Area Related to Circles Test

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Area Related to Circles Test - 22

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Weekly Quiz Competition

Question 1
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If the circumference of a circle is $$\displaystyle \frac{30}{\pi }$$ then the diameter of the circle is

A
$$\displaystyle 60\pi $$

B
$$\displaystyle \frac{15}{\pi }$$

C
$$\displaystyle \frac{30}{\pi^{2} }$$

D
$$30$$

Solution

$$\displaystyle 2\pi r=\frac{30}{\pi }$$
$$\displaystyle 2 r=\frac{30}{\pi^{2} }$$
Question 2
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The radius of a circle is $$14$$ m, then the circumference of a circle is

A
$$616$$ m

B
$$88$$ m

C
$$154$$ m

D
None of the above

Solution

$$r = 14 $$m
Circumference $$ = 2\displaystyle \pi r$$
$$\displaystyle =2\times \frac{22}{7}\times 14= 88m$$
Question 3
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The area of the sector of a circle whose radius is 6 m when the angle at the centre is $$\displaystyle 42^{\circ}$$ is

A
$$\displaystyle 13.2\:m^{2}$$

B
$$\displaystyle 14.2\:m^{2}$$

C
$$\displaystyle 13.4\:m^{2}$$

D
$$\displaystyle 14.4\:m^{2}$$

Solution

Area of sector $$\displaystyle =\frac{42}{360}\times \pi r^{2}$$
$$\displaystyle =\frac{42}{360}\times \frac{22}{7}\times6\times6=13.2m^{2}$$
Question 4
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Directions For Questions

In the given figure, $$ABCD$$ is a rectangle inscribed in a circle. If two adjacent sides of the rectangle be $$8\ cm$$ and $$6\ cm$$, calculate
[Take $$\pi \, =\, 3.14$$]

...view full instructions

The area of the shaded region

A
$$29\ cm^2$$

B
$$30.5\ cm^2$$

C
$$25\ cm^2$$

D
$$32.5\ cm^2$$

Solution

Given,
$$AB=8$$ cm and $$BC=6$$ cm
The diagonal of the inscribed rectangle will be the diameter of the circle.
Thus, by Pythagoras Theorem, we get
$$(AC)^2=(AB)^2+(AC)^2$$
$$=8^2+6^2$$
$$=64+36$$
$$=100$$
$$=>AC=\sqrt{100}$$
$$=>AC=10$$
Thus, radius $$= \frac{10}{2}$$
$$=5$$ cm
Area of the rectangle ABCD $$=8\times 6$$
$$=48cm^2$$
Area of the circle $$=\pi (5)^2$$
$$=3.14\times 25$$
$$=78.5 cm^2$$
Thus, the area of the shaded portion is $$=(78.5-48)cm^2$$
$$=30.5 cm^2$$
Question 5
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The area of a segment of a circle of radius $$21$$ cm if the arc of the segment has a measure of $$60^{\circ}$$ is
(Take $$\sqrt 3\, =\, 1.73$$)

A
$$45.27$$ sq. cm

B
$$40.27$$ sq. cm

C
$$40.8$$ sq. cm

D
none of these

Solution

Area of sector $$OAB$$ $$=\, \displaystyle \frac {x}{360}\, \times\, \pi r^2$$
$$= \displaystyle \frac {60}{360}\, \times\, \displaystyle \frac {22}{7}\, \times\, 21\, \times\, 21$$
$$=231$$ sq. cm
Here, the triangle ABC is an equilateral triangle, so the area of $$\triangle\, OAB\, =\, \displaystyle \frac {\sqrt 3}{4}\, r^2$$
$$= \displaystyle \frac {\sqrt 3}{4}\,\times\, 21\, \times\, 21$$
$$=190.73$$ sq. cm
Therefore, area of shaded region $$=\, 231\, -\, 190.73$$
$$=\, 40.27$$ sq. cm
Question 6
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The sides of a triangle are $$5$$, $$12$$ and$$ 13$$ units. A rectangle of width $$10$$ units is constructed equal in area to the area of the triangle. Then the perimeter of the rectangle is

A
30 units

B
26 units

C
13 units

D
15 units

Solution

By Pythagoras theorem, we find that the given triangle is a right angled triangle with $$12$$ as height and $$5$$ as base.
$$\displaystyle \therefore $$ Area of the triangle $$\displaystyle =\frac{1}{2}\times12\times5sq. units$$
$$=30$$ sq. units
$$\displaystyle \therefore $$ Area of the rectangle $$= length \times breadth = 30$$
$$\displaystyle \Rightarrow Length =\frac{30}{breadth}=\frac{30}{10}=3\, units$$
$$\displaystyle \therefore$$ Perimeter of the rectangle $$ = 2 \times ( 10 + 3 ) $$Units
$$= 26$$ units.
Question 7
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A square sheet of paper is converted into a cylinder by rolling it along its length. What is the ratio of the base radius to side of the square ?

A
$$\displaystyle \frac{1}{2\pi}$$

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

C
$$\displaystyle \frac{1}{\sqrt{2\pi }}$$

D
$$\displaystyle \frac{1}{\pi}$$

Solution

Let length of each side of the square $$=a$$
Base radius of cylinder $$=r$$
Surface area of sheet $$={ a }^{ 2 }$$
Surface area of cylinder $$=2\pi rh$$
But height $$h=a$$, because the square sheet is rolled it along its length.
Thus, surface area of cylinder $$=2\pi ra$$
Therefore, $${ a }^{ 2 }=2\pi ra\Rightarrow a=2\pi r\Rightarrow \dfrac { r }{ a } =\dfrac { 1 }{ 2\pi } $$
Question 8
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In the given figure , AC is diameter of a circle with radius 5cm. If AB = BC and CD = 8 cm, the area of the shaded region to the nearest whole number is

A
$$\displaystyle 28 \ cm^{2}$$

B
$$\displaystyle 29\ cm^{2}$$

C
$$\displaystyle 30\ cm^{2}$$

D
$$\displaystyle 45\ cm^{2}$$

Solution

Given that, AC is the diameter of a circle with radius $$5 \ cm$$
and $$AB=BC, CD=8\ cm$$
$$\therefore \ AC=10\ cm$$
Area of the circle $$=\pi (5)^2$$$$=25\pi \ cm^2$$ $$[\because \ \text{Area of a circle}=\pi r^2 \text{ square units}]$$

We know that, angle subtended by the diameter of a circle at any point on the circle is a right angle.
Hence, $$\triangle ABC$$ is a right angled triangle with $$\angle B=90^o$$
$$\therefore AC^2=AB^2+BC^2$$ [Pythagoras theorem]
$$\Rightarrow100=2AB^2$$
$$\Rightarrow AB^2=50$$
$$\Rightarrow AB=\sqrt{50}$$
$$\Rightarrow AB=5\sqrt2$$
$$\therefore AB=BC=5\sqrt2$$
Area of $$\triangle ABC$$$$=\cfrac{1}{2}\times AB\times BC$$
$$\Rightarrow \cfrac{1}{2}\times 5 \sqrt2 \times 5 \sqrt2$$$$=25 \ cm^2$$
Similarly,
$$\triangle ADC$$ is a right angled triangle with $$\angle D=90^o$$
$$\therefore AC^2=AD^2+DC^2$$
$$\Rightarrow 10^2=AD^2+8^2$$
$$\Rightarrow 100=AD^2+64$$
$$\Rightarrow AD^2=36$$
$$\Rightarrow AD=6$$
Area of $$\triangle ADC$$$$=\cfrac{1}{2}\times AD\times DC$$
$$\Rightarrow \cfrac{1}{2}\times 6\times 8$$$$=24 \ cm^2$$
Area of shaded region $$=25\pi-25-24$$
$$\Rightarrow 25\pi -49=29.5 \ cm^2$$
The nearest whole number of $$29.5$$ is $$30$$.
Hence, the area of the shaded region to the nearest whole number is $$30 \ cm^2$$
Question 9
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A right angled isosceles triangle is inscribed in a circle of radius $$ r$$. What is the area of the remaining portion of the circle?

A
$$\displaystyle \frac{\pi r^{2}}{2}$$

B
$$\displaystyle \left ( \pi -\frac{1}{2} \right )r^{2}$$

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

D
$$\left ( \pi -2 \right )r^{2}$$

Solution

Let the sides containing the right angle be $$a$$ units each and $$r$$ be the radius of the circle.
The hypotenuse of the right-angled triangle is the diameter of the circle. So, by applying Pythagoras theorem,
$$a^{2}+a^{2}=\left ( 2r \right )^{2}$$
$$\displaystyle \Rightarrow 2a^{2}=4r^{2}$$
$$\Rightarrow a=\sqrt{2} r $$

$$\therefore $$ Remaining area $$=$$ Area of circle $$-$$ Area of triangle
$$A=\pi r^{2}-\dfrac{1}{2}\left ( r\sqrt{2} \right )\left ( r\sqrt{2} \right )$$
$$=\pi r^{2}-r^{2}$$
$$=r^{2}\left ( \pi - 1 \right )$$

Hence, area of the remaining region is equal to $$(\pi-1)r^2.$$
Question 10
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In the diagram shown, $$ABCD$$ is a parallelogram and $$CDEF$$ is a rectangle. The total area of the whole figure is

A
$$35\, cm^2$$

B
$$30\, cm^2$$

C
$$25\, cm^2$$

D
$$20\, cm^2$$

Solution

Total area of the whole diagram $$=$$ Area of parallelogram $$ABCD +$$ Area of rectangle $$CDEF$$

Area of parallelogram$$ABCD=$$ base$$\times $$height
$$=5\times 3\ cm^2$$
$$=15\ cm^2$$

Area of rectangle $$CDEF=2\times 5\ cm^2$$
$$=10\ cm^2$$

Let $$A$$ be the total area of figure,

$$A=(15+10)\ cm^2$$
$$=25\ cm^2$$

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Re-Attempt Weekly Quiz Competition

Area Related to Circles Test - 22

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Area Related to Circles Test - 22

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Question 1

$$\displaystyle 60\pi $$

$$\displaystyle \frac{15}{\pi }$$

$$\displaystyle \frac{30}{\pi^{2} }$$

$$30$$

Solution

Question 2

$$616$$ m

$$88$$ m

$$154$$ m

None of the above

Solution

Question 3

$$\displaystyle 13.2\:m^{2}$$

$$\displaystyle 14.2\:m^{2}$$

$$\displaystyle 13.4\:m^{2}$$

$$\displaystyle 14.4\:m^{2}$$

Solution

Question 4

$$29\ cm^2$$

$$30.5\ cm^2$$

$$25\ cm^2$$

$$32.5\ cm^2$$

Solution

Question 5

$$45.27$$ sq. cm

$$40.27$$ sq. cm

$$40.8$$ sq. cm

none of these

Solution

Question 6

30 units

26 units

13 units

15 units

Solution

Question 7

$$\displaystyle \frac{1}{2\pi}$$

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

$$\displaystyle \frac{1}{\sqrt{2\pi }}$$

$$\displaystyle \frac{1}{\pi}$$

Solution

Question 8

$$\displaystyle 28 \ cm^{2}$$

$$\displaystyle 29\ cm^{2}$$

$$\displaystyle 30\ cm^{2}$$

$$\displaystyle 45\ cm^{2}$$

Solution

Question 9

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

$$\displaystyle \left ( \pi -\frac{1}{2} \right )r^{2}$$

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

$$\left ( \pi -2 \right )r^{2}$$

Solution

Question 10

$$35\, cm^2$$

$$30\, cm^2$$

$$25\, cm^2$$

$$20\, cm^2$$

Solution

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