Area Related to Circles Test

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Area Related to Circles Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

The geometric shape that has no corners is _____.

A
square

B
circle

C
rectangle

D
hexagon

Solution

The geometric shape which has no corners is circle.
Question 2
1 / -0

A shuttlecock used for playing badminton has the shape of a combination of

A
a cone and a sphere

B
frustum of a cone and a hemisphere

C
a cylinder and a hemisphere

D
a cone and a cylinder

Solution

A shuttlecock used for playing badminton has the shape of a combination of frustum of a cone and a hemisphere
Question 3
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The figure below shows two concentric circles with centre $$O$$. $$PQRS$$ is a square inscribed in the outer circle. It also circumscribes the inner circle, touching it at point $$B, C, D$$ and $$ A$$. The ratio of the perimeter of the outer circle to that of polygon $$ABCD$$ is

A
$$\displaystyle \frac{\pi}{4}$$

B
$$\displaystyle \frac{3\pi}{4}$$

C
$$\displaystyle \frac{\pi}{2}$$

D
$$\pi$$

Solution

Let the radius of the outer circle be r.
$$\therefore$$ perimeter of outer circle = $$2\pi r$$
But $$OQ = BC = r$$ [diagonals of the square BQCO]
$$\therefore$$ Perimeter of Square ABCD $$= 4r $$
Hence, required ratio $$= \dfrac{2\pi r}{4r}=\dfrac{\pi}{2}$$
Question 4
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If a sector of a circle of diameter $$21\ cm$$ subtends an angle of $$120^{\circ}$$ at the centre, then what is its area ?

A
$$115.5\ cm^2$$.

B
$$84\ cm^2$$.

C
$$85.5\ cm^2$$.

D
$$78\ cm^2$$.

Solution

Area of sector $$=\cfrac{\theta}{360} \times \pi \times r^2$$

$$=\cfrac{120}{360} \times \pi \times (\cfrac{21}{2})^2$$

Thus area $$=\cfrac{1}{3} \times \cfrac{22}{7} \times \cfrac{441}{4} = 115.5\ cm^2$$
Question 5
1 / -0

Four circular cardboard pieces of radii $$7 cm$$ are placed on a paper in such a way that each piece touches other two pieces. The area of the region enclosed between these pieces is

A
$$42$$ $$cm^2$$

B
$$21$$ $$cm^2$$

C
$$84$$ $$cm^2$$

D
$$96$$ $$cm^2$$

Solution

The diameter of circle =$$2\times 7=14$$ cm
So, the side of square is 14cm.
Then area of square =$$a^{2}=(14)^{2}= 196 cm^{2}$$
& Area of quadrant $$ = \dfrac{1}{4} $$ × area of circle
Area of space enclosed by the circles $$=$$ Area of square of side $$14$$ cm $$– 4$$(Area of quadrant of radius $$7$$ cm)

$$=14^2-4\times\dfrac{1}{4}\times \dfrac{22}{7}\times 7\times 7$$

$$=196-154=42$$ $$cm^2$$
Question 6
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What is the area of the sector A' OB' if radius of circle is equal to $$6$$ cm and $$\angle AOB =60^{\circ}$$.

A
$$6 \pi cm^2$$

B
$$\pi cm^2$$

C
$$6cm^2$$

D
$$(6+\pi)cm^2$$

Solution

$$\angle AOB =60^{\circ}$$
Area of the sector A'OB $$\displaystyle =\dfrac {\theta }{360} \pi r^2=\frac{60}{360}\pi (6)^2 =6 \pi cm^2$$
Question 7
1 / -0

The area of the shaded region in the given

figure is:

A
$$ 10-2\pi$$

B
$$5-\frac{\pi}{2}$$

C
$$ 5-2\pi$$

D
$$ 10-\frac{\pi}{2}$$
Question 8
1 / -0

State true or false:
A circle of radius $$3$$cm can be drawn through two points $$A, B$$ such that $$AB = 6$$ cm.

A
False

B
True

C
Cannot be determined

D
None of the above

Solution

We know, diameter of a circle $$=2\times$$ radius of the circle.
A circle with $$AB=6cm$$ as diameter will have its radius equal to $$3cm$$.
So option $$B$$ is the right answer.
Question 9
1 / -0

Find the area of a sector of a circle of radius $$28\ cm$$ and central angle $$45^{\circ}$$.

A
$$616\ cm^{2}$$

B
$$308\ cm^{2}$$

C
$$508\ cm^{2}$$

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

Solution

Radius of sector $$r=28\ cm$$
Central angle $$\theta=45^{ \circ }$$
Area of sector $$=\cfrac { \theta }{ 360° } \times \pi { r }^{ 2 }$$
$$=\cfrac { 45° }{ 360° } \times \cfrac { 22 }{ 7 } \times 28\times 28\\ =308\quad { cm }^{ 2 }$$
Question 10
1 / -0

If the sector of a circle of diameter $$10\ \text{cm}$$ subtends an angle of $$144^{\circ}$$ at the centre, then the length of the arc of the sector is

A
$$2\pi \ \text{cm}$$

B
$$4\pi \ \text{cm}$$

C
$$5\pi \ \text{cm}$$

D
$$6\pi \ \text{cm}$$

Solution

Given, diameter $$=10$$ cm, $$\theta=144^o$$
Length of an arc of a circle $$=\dfrac { \theta }{ 360 } \times 2\pi { r }=\dfrac { 144 }{ 360 } \times 2\pi \times \dfrac { 10 }{ 2 } =4\pi $$ cm
Hence, option B is correct.

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Re-Attempt Weekly Quiz Competition

Area Related to Circles Test - 18

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Area Related to Circles Test - 18

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SHARING IS CARING

Question 1

square

circle

rectangle

hexagon

Solution

Question 2

a cone and a sphere

frustum of a cone and a hemisphere

a cylinder and a hemisphere

a cone and a cylinder

Solution

Question 3

$$\displaystyle \frac{\pi}{4}$$

$$\displaystyle \frac{3\pi}{4}$$

$$\displaystyle \frac{\pi}{2}$$

$$\pi$$

Solution

Question 4

$$115.5\ cm^2$$.

$$84\ cm^2$$.

$$85.5\ cm^2$$.

$$78\ cm^2$$.

Solution

Question 5

$$42$$ $$cm^2$$

$$21$$ $$cm^2$$

$$84$$ $$cm^2$$

$$96$$ $$cm^2$$

Solution

Question 6

$$6 \pi cm^2$$

$$\pi cm^2$$

$$6cm^2$$

$$(6+\pi)cm^2$$

Solution

Question 7

$$ 10-2\pi$$

$$5-\frac{\pi}{2}$$

$$ 5-2\pi$$

$$ 10-\frac{\pi}{2}$$

Question 8

False

True

Cannot be determined

None of the above

Solution

Question 9

$$616\ cm^{2}$$

$$308\ cm^{2}$$

$$508\ cm^{2}$$

$$154\ cm^{2}$$

Solution

Question 10

$$2\pi \ \text{cm}$$

$$4\pi \ \text{cm}$$

$$5\pi \ \text{cm}$$

$$6\pi \ \text{cm}$$

Solution

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