Area Related to Circles Test

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Area Related to Circles Test - 28

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Weekly Quiz Competition

Question 1
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Consider the above figure. Find the area in the given rectangle excluding the $$6$$ identical semicircles.

A
$$144\ cm^{2}$$

B
$$126\ cm^{2}$$

C
$$195\ cm^{2}$$

D
$$243\ cm^{2}$$

Solution

Required area$$ =$$ Area of rectangle $$- 6$$ $$\times $$ area of $$1$$ semi-circle
$$=$$ $$42\times 14-6\times \dfrac { 1 }{ 2 } \times \dfrac { 22 }{ 7 } \times 7\times 7$$

$$ = 588 - 462 = 126$$$${ cm }^{ 2 }$$
Question 2
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What is the length of an arc of a circle with a radius of $$5$$ if it subtends an angle of $${60}^{o}$$ at the center?

A
$$3.14$$

B
$$5.24$$

C
$$10.48$$

D
$$2.62$$

Solution

Given:
Radius (r)$$=5$$
Angle $$=60^o$$

Arc length for a particular angle we can write as -
$$=\dfrac{\theta}{360}\times (2\pi r)$$

$$=\dfrac{60}{360}\times 2\pi \times 5$$

$$=\dfrac{10\pi}{6}=5.24$$

Option 'B'.
Question 3
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When the figure above is spun around its vertical axis, the total surface area of the solid formed will be

A
$$144\pi$$

B
$$108\pi$$

C
$$72\pi$$

D
$$36\pi$$

E
$$9\pi + 12$$

Solution

Figure formed when the figure is rotated around the vertical axis is a hemisphere.
$$\therefore$$ The total surface area of the figure is the area of the hemisphare.
Given that the radius $$r=6$$
area of the solid (A) $$=2\pi r^2+\pi ^2$$
$$3\pi r^2$$
$$3\times \pi \times 36=108 \pi$$
Question 4
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A memento is made as shown in the figure. Find the area of the region in the triangle excluding the part cut by the circle.

A
$$138.20\ cm^{2}$$

B
$$161.5\ cm^{2}$$

C
$$133.567\ cm^{2}$$

D
$$108.97\ cm^{2}$$

Solution

Required area $$=$$ Area of triangle $$-$$ Area of sector
$$=\dfrac { 1 }{ 2 } \times 20\times 20-\dfrac { { 90 }^{ 0 } }{ { 360 }^{ 0 } } \times \dfrac { 22 }{ 7 } \times 7\times 7$$

$$=\dfrac { 400 }{ 2 } -\dfrac { 77 }{ 2 } =\dfrac { 323 }{ 2 } $$

$$= 161.5$$ $${ cm }^{ 2 }$$
Question 5
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ASB is a quarter circle. PQRS is a rectangle with sides PQ$$=8$$ and PS$$=6$$. What is the length of the arc AQB?

A
$$5\pi$$

B
$$10\pi$$

C
$$25$$

D
$$14$$

E
$$28$$

Solution

Given, $$PQ=8$$ $$PS=6$$
From figure, $$\triangle PQS$$ is a right angled triangle.
So, using Pythagoras theorem,
$$PS^2=PQ^2+PS^2$$
$$\Rightarrow PS^2=8^2+6^2=64+36=100$$
$$PS\sqrt{100}=10$$
NOw, QS is also the radius of are AQB.
We known that formula for length of are $$=2\pi r\left(\dfrac{\theta}{360}\right)$$
where $$r=radius \, of \, arc$$
$$\theta=$$ angle subtended by the arc.
Here, $$r=10$$ and $$\theta=90$$
$$\therefore$$ Arc length $$AQB=2\times \pi\times 10\times \left(\dfrac{90}{360}\right)$$
$$2\times\pi \times 10\times\dfrac{1}{4}=5\pi$$
Hence the length of arc is $${5\pi}$$
Question 6
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A show-piece, as shown in the figure, is made of a cube and a hemisphere. If the measure of the total surface area of the cube is represented by $$A$$, the curved surface area of the hemisphere is represented by $$B$$ and the area of the base of the hemisphere is represented by $$C$$, then ................ is true for the total surface area of the show-piece.

A
$$A+B+C$$

B
$$A+B-C$$

C
$$B+C-A$$

D
$$A+C-B$$

Solution

We know that the total surface area of the showpiece is obtained by adding the surface area of every part.

The surface area of the cube and hemisphere will include the area of the base of the hemisphere which is extra.

So, to get the total surface area, we need to subtract the area of the base of the hemisphere from that of the hemisphere.

$$\therefore$$ Total surface Area $$=$$ TSA of cube $$+$$ CSA of hemisphere $$-$$ (Area of base of the hemisphere)

$$=A+B-C$$
Question 7
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$$ABCD$$ is a rectangle having length $$30\ cm$$ and breadth $$25\ cm, P, Q, R, S$$ are midpoints of $$AB, BC, CD$$ and $$AD$$ respectively. Find the area of the shaded region of the figure.

A
$$375\ cm^{2}$$

B
$$370\ cm^{2}$$

C
$$475\ cm^{2}$$

D
$$450\ cm^{2}$$

Solution

Area of shaded region $$=$$ Area of rectangle $$-$$ Area of all $$4$$ triangles

A$$_{1} =$$ Area of rectangle $$= 30\times25 = 750\ cm^{2}$$
A$$_{2} =$$ Area of triangle $$\triangle SDR = 93.75\ cm^{2}$$
A$$_{3} =$$ Area of triangle $$\triangle RCQ = 93.75\ cm^{2}$$
A$$_{4} =$$ Area of triangle $$\triangle QBP = 93.75\ cm^{2}$$
A$$_{5} =$$ Area of triangle $$\triangle PAS = 93.75\ cm^{2}$$

Area of shaded region $$= 750-4\times93.75 = 375cm^{2}$$
Question 8
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Which of the following figures has $$10$$ vertices?

A
$$U$$

B
$$V$$

C
$$W$$

D
$$X$$

Solution

Number of vertices in figure $$U = 5$$
Number of vertices in figure $$V = 6$$
Number of vertices in figure $$W = 8$$
Number of vertices in figure $$X = 10$$.
Question 9
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Find the area of the following figure (not drawn to scale) by splitting them into rectangles and squares.

A
$$14$$ sq. cm

B
$$15$$ sq. cm

C
$$17$$ sq. cm

D
$$16$$ sq. cm
Question 10
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Match the figure is Column-I with their shaded areas given the Column-II.

A
$$(P) \rightarrow (3), (Q) \rightarrow (1), (R) \rightarrow (2), (S)\rightarrow (4)$$

B
$$(P) \rightarrow (1), (Q) \rightarrow (3), (R) \rightarrow (3), (S)\rightarrow (4)$$

C
$$(P) \rightarrow (2), (Q) \rightarrow (3), (R) \rightarrow (1), (S)\rightarrow (4)$$

D
$$(P) \rightarrow (3), (Q) \rightarrow (1), (R) \rightarrow (4), (S)\rightarrow (2)$$

Solution

(P) Area of $$\triangle ABC$$
$$= \dfrac {1}{2}\times 25\times 18 = 225\ cm^{2}$$
$$=$$ Area of $$\triangle BDC = \dfrac {1}{2}\times 18 \times 6 = 54\ cm^{2}$$
$$\therefore$$ Area of shaded region $$= 225 - 54$$
$$= 171\ cm^{2}$$

(Q) Area of shaded region
$$= \dfrac {1}{2}\times AB \times AE + \dfrac {1}{2}\times AB \times DE$$ {Because ED parallel to BC therefore height is AB}
$$= \dfrac {1}{2}\times AB \times (AE + DE) = \dfrac {1}{2} \times 6\times 18 = 54\ cm^{2}$$

(R) Area of shaded region$$=$$ Area of $$\triangle AED +$$ Area of square $$ABCD -$$ Area of circle
$$= \dfrac {1}{2}\times AD \times ED + DC \times DC - \pi \times (7)^{2}$$
$$= \dfrac {1}{2} \times 24\times 24 + (24)^{2} - \dfrac {22}{7} \times (7)^{2}$$
$$288 + 576 - 154 = 710\ cm^{2}$$

(S) Shaded area $$=$$ Area of rectangle $$ABCD -$$ Area of rectangle $$EFGH$$
$$= AB\times AD - EF \times FG$$
$$= (18 + 3) \times (9 + 3) - 18\times 9$$
$$= 252 - 162 = 90\ cm^{2}$$.