Area Related to Circles Test

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Area Related to Circles Test - 21

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Question 1
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A square is inscribed in a circle of radius $$7\: \text{cm}$$. Find area of the square.

A
$$98 \: \text{cm}^{2}$$

B
$$97 \: \text{cm}^{2}$$

C
$$91 \: \text{cm}^{2}$$

D
$$90 \: \text{cm}^{2}$$

Solution

Given: Radius of the circle $$=7\:\text{cm}.$$
Let the side of the square be $$a\:\text{cm}.$$
A square when inscribed in a circle then the diameter of the circle must be equal to diagonal of the square.
Therefore,
$$\begin{aligned}{}\text{Diagonal of square } &= \sqrt {{a^2} + {a^2}} \\& = a\sqrt 2 \end{aligned}$$

Now,
$$\begin{aligned}{}\text{Diagonal of square } &= 2\times 7 \\& = 14 \text{ cm}\end{aligned}$$

$$ \Rightarrow \sqrt 2 \cdot a=14$$
$$ \Rightarrow a=\dfrac{14}{\sqrt 2}$$
$$ \Rightarrow a=7\sqrt 2\: \text{cm}$$

Therefore,
$$\begin{aligned}{}\text{Area of square }& = {a^2}\\ &= {(7\sqrt 2\; {\rm{cm}})^2}\\ &= (7\sqrt 2\; {\rm{cm}})(7\sqrt 2\; {\rm{cm}})\\& = 98\,{\rm{c}}{{\rm{m}}^2}\end{aligned}$$
Question 2
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In a circle of radius 21 cm, an arc subtends an angle of $$60^{\circ}$$ at the centre. The area of the segment formed by the corresponding chord of the arc is :

A
42 $$cm^2$$

B
421.73 $$cm^2$$

C
429.43 $$cm^2$$

D
40 $$cm^2$$

Solution

Radius $$= 21\ cm$$
Angle subtended by the arc = $$60^{\circ}$$
Area of the sector = $$\dfrac{\theta}{360^{\circ}} \times \pi r^2$$

Area of the sector = $$\dfrac{60}{360} \pi (21)^2$$
Area of the sector = $$231$$ $$cm^2$$

Now, area of the triangle formed between the chord and the radius. Since, the angle subtended is $$60^{\circ}$$ and the other two angles are equal, (ngles opposite to he radius), which will be gain $$60^{\circ}$$. The triangle thus formed is an equilateral triangle.
Hence, area = $$\dfrac{\sqrt{3}}{4} s^2$$
Area of triangle = $$\dfrac{\sqrt{3}}{4} (21)^2$$ = $$190.95$$ $$cm^2$$
Thus, area of the segment = area of the sector - area of triangle
Area of segment = $$231- 190.95=$$ $$40 cm^2$$
Question 3
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In the given figure, the area of the shaded portion APB is ?

A
$$\displaystyle \frac{1}{4}\, \pi r^2$$

B
$$\displaystyle \frac{1}{4}\, (\pi\, -\,2) r^2$$

C
$$\displaystyle \frac{1}{4}\, (\pi\, -\,1) r^2$$

D
None of these

Solution

Angle subtended by the arc = $$90^{\circ}$$
Area of the sector = $$\dfrac{\theta}{360^{\circ}} \times \pi r^2$$

Area of the sector = $$\dfrac{90}{360} \pi r^2$$

Area of the sector = $$\dfrac{\pi r^2}{4}$$ $$cm^2$$

Area of triangle = $$\dfrac{1}{2} OA \times OB$$

Area of triangle = $$\dfrac{1}{2} r \times r$$ = $$\dfrac{r^2}{2}$$
Thus, area of the segment = area of the sector - area of triangle

Area of segment = $$\dfrac{\pi r^2}{4} - \dfrac{r^2}{2}$$ = $$\displaystyle \dfrac{1}{4}\, (\pi\, -\,2) r^2$$.
Question 4
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In the figure, an equilateral triangle of side 6 cm and its circumcircle is shown. Find the area of shaded portion. Take $$\displaystyle (\pi = 3.14, \sqrt3 = 1.73)$$

A
$$22.11 \ cm^2$$

B
$$22 \ cm^2$$

C
$$21.11 \ cm^2$$

D
$$23.11\ cm ^2$$
Question 5
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In the given figure, Find the shaded area (use $$\pi=\frac{22}{7}$$)

A
$$205.03\, cm^2$$

B
$$205.04\, cm^2$$

C
$$205.33\, cm^2$$

D
$$205.25\, cm^2$$

Solution

Radius = $$14\ cm$$
Angle subtended by the arc = $$120^{\circ}$$
Area of the sector = $$\dfrac{\theta}{360^{\circ}} \times \pi r^2$$

Area of the sector = $$\dfrac{120}{360} \pi (14)^2$$

Area of the sector = $$205.33$$ $$cm^2$$.
Question 6
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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 radius of the circle

A
$$6\ cm$$

B
$$7.5\ cm$$

C
$$4\ cm$$

D
$$5\ cm$$

Solution
Question 7
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A garden is in the shape of a square with a semicircle on one of its sides. The side of the square is $$7$$ m. What is the area of the garden in sq. m.?

A
$$\displaystyle 68\frac{1}{4}$$

B
$$49$$

C
$$\displaystyle 58\frac{5}{8}$$

D
$$70$$

Solution

The side of the square is $$7m$$.
Hence, the area of the square part of the garden $$=7^2 = 49$$ sq. m.

The semicircle lies on one of the sides of the square part of the garden.
So, its radius $$=\dfrac 72 = 3.5m$$

Area of the semi-circular part of garden $$=\dfrac 12 \times \pi r^2$$

$$=\dfrac 12 \times \dfrac {22}{7} \times \dfrac 72 \times \dfrac 72 $$

$$= \cfrac {77}{4}$$ sq. m.

Area of garden $$=49+\cfrac {77}{4} = 68\cfrac 14$$ sq. m
Question 8
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Which of the following statement is false?

A
If we join any two points on a circle we get a diameter of the circle

B
A diameter of a circle contains the centre of the circle

C
A semicircle is an arc

D
The length of a circle is called its circumference

Solution

$$\textbf{Step - 1: Verify, If we join any points on a circle we get a diameter of the circle}$$

$$\text{A Chord is a line segment that joins any two points of the circle.}$$

$$\text{The endpoints of this line segments lie on the circumference of the circle.}$$

$$\therefore \text{Option (A) is false}$$

$$\textbf{Step - 2: Verify, A diameter of a circle contains the center of the circle}$$

$$\text{Any interval joining two points on the circle and passing through the center is called a}$$
$$\text{diameter of the circle.}$$

$$\therefore \text{Option (B) is True}$$

$$\textbf{Step - 3: Verify, A semicircle is an arc}$$

$$\text{The arc of a circle consists of two points on the circle and all of the points on the circle that lie}$$
$$\text{between those two points.}$$

$$\text{It's like a segment that was wrapped partway around a circle.}$$

$$\text{An arc whose measure equals 180 degrees is called a semicircle since it divides the circle in two}$$

$$\therefore \text{Option (C) is True}$$

$$\textbf{Step - 4: Verify, the length of a circle is called its circumference}$$

$$\text{A Circle is a round closed figure where all its boundary points are equidistant from a fixed point}$$
$$\text{called the center.}$$

$$\text{The two important metrics of a circle is the area of a circle and the circumference of a circle.}$$

$$\therefore \text{Option (D) is True}$$

$$\textbf{Hence, option A is correct as it is false}$$
Question 9
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A chord of a circle of radius 10 cm subtends a right angle at its centre.The length of the chord (in cm) is.

A
$$5 \sqrt 2$$

B
$$10 \sqrt 2$$

C
$$ \frac {5}{\sqrt 2}$$

D
$$10 \sqrt 3$$

Solution

Consider a circle with centre at O, radius 10 cm. AB is the chord which subtends $$90^{\circ}$$ at the centre.
Now, In $$\triangle OAB$$,
$$OA^2 + OB^2 = AB^2$$ (Using Pythagoras Theorem)
$$10^2 + 10^2 = AB^2$$
$$AB^2 = 200$$
$$AB = 10\sqrt{2}$$
Question 10
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The perimeter of a square whose area is equal to that of a circle with perimeter $$\displaystyle 2\pi x$$ is

A
$$\displaystyle 2\pi x$$

B
$$\displaystyle \sqrt\pi x$$

C
$$\displaystyle 4\sqrt{\pi x}$$

D
$$\displaystyle 4x\sqrt\pi$$