Area Related to Circles Test

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Area Related to Circles Test - 27

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Question 1
1 / -0

What is the area of the shaded segment?

A
$$1$$ $$m^2$$

B
$$2$$ $$m^2$$

C
$$3$$ $$m^2$$

D
$$4$$ $$m^2$$

Solution

From figure, we have
$$A_{sector}= \dfrac{n}{360}\pi r^2$$
$$= \dfrac{30}{360}\pi 4^2$$
$$= 4.18$$ $$cm^2$$
$$A_{triangle}=\dfrac{1}{2}bh$$
$$b=sin60^0*4=$$$$2\sqrt{3}$$
$$h=cos60^0*4=$$$$2$$
$$A_{triangle}=\dfrac{1}{2}bh$$
$$=$$ $$\dfrac{1}{2}2\sqrt{3}\times 2$$
$$=$$ $$3.46$$
$$A_{segment}=4.18-3.46 \approx 1\space\ m^2$$
Question 2
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Find the area of the shaded segment.

A
$$11$$

B
$$12$$

C
$$13$$

D
$$14$$

Solution

$$A_{sector}= \dfrac{n}{360}\pi r^2$$
$$= \dfrac{60}{360}\pi 6^2$$
= $$18.84\ cm^2$$
$$A_{triangle}=\frac{1}{2}bh$$
$$b=sin60^0*6=$$$$3\sqrt{3}$$
$$h=cos60^0*6=$$$$3$$
$$A_{triangle}=\dfrac{1}{2}bh$$
= $$\dfrac{1}{2}3\sqrt{3}\times 3$$
= $$7.79$$
$$A_{segment}=18.84-7.79 \approx 11 \space\ cm^2$$
Question 3
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In figure, if the area of square $$ABCD$$ is $$5$$, calculate the area of square $$BEFD$$.

A
$$7.07$$

B
$$8.25$$

C
$$10.00$$

D
$$12.50$$

E
$$25.00$$

Solution

Given, the area of square $$ABCD$$ is $$5$$
Then side of $$ABCD$$ square $$=$$ $$\sqrt{5}$$
The side of square $$BDEF$$ is the hyptenuse of triangle $$BCD$$
Then side $$BD =$$ $$\sqrt{(\sqrt{5})^{2}+(\sqrt{5})^{2}}=\sqrt{5+5}=\sqrt{10}$$
Then area of square $$BDEF=$$ $$($$ $$\sqrt{10}$$ $$)^{2}= 10$$
Question 4
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In Figure 5, rectangle $$ABCD$$ is inscribed in a circle. If the radius of the circle is $$2$$ and $$\overline{CD} = 2$$, find the area of the shaded region.

A
$$0.362$$

B
$$0.471$$

C
$$0.577$$

D
$$0.707$$

E
$$0.866$$

Solution

Given, $$CD=2=AB$$ and
radius is $$2$$
Let $$O$$ be the center of circle.
Area of shaded region $$=$$ Area of sector made by arc $$AB$$ with center - area of triangle $$AOB$$.
Area of shaded region $$=\dfrac{\text{Length of arc CD}}{2\pi r}\times \pi r-\dfrac{\sqrt 3}{4} a^2$$

Area of shaded region $$=$$ $$\dfrac { 2 }{ 2\pi \times 2 } \times \dfrac { 22 }{ 7 } \times 2\times 2 - \dfrac { 1 }{ 2 } \times 2\times 2\times \dfrac { \sqrt { 3 } }{ 2 } $$

$$=\dfrac{88}{42}-\sqrt { 3 } $$

$$=2.095-1.732 $$
$$= 0.362$$ sq. units
Question 5
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The geometric figure consists of a right triangle and $$2$$ semicircles. Find the area of the triangle if the diameter of the two circles are $$ 25 \ m $$ and $$ 12 \ m$$ respectively.

A
$$150 \ m^2$$

B
$$140 \ m^2$$

C
$$142 m^2$$

D
$$117 m^2$$

E
$$15 m ^2$$

Solution

If the the diameters of the circles are $$25$$ m and $$12$$ m,
then sides of the triangle are $$25$$ m and $$12$$ m

Hence area of the triangle $$=\dfrac{1}{2} \times $$ height $$\times $$ base
$$=\quad \dfrac{1}{2}\times 25\times 12=150$$ m$$^2$$
Question 6
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The given figure is made up of a rectangle and $$2$$ identical circles. If the length of the rectangle is $$28\ cm$$, what is the area of the region in the rectangle excluding the areas occupied by the circle? $$\left (Use\ \pi = \dfrac {22}{7}\right )$$

A
$$84\ cm^{2}$$

B
$$105\ cm^{2}$$

C
$$152\ cm^{2}$$

D
$$123\ cm^{2}$$

Solution

Radius of the circle $$=\dfrac{\text{length of rectangle}}{4}$$
Thus radius of the circle and breath of the rectangle are $$7$$ and $$14$$ respectively.

Required Area $$=$$ Area of rectangle $$- 2$$ $$\times $$ area of circle( radius of circle
$$ = 28$$$$\times $$$$14 - 2$$$$\times $$$$\pi { r }^{ 2 }$$
$$=392-2\times \dfrac { 22 }{ 7 } \times 7\times 7$$
$$= 392 - 308$$
$$= 84$$ $${ cm }^{ 2 }$$
Question 7
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Figure $$ABCD$$ is a rectangle whose length is twice its width. $$\widehat {FC}$$ and $$\widehat {AE}$$ are arcs of circles centered at B and D respectively. If the length of $$\overline {AD}$$ is x, then the area of the shade region is

A
$$\left (\dfrac {2 - \pi}{4}\right )x^{2}$$

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

C
$$\left (\dfrac {\pi - 2}{4}\right )x^{2}$$

D
$$\left (\dfrac {2}{\pi}\right )x^{2}$$

E
$$\left (\dfrac {4}{\pi}\right )x^{2}$$

Solution

Solution:
Given that:
$$AD=x$$ and $$AB=2AD$$
To Find:
Area of the shaded region$$=?$$
Solution:
$$AB=2AD=2x$$
Area of the shaded region $$=$$ Area of the rectangle $$ABCD-$$ Area of the quadrant $$AED$$- Area of the quadrant $$FCB$$
$$=AD\times AB-\cfrac{\pi}4r^2-\cfrac{\pi}{4}r^2$$
$$=x\times2x-2\times \cfrac{\pi}{4}x^2$$
$$=2x^2-\cfrac{\pi}2x^2$$
$$=\left(\cfrac{4-\pi}2\right){x^2}$$
Hence, B is the correct option.
Question 8
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The figure shows a semicircle on top of an isosceles right triangle. If the length of semicircle $$\overline {AB}$$ is $$16\pi$$, what is the approximate length of $$\overline {BC}$$?

A
$$2.8$$

B
$$4.0$$

C
$$5.7$$

D
$$11.3$$

E
$$22.6$$

Solution

Let the length of $$AB$$ be $$2r$$ , where $$r$$ is radius of semicircle
Given that $$\pi r=16\pi$$
$$\Rightarrow r=16$$
The length of $$BC$$ is equal to $$\dfrac{AB}{\sqrt2} = \dfrac{32}{\sqrt2}=22.6$$
The correct option is $$E$$
Question 9
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A lawn of circular shape of radius $$r$$ contains a pond . The pond is a square inscribed in it. The owner of lawn wants to plant tress in the remaining area. Find the remaining area.

A
$${ r }^{ 2 }(\pi -2)$$

B
$${ r }^{ 2 }(\pi -1 )$$

C
$${ r }^{ 2 }(\pi -\cfrac { 1 }{ 2 } )$$

D
None of these

Solution

Area of square $$\sqrt { 2 } r\times \sqrt { 2 } r$$
$$=2{ r }^{ 2 }$$
Remaining area$$=\pi { r }^{ 2 }-2{ r }^{ 2 }$$
$$=(\pi -2){ r }^{ 2 }$$
Question 10
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When the figure given above is spun around its vertical axis, the volume of the solid formed will be:

A
$$9\pi$$

B
$$36\pi$$

C
$$72\pi$$

D
$$144\pi$$

E
$$288\pi$$

Solution

The solid formed after spin will be a Hemisphere whose radius is $$6$$ cm.
Volume of hemisphere $$=$$ $$\dfrac{2}{3}\pi r^3$$
$$\therefore$$ volume $$=\dfrac{2}{3}\pi\times 6^3$$

$$= \dfrac{2}{3}\pi \times 6\times 6\times 6=144\pi$$