Area Related to Circles Test

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Area Related to Circles Test - 24

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Weekly Quiz Competition

Question 1
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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 $$\pi=\displaystyle\frac{22}{7}$$]

A
$$12843m^2$$

B
$$11284m^2$$

C
$$9384m^2$$

D
$$7382m^2$$

Solution

Area of remaining park $$=$$ area of park $$- 4$$$$\times $$area of $$1$$ quadrant

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

$$ = 10000 -$$ $$\dfrac { 22 }{ 7 } \times 14\times 14$$

$$ = 10000 - 616$$

$$= 9384$$ $${ m }^{ 2 }$$
Question 2
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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 $$20\ cm$$.

A
$$35 \displaystyle \ cm^{2} $$

B
$$45 \displaystyle \ cm^{2} $$

C
$$55 \displaystyle \ cm^{2} $$

D
$$65 \displaystyle \ cm^{2} $$

Solution

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 $$55\ cm^2$$.
Question 3
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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 $$\displaystyle \pi =\frac{22}{7} $$]

A
$$40 cm$$

B
$$44 cm$$

C
$$35 cm$$

D
$$28 cm$$

Solution

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 = $$\dfrac { \theta }{ { 360 }^{ 0 } } \times 2\pi r=\dfrac { { 120 }^{ 0 } }{ { 360 }^{ 0 } } \times 2\times \dfrac { 22 }{ 7 } \times 21=44cm$$
Question 4
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If one side of a square is 2.4 m. Then what will be the area of the circle inscribed in the square?

A
$$1.44 \displaystyle\, m^{2} $$

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

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

D
None of these

Solution

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 $$\displaystyle =1\frac{11}{25}\pi m^{2}$$
Question 5
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$$ABCD$$ 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:

A
$$a^2(1 -\pi )$$

B
$$a^2\left ( \dfrac{4-\pi}{4} \right )$$

C
$$a^2(\pi -1)$$

D
$$\dfrac {\pi a^2}{4}$$

Solution

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.
Question 6
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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

A
$$384$$

B
$$32$$

C
$$24$$

D
$$216$$

Solution

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

$$\displaystyle =\dfrac{12\times 12\times 18\times 12}{9\times 9}=384$$
Question 7
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The portion of a circle between two radii and an arc is called

A
Sector

B
Segment

C
Chord

D
Secant

Solution

Sector of a circle is the portion of a disk enclosed by two radii and an arc. The smaller area is known as the minor sector and the larger as the major sector.
So option A is the correct answer.
Question 8
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Find the area of portion between the 2 semicircles.

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

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

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

D
None

Solution

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 $$\displaystyle =\frac { \pi }{ 2 } { \left( 28 \right) }^{ 2 }-\frac { \pi }{ 2 } { \left( \frac { 28 }{ 2 } \right) }^{ 2 }$$
$$\displaystyle =\frac { \pi }{ 2 } \left( { 28 }^{ 2 }-\frac { { 28 }^{ 2 } }{ 4 } \right) $$
Question 9
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Arc of a sector is equal to-

A
Length of arc $$\times$$ radius

B
$$ \displaystyle \frac{sector angle}{360^{\circ}}\times circumference of circle $$

C
$$ \displaystyle \frac{sector angle}{360^{\circ}}\times (area of circle) $$

D
None of these

Solution

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.
Question 10
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Find the area of the figure shown below

A
9 sq cm

B
10 sq cm

C
11 sq cm

D
15 sq cm

Solution

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