Area Related to Circles Test

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Area Related to Circles Test - 20

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Weekly Quiz Competition

Question 1
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In the figure, if the $$\angle AOB=60^{\circ}$$ and radius is $$12 $$cm, then find the area of the segment AXB.$$(\pi =3.14,\:\sqrt{3}=1.73)$$

A
$$14$$

B
$$13.08$$

C
$$12$$

D
$$3.14$$

Solution

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 $$ = \dfrac {60}{360} \times 3.14 \times {12}^{2} - \dfrac{1.73}{4} \times {12}^{2} = 75.36 - 62.28 = 13.08 sq cm $$
Question 2
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The radius of a circle is $$7 cm$$, then area of the sector of this circle if the corresponding angle is:
$$3$$ right angles.

A
$$120. cm^2$$

B
$$115.5 cm^2$$

C
$$150. cm^2$$

D
$$130. cm^2$$

Solution

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} $$
Question 3
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The radius of a circle is $$7 cm$$, then area of the sector of this circle if the corresponding angle is:$$210^{\circ}$$ is

A
$$88.83 \,cm^2$$

B
$$87.83 \,cm^2$$

C
$$89.83 \,cm^2$$

D
$$86.83 \,cm^2$$

Solution

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} $$
Question 4
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The diameter of a circle is $$10$$ cm, then find the length of the arc, when the corresponding central angle is $$180^{\circ}$$. $$(\pi =3.14)$$

A
$$15.7$$

B
$$16$$

C
$$3.14$$

D
$$18$$

Solution

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
Question 5
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The diameter of a circle is $$10$$ cm, then find the length of the arc, when the corresponding central angle is $$144^{\circ}$$.$$(\pi =3.14)$$

A
$$44$$ cm

B
$$12.56$$ cm

C
$$12$$ cm

D
$$88$$ cm

Solution

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 $$ = \dfrac {144}{360} \times 2 \times 3.14 \times 5\ cm$$
$$=\dfrac{144}{360}\times 31.7\ cm$$
$$=12.56\ cm$$
Question 6
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The radius of a circle is $$7 cm$$, then area of the sector of this circle if the corresponding angle is $$30^{\circ}$$ is

A
$$12.83 \,cm^2$$

B
$$11.83 \,cm^2$$

C
$$12.25 \,cm^2$$

D
None of these

Solution

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 $$ = \dfrac { 30 }{ 360 } \times \dfrac { 22 }{ 7 } \times 7 \times 7 = 12.83 \ \text{cm}^{2} $$
Question 7
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In the figure, $$\angle POQ=30^o$$, $$PM \perp OQ$$ and radius $$OP =12 \ cm$$, then find the area of segment $$PRQ$$ (Given $$\pi= 3.14$$).

A
$$1.68 cm^2$$

B
$$2.46 cm^2$$

C
$$0.68 cm^2$$

D
none of these

Solution

Radius $$OP=OQ=r=12$$ cm

$$\theta =30^o$$

and $$PM$$ is perpendicular to $$OQ$$

Area of a sector $$OPQR$$ $$\ \ \ =\pi r^2 \left (\dfrac{\theta}{360^o} \right )$$

$$=3.14\times (12)^2\times \left (\dfrac{30}{360^o} \right )$$

$$=37.68 \ cm^2$$

In $$\triangle OMP , \sin (30^o)=\dfrac{PM}{OP}...........(1)$$

Area of $$\triangle OQP=\dfrac{1}{2} \times OQ\times PM$$

$$=\dfrac{1}{2} \times 12\times OP\times \sin 30^0$$ [from $$(1)$$]

$$=\dfrac{1}{2}\times12\times 12\times \dfrac{1}{2}$$

$$=36 \ cm^2$$

Area of segment $$PRQ =$$ (Area of sector $$OPQR$$)$$ -$$ (Area of $$\triangle OQP)$$

$$=(37.68-36)\ cm^2$$

$$=1.68 \ cm^2$$
Question 8
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In the given figure, $$ABCD$$ is a square of side $$5\ cm$$ inscribed in a circle. Find the area of the shaded region
[Take $$\pi \, =\, 3.14$$]

A
$$14.25\ cm^2$$

B
$$16.75\ cm^2$$

C
$$25\ cm^2$$

D
$$15.25\ cm^2$$

Solution

Diagonal $$AC$$ is the diameter of the circle with centre $$O$$
$$\therefore OC=OA=$$ radius of the circle.

Given,
$$AB=AC=5\ cm$$
$$\triangle ABC$$ is a right-angled triangle.
$$\therefore AC^2=AB^2+AC^2$$
$$=5^2+5^2$$
$$=25+25$$
$$=50 cm$$
$$\Rightarrow AC=5\sqrt 2$$

$$\therefore OA=OC=\cfrac{AC}{2}$$
Radius $$=\cfrac{5\sqrt 2}{2}$$

Area of square $$=AC^2$$
$$=5^2$$
$$=25$$

Area of circle $$=\pi r^2$$
$$=3.14\times \left(\cfrac{5\sqrt 2}{2}\right)^2$$
$$=3.14\times \cfrac{50}{4}$$
$$=39.25$$

$$\therefore$$Area of the shaded region $$=$$ Area of circle $$-$$ Area of square
$$=39.25-25$$
$$=14.25\ cm^2$$
Question 9
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In the figure, m$$\angle POQ = 30^{\circ}$$ and radius OP =12 cm. Find the given area of sector O-PRQ (Given $$\pi$$ = 3.14)

A
$$37.68 cm^2$$

B
$$57.68 cm^2$$

C
$$47.68 cm^2$$

D
none of these

Solution

Radius $$OP=OQ=r=12$$ CM

$$\theta =30$$

and PM is perpendicular to OQ

Area of a sector of a circle $$=\pi r^2 \left (\dfrac{\theta}{360} \right )$$

$$=3.14\times (12)^2\times \left (\dfrac{30}{360} \right )$$

$$=37.68 cm^2$$
Question 10
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The circumference of a given circular park is $$55$$ m. It is surrounded by a path of uniform width $$35$$ cm. Find the area of the path.

A
$$23.125 \: m^{2}$$

B
$$19.635 \: m^{2}$$

C
$$17.245 \: m^{2}$$

D
$$27.155 \: m^{2}$$

Solution

Circumference of circular park $$=55$$ m
$$2\pi r = 55$$ m
$$\Rightarrow r = \dfrac{7}{2\times 22}\times 55 = \dfrac{35}4$$
Area of this field:
$$\pi\times r^2 = \pi\times \left(\dfrac{35}4\right)^2 = 240.625\ m^2$$
A uniform width path is there in surrounding this circular field.
Width $$= 0.35$$ m
Radius increased by $$0.35$$ m
Now new radius will be $$R = 0.35 + \dfrac{35}4 = \dfrac{910}{100}$$ cm
New Area $$= \pi\times\left(\dfrac{910}{100}\right)^2 = 260.26\ m^2$$
Area of extended path:
$$A = 260.26-240.625 = 19.635\ m^2$$