi R^2$
0%
$\displaystyle \frac{p}{720} \times 2 \pi R$
0%
$\displaystyle \frac{p}{720} \times 2 \pi R^2$

Explanation

${\textbf{Step-1: Write the formula of area of a sector.}}$

${\text{As per the given condition}}$

$\theta = p\ (\text {in degree).}$

${\text{Angle}}$

$= \dfrac{p}{360}.$

${\text{Area of sector}}$

$= \pi R^2$

${\textbf{Step-2: For angle P.}}$

${\text{Area of sector of angle P}}$

$= \dfrac{p}{360} \pi R^2$

${\text{On multiplying and dividing by 2 simultaneously.}}$

${\text{Area of sector of angle P}}$

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

${\text{Area of sector of angle P}}$

$= \dfrac{p}{720} 2 \pi R^2$

${\textbf{Hence option (D) is correct. Area of sector of angle P is}}$ $\mathbf{\dfrac{p}{720} 2 \pi R^2}$

If an arc of a circle of radius

$14\ cm$ subtends an angle of

$60^{\circ}$ at the centre, then the length of the arc is

$\displaystyle \dfrac{44}{3}\ cm$ .

0%
True
0%
False
0%
Niether
0%
Either

In a circle of radius 21 cm, an arc subtends an angle of

$60^{\circ}$ at the centre the length of the arc is 22 cm.

0%
True
0%
False
0%
Neither
0%
Either

Explanation

Arc length =

$\frac{\theta}{360} \times 2 \pi r = \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = 22 cm.$

Identify which among the pieces given below will not be requires to complete the square.

0%
K
0%
H
0%
J
0%
I

Explanation

Among the following we can see that,

$J$ is not required to complete the square.

By rotating and arranging the pieces

$H, I$ and

$K$ we can form a square.

From the figure, identify the center of the circle.

0%
$A$
0%
$D$
0%
$B$
0%
$C$

Explanation

Center of a circle is a point inside the circle and is at an equal distance from all of the points on its circumference.

In the given figure,

$C$ is such point.
So,

$C$ is the center of the given circle.

Hence, option

$D$ is correct.

Find the difference of the areas of two segments of a circle formed by a chord of length

$5\ cm$ . Subtending an angle of

$90^{o}$ till the centre.

0%
$32.14$ sq. cm
0%
$35.42$ sq. cm
0%
$38.96$ sq. cm
0%
$42.43$ sq. cm

Explanation

Let

$r$ be the radius of circle and

$AB$ be the chord, which makes

$90^o$ angle at centre.

$AB=5cm$

In right

$\triangle OAB$ , using pythagoras theorem

$OA^2+OB^2=AB^2 \\ \Rightarrow r^2+r^2=5^2 \\ \Rightarrow 2r^2=25 \\ \Rightarrow r=\cfrac{5}{\sqrt{2}}$

Area of circle

$=\pi r^2 =\cfrac{22}{7} \times \cfrac{5}{\sqrt{2}} \times \cfrac{5}{\sqrt{2}}=39.28cm^2$

Area of minor segment

$=Area\; of\; sector - Area \;of\; \triangle OAB \\ =\cfrac{90^o}{360^o}\times \pi r^2 - \cfrac{1}{2} \times \cfrac{5}{\sqrt{2}} \times \cfrac{5}{\sqrt{2}} \\ =\cfrac{1}{4}\times 39.28 - \cfrac{25}{4}=\cfrac{14.28}{4}=3.57cm^2$

Area of major segment

$=39.28-3.57=35.71 cm^2$

Required difference

$=35.71-3.57=32.14cm^2$

973859_1040120_ans_3f22c2d0eaf4451e92712d88b310598a.png

A shuttlecock used for playing badminton has the shape of a combination of

0%
a cone and a sphere
0%
frustum of a cone and a hemisphere
0%
a cylinder and a hemisphere
0%
a cone and a cylinder

The radius of a circle is increased by

$1$ cm, then the ratio of the new circumference to the new diameter is

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

Explanation

$\frac{Circumference}{Diameter}=\frac{2 \pi (r+1)}{2(r+1)}$

$\frac{Circumference}{Diameter}=\pi$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Explanation

Both assertion and reason are correct. Also, Reason is the correct explanation of the assertion.
Area of the path

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

$=\pi \left [ \left ( \dfrac{10}{2} \right )^2 - \left ( \dfrac{6}{2} \right )^2 \right ]$

$=\pi (25-9)$

$=16 \pi \ m^2$

A horse is placed for grazing inside a square field 12 cm long and is tethered to one corner by a rope 8 cm long. The area it can graze is

0%
50.18 sq cm
0%
50.28 sq cm
0%
50.38 sq cm
0%
50.48 sq cm

Explanation

Let the grazed portion is shown by shaded quadrant ABC,
Hence, area which horse can graze

$= \frac{1}{4}\pi (8)^2 = \frac{1}{4}\times \frac{22}{7}\times 64$

$= 50.28$ sq cm

A circular disc of radius 10 cm is divided into sectors with angles

$120^{\circ}$ and

$150^{\circ}$ then the ratio of the area of two sectors is

0%
4 : 5
0%
5 : 4
0%
2 : 1
0%
8 : 7

Explanation

Area of sector formed from angle

$\theta$ =

$\frac{\theta}{360} \pi r^2$ , where r is the radius of the circle
Now, if angle is 120, 150 then the ratio of area of sector will be:
=

$\dfrac{\dfrac{120}{360} \pi r^2}{\dfrac{150}{360} \pi r^2}$
=

$\dfrac{120}{150} = 4:5$

If circumference of a circle is

$110\ cm$ , then its diameter is

0%
$\cong \ 25 \ cm$
0%
$\cong 15\ cm$
0%
$\cong 20 \ cm$
0%
$35 \ cm$

Explanation

$\Rightarrow$ The circumference of a circle

$=110\,cm$

$\Rightarrow$

$2\pi r=110$

$\Rightarrow$

$2\times \dfrac{22}{7}\times r=110$

$\Rightarrow$

$r=110\times \dfrac{7}{22}\times \dfrac{1}{2}$

$\Rightarrow$

$r=\dfrac{35}{2}$

$\Rightarrow$

$r=17.5\,cm$

$\Rightarrow$ Diameter of circle

$(d)=2r=2\times 17.5\,cm=35\,cm$

Semi-circular lawns are attached to the edges of a rectangular field measuring

$42 \ m \times 35\ m$ . The area of the total field is:

0%
$3818.5 m^{2}$
0%
$8318 m^{2}$
0%
$5813 m^{2}$
0%
$1358 m^{2}$

Explanation

Given, length of field

$= 42 \ m$
Breadth of field

$= 35 \ m$
Radius of semi circles along length

$= \cfrac{42}2 \ m = 21\ m$
Radius of semi circles along breadth

$= \cfrac{35}2\ m = 17.5\ m$
Now,

Total area of the field

$=$ Area of rectangle

$+$ Area of circle formed along length

$+$ Area of circle formed along breadth

$= l\times b + \pi (21)^2 + \pi( 17.5)^2 \ m^2$

$= (42\times 35 + 1386 + 962.5)\ m^2$ (take

$\pi = \dfrac {22}{7}$ )

$= 3818.5 \ m^2$

So, Total Area of the field

$= 3818.5 m^2$

The figure is made up of an isosceles triangle and a semi-circle. Find the perimeter of the figure.

0%
$2(7-2\pi ) cm.$
0%
$(7+2\pi ) cm.$
0%
$2(7+2\pi ) cm.$
0%
$(7-2\pi ) cm.$

Explanation

Perimeter of semicircular arc =

$\pi \times \dfrac d2$

$=\pi \times \dfrac82$

$=4\pi\ cm$

Total Perimeter = Perimeter of semi-circular arc

$+$ Sum of outer sides of triangle

$ABC$

Total Perimeter

$= l(BOC) + CA+AB$

$= 7 + 7 + 4\pi$

$=14+4\pi$

$=2(7+2\pi)$ cm

The number of revolutions made by a wheel of diameter

$56 cm$ in covering a distance of

$1.1 km$ is

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

0%
$31.25$
0%
$56.25$
0%
$625$
0%
$62.5$

Explanation

Total distance covered by the wheel

$= 1.1 km = 1.1\times 100000cm = 110000 cm$
diameter of wheel

$= 56 cm$
So,

$r = \dfrac {56}{2}$

$r = 28 cm$
Now,Circumference of wheel = Distance covered in one round

$= 2\pi\times r = 2\pi\times 28$
So, distance covered in one round

$= 176 cm$
Number of revolutions

$= \dfrac {total\ distance\ covered}{Circumference}$

$= \dfrac {110000}{176}$
So, number of revolutions covered by wheel

$= 625$

Find the area enclosed between the circles. Given that each circle is having a radius of

$2\ cm.$

0%
$4\sqrt{3}\ cm^{2}$
0%
$(2\sqrt{3}-\pi )\ cm^{2}$
0%
$2\sqrt{3}-2\pi )\ cm^{2}$
0%
$2(2\sqrt{3}-\pi )\ cm^{2}$

Explanation

The radii of the circles with center

$A$ and

$B$ is equal to

$r=2\ cm.$ Since they touch each other

$AB$ will pass through the point of contact of those circles and,

$AB=2r$

$=2\times 2$

$=4\ cm$

Similarly,

$BC=4\ cm$

$AC=4\ cm$

$\therefore\ \triangle ABC$ is an equilateral triangle. So,

$ar\left( {\triangle ABC} \right) = \dfrac{{\sqrt 3 }}{4}{a^2}$

$= \dfrac{{\sqrt 3 }}{4} \times 4 \times 4$

$= 4\sqrt3 \ cm^2$

$\triangle ABC$ crops

$3$ sectors with each of them having radius equal to

$2\ cm$ and central angle equal to respective vertex angle of the

$\triangle ABC$ that is equal to

$60^o.$ So, the three sectors are equal in area.

Sum of area of

$3$ sectors

$=$

$3$

$\times$ Area of

$1$ sector

$=3\times \dfrac{\theta}{360^o}\times\pi r^2$

$=3\times \dfrac{60^o}{360^o}\times\pi \times2^2$

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

Now, area between circles

$=$ Area of

$\triangle ABC$

$-$ Area of

$3$ sectors

$=4\sqrt3-2\pi$

$=2\left({2\sqrt3-\pi}\right)\ {cm}^2$

In the figure, if the

$\angle AOB=60^{\circ}$ and radius is

$12$ cm, then find the area of the segment AXB.

$(\pi =3.14,\:\sqrt{3}=1.73)$

0%
$14$
0%
$13.08$
0%
$12$
0%
$3.14$

Explanation

Area of the segment AXB

$=$ Area of the sector OAXB

$-$ Area of triangle OAB.

The triangle OAB is an equilateral triangle since the length of sides are equal to the radius of the circle and an angle is

${60}^{0}$

Area of an equilateral triangle with length of side

$a = \dfrac { \sqrt { 3 } }{ 4 } { a }^{ 2 }$

Area of a sector of a circle of radius 'r' and angle

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

Hence, Area of segment AXB

$= \dfrac {60}{360} \times 3.14 \times {12}^{2} - \dfrac{1.73}{4} \times {12}^{2} = 75.36 - 62.28 = 13.08 sq cm$

The radius of a circle is

$7 cm$ , then area of the sector of this circle if the corresponding angle is:

$3$ right angles.

0%
$120. cm^2$
0%
$115.5 cm^2$
0%
$150. cm^2$
0%
$130. cm^2$

Explanation

3 right angles $= 3 \times {90}^{0} = {270}^{0}$

Area of a sector of a circle of radius 'r' and angle

Area of a sector = $\dfrac { \theta }{ 360 } \pi {r}^{2}$

Hence, area of the sector of the circle of radius $7$ cm and angle ${ 270 }^{ 0 } = \dfrac { 270 }{ 360 } \times \dfrac { 22 }{ 7 } \times 7 \times 7\quad = 115.5 {cm}^{2}$

The radius of a circle is

$7 cm$ , then area of the sector of this circle if the corresponding angle is:

$210^{\circ}$ is

0%
$88.83 \,cm^2$
0%
$87.83 \,cm^2$
0%
$89.83 \,cm^2$
0%
$86.83 \,cm^2$

Explanation

Area of a sector of a circle of radius 'r' and angle

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

Hence,area of the sector of the circle of radius

$7$ cm and angle

${ 210 }^{ 0 } = \dfrac { 210 }{ 360 } \times \dfrac { 22 }{ 7 } \times 7 \times 7\quad = 89.83 {cm}^{2}$

The diameter of a circle is

$10$ cm, then find the length of the arc, when the corresponding central angle is

$180^{\circ}$ .

$(\pi =3.14)$

0%
$15.7$
0%
$16$
0%
$3.14$
0%
$18$

Explanation

Radius of the circle

$= \dfrac {\text{Diameter}}{2} = 5$ cm

Length of an arc subtending an angle

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

$R$ is the radius of the circle.

So, length of the arc

$= \dfrac {180^o}{360^o} \times 2 \times 3.14\times 5 = 15.7$ cm

The radius of a circle is

$7 cm$ , then area of the sector of this circle if the corresponding angle is

$30^{\circ}$ is

0%
$12.83 \,cm^2$
0%
$11.83 \,cm^2$
0%
$12.25 \,cm^2$
0%
None of these

Explanation

Area of a sector of a circle of radius '

$r$ ' and angle

$= \dfrac { \theta }{ 360 } \pi {r}^{2}$
Hence, area of the sector of the circle of radius

$7$ cm and angle

$= \dfrac { 30 }{ 360 } \times \dfrac { 22 }{ 7 } \times 7 \times 7 = 12.83 \ \text{cm}^{2}$

The diameter of a circle is

$10$ cm, then find the length of the arc, when the corresponding central angle is

$144^{\circ}$ .

$(\pi =3.14)$

0%
$44$ cm
0%
$12.56$ cm
0%
$12$ cm
0%
$88$ cm

Explanation

Radius of the circle

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

Length of an arc subtending an angle

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

$R$ is the radius of the circle.

So, length of the arc

$= \dfrac {144}{360} \times 2 \times 3.14 \times 5\ cm$

$=\dfrac{144}{360}\times 31.7\ cm$

$=12.56\ cm$

Find the perimeter and area of the shaded region shown in the figure. The four corners are circle quadrants and at the centre, there is a circle.
[Take

$\pi \, =\, 3.14$ ]

0%
Perimeter = 20.56 cm
0%
Area = 9.72 $cm^2$
0%
Area = 20.56 $cm^2$
0%
Perimeter = 9.72 cm

Explanation

Area of square

$= Side^2 = 4^2 = 16cm^2$
Area of cut portion = Area of two circles = 1 circle + 4quadrants

$= 2 \times\ pi \times 1^2$

$= 6.28 cm^2$
So, area of remaining portion of square

$= (16- 6.28) = 9.72cm^2$
Perimeter of the shaded region =4 quadrant + 1 circle+portion of the square

$=4\times \frac{2\pi r}{4}+2\pi r+4(2)$

$=4\times \frac{2\pi (1)}{4}+2\pi (1)+8$

$=4\pi +8$

$=20.56 cm$

A garden is in the shape of a square with a semicircle on one of its sides. The side of the square is

$7$ m. What is the area of the garden in sq. m.?

0%
$\displaystyle 68\frac{1}{4}$
0%
$49$
0%
$\displaystyle 58\frac{5}{8}$
0%
$70$

Explanation

The side of the square is

$7m$ .

Hence, the area of the square part of the garden

$=7^2 = 49$ sq. m.

The semicircle lies on one of the sides of the square part of the garden.

So, its radius

$=\dfrac 72 = 3.5m$

Area of the semi-circular part of garden

$=\dfrac 12 \times \pi r^2$

$=\dfrac 12 \times \dfrac {22}{7} \times \dfrac 72 \times \dfrac 72$

$= \cfrac {77}{4}$ sq. m.

Area of garden

$=49+\cfrac {77}{4} = 68\cfrac 14$ sq. m

The radius of the circle

0%
$6\ cm$
0%
$7.5\ cm$
0%
$4\ cm$
0%
$5\ cm$

CBSE Questions for Class 10 Maths Areas Related To Cricles Quiz 1 - 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 1 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

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