Area Related to Circles Test

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Area Related to Circles Test - 29

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Weekly Quiz Competition

Question 1
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Area of figure as shown

A
$$12\ \text{m}^{2}$$

B
$$13\ \text{m}^{2}$$

C
$$14\ \text{m}^{2}$$

D
$$15\ \text{m}^{2}$$

Solution

Area of rectangle $$( ABCD)=5\times 1=5m^{2}$$
Area of rectangle $$( EFGH)=7\times 1=7m^{2}$$
Total = $$7+5=12m^{2}$$
Question 2
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In the given figure, $$ABC$$ is a quadrant of a circle of radius $$14\ cm$$ and a semicircle is drawn with $$BC$$ as diameter. Find the area of the shade region.

A
$$98cm^2$$

B
$$38cm^2$$

C
$$48cm^2$$

D
$$58cm^2$$

Solution

Area of shaded region=area of semicircle of diameter BC-{area of quadrant of radius AB/AC- area of $$\triangle$$ABC}
$$\because$$ BC is hypotenuse of right angle $$\triangle ABC$$
here $$AB=BC=14$$

So, $$BC=14\sqrt2=2\times radius\Rightarrow radius=7\sqrt2$$

So, Area of semicircle of diameter BC$$=\dfrac{\pi r^2}{2}$$

$$=\dfrac{1}{2}\times\dfrac{22}{7}\times(7\sqrt2)^2$$

$$=154cm^2$$

Area of quadrant of radius AB/AC$$=\dfrac{\pi r^2}{4}$$

$$=\dfrac{1}{4}\times\dfrac{22}{7}\times14\times14$$

$$=154cm^2$$

Area of $$\triangle ABC=\dfrac{1}{2}\times h\times b$$

$$=\dfrac{1}{2}\times14\times14=98cm^2$$

Now, area of shaded region$$=154-\{154-98\}=98cm^2$$
Hence, area of shaded region$$=98cm^2$$
Question 3
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From the above figure area of shade region is

A
$$42\ cm^{2}$$

B
$$196\ cm^{2}$$

C
$$154\ cm^{2}$$

D
$$350\ cm^{2}$$

Solution

Area of $$ABCD$$ $$={14}^{2}=196$$ sq.cm

Area of $$APBA$$

$$\dfrac{\pi{d}^{2}}{{8}}=\dfrac{\pi{(14)}^{2}}{8}$$ sq.cm

$$=76.97$$ sq.cm

Area of $$CPDC$$ $$=$$ Area of $$APBA$$

$$=76.97$$ sq.cm

$$\therefore$$ Area of the shaded region

$$=196-\left(2\times76.97\right)$$ sq.cm

$$=42.06$$ sq.cm

$$\sim 42$$ sq.cm

$$Note : \left[sq.cm={cm}^{2}\right]$$
Question 4
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The area of small square is $$100\ sq.\ cm$$. The area of the large square in $$sq.\ cm$$ is

A
$$144$$

B
$$165$$

C
$$200$$

D
$$400$$

Solution

Small square area $$=100cm^2$$
$$s^2=100\ cm^2$$
$$s=10\ cm$$
Diagonal of square $$=10\sqrt{2}$$
Half the diagonal length$$=$$radius of circle $$=\dfrac{10\sqrt{2}}{2}$$
$$=5\sqrt{2}$$
Diameter of circle$$=$$length of side of big square=diagonal of smaller square =$$ 10\sqrt{2}$$
Area of big square$$=(10\sqrt{2})^2$$
$$=200\ cm^2$$.
Question 5
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If the perimeter and the area of a circle are numerically equal, then the radius of the circle is

A
$$2\ units$$

B
$$p\ units$$

C
$$4\ units$$

D
$$7\ units$$

Solution

Given,
Numerically perimete of the circle is equal to area of circle

$$\Rightarrow 2\pi r=\pi r^2$$

$$\Rightarrow 2=r$$

$$\therefore r=2$$
Question 6
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In the given figure $$ABCDE$$ is a pentagonal park in which $$DE=DC,AB=BC=CE=EA=25m$$ and its total height is $$41m$$. Find the area of the park

A
$$225cm^2$$

B
$$525cm^2$$

C
$$625cm^2$$

D
$$825cm^2$$

Solution

Height of triangle $$=41-25 =16cm $$
Area of park $$=$$ area of square $$ABCE \,+$$ area of $$\Delta DEC$$
$$= 25\times 25+\dfrac{1}{2}\times 25 \times 16 $$
$$=625+200$$
$$=825cm^{2}$$
Question 7
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The length of the are that subtends an angle $${30}^{o}$$ at the centre of a circle of radius $$4\ cm$$:

A
$$\dfrac {2\pi}{3}\ cm$$

B
$$\dfrac {4\pi}{8}\ cm$$

C
$$\dfrac {5\pi}{4}\ cm$$

D
$$\dfrac {\pi}{3}\ cm$$

Solution

Length of arc $$ = (\dfrac{\theta }{360}\times 2\pi r)\,cm $$
$$ \therefore $$ Length $$ = (\dfrac{30}{360}\times 2\times \dfrac{22}{7}\times 4)\,cm $$
$$ \therefore $$ Length $$ = (\dfrac{1}{6}\times \dfrac{22\times 4}{7})\,cm $$
$$ \therefore $$ Length $$ = \dfrac{44}{21}\,cm $$
$$ \therefore $$ Length $$ = 2.09\,cm $$
$$ = (\dfrac{2\pi }{3})\,cm $$
Question 8
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Find the area of the shaded region.

A
$$5634cm^2$$

B
$$1004cm^2$$

C
$$1034cm^2$$

D
$$1134cm^2$$

Solution

$$r+2r+r=42$$

$$4r=42$$

$$r=\dfrac{21}{2}cm$$

Area of shaded region = $$4\times \dfrac{\pi r^2}{2}+(2r)(2r)$$

$$=2\pi r^2+4r^2$$

$$2\times \dfrac{22}{7}\times \dfrac{21}{2}\times \dfrac{21}{2}+4\times\dfrac{21}{2}\times \dfrac{21}{2}$$

$$=33\times 21+21\times 21$$

$$=54\times 21$$

$$=1134 cm^2$$
Question 9
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The area of shaded region of the figure given below is __________. [Take $$\pi = \dfrac {22}{7}$$ and $$\sqrt {3} = 1.732]$$.

A
$$4.28\ cm^{2}$$

B
$$5.76\ cm^{2}$$

C
$$3.27\ cm^{2}$$

D
$$18.84\ cm^{2}$$

Solution

Area of sector $$OAB = \dfrac{\theta}{360^{\circ} } \times \pi r^2$$

where $$\theta = \angle AOB$$

Since $$\Delta AOB $$ is an equilateral triangle
$$\therefore \angle AOB = 60^{\circ}$$

$$\therefore$$ Area of sector $$OAB = \dfrac{60^{\circ}}{360^{\circ}} \times \pi \times 6^2$$

$$= \dfrac{1}{6} \times \dfrac{22}{7} \times 6^2$$

$$= 18.85 cm^2$$

Area of equilateral triangle $$OAB$$

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

$$= \dfrac{\sqrt{3}}{4} \times 6 \times 6$$

$$= 9 \sqrt{3} cm^2$$

$$= 15.58 cm^2$$

Area of shaded region $$= (18.85 - 15.58) cm^2$$
$$= 3.27 cm^2$$

Answer: $$C) \, 3.27 cm^2$$
Question 10
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A chord of a circle of radius $$6\ cm$$ subtends a right angle at centre. The area of the minor segment $$(\pi=3.14)$$ is____.

A
$$9.82\ cm^{2}$$

B
$$10.14\ cm^{2}$$

C
$$10.26\ cm^{2}$$

D
$$12.35\ cm^{2}$$

Solution

$$\text{Area of the minor segment}$$ $$= \text{Area}(Sector\; OACB) - \text{Area}(\triangle AOB)$$

$$= \dfrac{90^{\circ}}{360^{\circ}} \times \pi \times r^2 - \dfrac{1}{2} \times OA\times OB$$

$$= \dfrac{90^{\circ}}{360^{\circ}} \times 3.14 \times6^2 - \dfrac{1}{2} \times 6\times 6$$

$$= 28.26 - 18$$

$$= 10.26 \text{ cm}^2 $$

Option C is correct.