Area Related to Circles Test

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Area Related to Circles Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

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

A
2 cm

B
3 cm

C
4 cm

D
5 cm

Solution

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
Question 2
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Area of shaded figure is

A
2400 sq m

B
48 sq m

C
50 sq m

D
98 sq m
Question 3
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Tick the correct answer in the following:
Area of a sector of angle $$\theta$$ (in degrees) of a circle with radius R is

A
$$\dfrac {\theta}{180}\times 2\pi R$$

B
$$\dfrac {\theta}{180}\times \pi R^{2}$$

C
$$\dfrac {\theta}{3600}\times 2\pi R$$

D
$$\dfrac {\theta}{720}\times 2\pi R^{2}$$

Solution

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
Question 4
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What is the area of a sector with a central angle of $$100$$ degrees and a radius of $$5$$? (Use $$\pi = 3.14$$)

A
$$21.80$$

B
$$11.56$$

C
$$12.46$$

D
$$15.75$$

Solution

Area = $$\dfrac{n}{360}\pi r^2$$
= $$\dfrac{100}{360}\pi 5^2$$
= $$6.944\pi$$
= 21.80
Question 5
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Find the area of a sector in radians whose central angle is $$45^o$$ and radius is $$2$$.

A
$$\dfrac{\pi}{3}$$

B
$$\dfrac{\pi}{4}$$

C
$$\dfrac{\pi}{2}$$

D
$$\dfrac{\pi}{6}$$

Solution

Given: $$\theta = 45^o $$ r =2

Area of sector $$=$$ $$\dfrac{\theta}{360}{\pi}r^2$$
$$\dfrac{45^o}{360}{\pi}2^2$$
= $$\dfrac{\pi}{2}$$
Question 6
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Find the area of the shaded segment(in sq. units)

A
$$88.5$$

B
$$90.5$$

C
$$86.5$$

D
$$89$$

Solution

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
Question 7
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What is the length of the arc?

A
$$20.445$$ ft

B
$$40.445$$ ft

C
$$60.5$$ ft

D
$$80.445$$ ft

Solution

Solution:
Length of the arc$$=r\theta$$
$$=11ft.\times\cfrac{315^\circ}{360^\circ}\times2\pi$$
$$=60.5ft.$$
Hence, C is the correct option.
Question 8
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Length of the arc $$DE$$ = ?

A
$$22.32 m$$

B
$$12.32 m$$

C
$$24.12 m$$

D
$$10.45 m$$

Solution

Arc length = $$\dfrac{\theta}{360}2 \pi r$$
= $$\dfrac{160}{360}\times 2 \times \pi \times 8$$
= $$22.32 m$$
Question 9
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What is the length of the given arc?

A
$$18.26$$ km

B
$$28.26$$ km

C
$$38.26$$ km

D
$$48.26$$ km

Solution

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.
Question 10
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The area of a sector is $$120\pi$$ and the arc measure is $$160^o$$. What is the radius of the circle?

A
$$16.43$$

B
$$11.43$$

C
$$12.23$$

D
$$10.43$$

Solution

$$A_{sector}= \dfrac{n}{360}\pi r^2$$
$$120\pi= \dfrac{160}{360}\pi r^2$$
$$270=r^2$$
$$r = 16.43$$