MCQs on Areas Related to Circles

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A circle is a collection of all points in a plane at a fixed distance from a fixed position. This fixed point is known as the centre of the circle. A circle may also be defined as a closed figure with zero eccentricity and two coincident foci. Some important terms related to circle are:

Radius: The constant distance from the centre of the circle to its boundary is called radius.
Diameter: A straight line segment that passes through the centre of the circle is known as diameter of the circle or the chord that passes through the centre of a circle is called a diameter. The length of diameter is twice the length of radius.
Chord: The part of a line that connects any two points in a circle is called a chord.
Tangent: A tangent is any line that touches the circle at only one point. The tangent to the circle is perpendicular to the radius at the point of contact.
Area of Circle: The area of a circle is the region occupied by it. Mathematically, area of circle = πr².
Circumference of Circle: The circumference is the perimeter of the circle, i.e, the length of its boundary of the circle. Mathematically, C = 2πr.

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MCQs

Ques. A circular perimeter with a radius of 5 cm is equal to:

31.4 cm
37 cm
78 cm
24 cm

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Ans. a) 31.4 cm

Explanation: Given that the radius of the circle is 5 cm.

The perimeter of a circle is equal to the circumference of a circle. Mathematically, Perimeter = 2πr

Substituting the value of r, we get,

Perimeter of circle = 2 x 3.14 x 5 = 31.4 cm.

Ques. A 5cm wide circular area is equal to:

78.5 sq.cm
56 sq.cm
78 sq.cm
48 sq.cm

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Ans. a) 78.5 sq.cm

Explanation: Given that the radius of the circle is 5 cm.

We know that area of the circle = πr².

Substituting the value of r, we get,

Area of circle = 3.14 x 5 x 5 = 78.5 sq.cm

Ques. The largest triangle is inscribed on the semi-circle of radius r, and the location of that triangle is:

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Ans. a) r²

Explanation: The height of the largest written triangle will be equal to the radius of the semi-circle and the base will be equal to the width of the semi-circle.

The area of the triangle = ½ x base x height = ½ x 2r x r = r²

Ques. If the perimeter of a circle and a square are equal, the average of their locations will be equal:

14/11
24/11
34/22
24/22

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Ans. a) 14/11

Explanation: Given, Circular perimeter = square perimeter

2pr = 4a

a = πr / 2

Square area = a² = (πr / 2)²

Circle = πr² / (πr / 2)² = 14/11

Ques. It is proposed to construct a single circular park equal to the area and the total area of two circular parks with a width of 16 m and 12 m in area. The radius of the new park will be

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Ans. b) 10 m

Explanation: The Radii of the two circular parks will be:

R₁ = 16/2 = 8 m

R₂ = 12/2 = 6 m

R must be the radius of the new circular park.

If the circular areas with radii R₁ and R₂ are equal to the circular area with radius R, then

R2 = R₁₂ + R₂₂ = (8)² + (6)² = 64 + 36 = 100

Hence, R = 10 m

Ques. The motorcycle wheel is 35 cm radius. The number of revolutions per minute that the wheel has to perform in order to maintain a top speed of 66 km / h will be

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Ans. a) 500

Explanation: Circumference of round wheel = 2πr = 2 × (22/7) × 35 = 220 cm

Wheel speed = 66 km / h

= (66 × 1000) / 60 m / min

= 1100 × 100 cm / min

= 110000 cm / min

Variability number in 1 minutes = 110000/220 = 500

Ques. The radio for the two circles is 19 cm and 9 cm respectively. Circular radiator with a circumference equal to the sum of the two circular cycles

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Ans. a) 28

Explanation: The radius of the two circles should be r1 and r2 and the area of the largest circle should be r.

∴ r₁ = 19 cm, r₂ = 9 cm

Dual circle = C1 + C2… (when C = circle)

= 2πr² + 2πr² = 2π × 19 + 2π × 9 = 38π + 18π = 56π

∴ Circle = 56π

⇒ 2πr = 56π

⇒ r = 28

∴ Round circle radius = 28 cm

Ques. The perimeter (in cm) of a square around a circle of radius a cm, is

7a cm
8a cm
5a cm
6a cm

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Ans. b) 8a cm

Explanation: The side of the square around the radius circle a cm = circle width = 2a cm

∴ Square perimeter

= 4 × 2a = 8a cm

Ques. If the circumference of a circle is equal to the number of its rotation, then the width of the circle is the same

6 units
8 units
9 units
3 units

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Ans. b) 8 units

Explanation: r² = 2πr × 2

⇒ r = 4

⇒ 2r = 8 units

Ques. The area of the circle can be marked with a 6 cm side square

5 π cm²
6 π cm²
9 π cm²
8 π cm²

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Ans. c) 9 π cm²

Explanation: Square side = 6 cm

Circle width = square side = 6 cm

Therefore, Radius of circle = 3 cm

Circle area = π r² = π (3)² = 9π cm²

Ques. The area of the square that can be inscribed in a circle of radius 8 cm is

256 cm²
128 cm²
642 cm²
64 cm²

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Ans. b) 128 cm²

Explanation: Radius of circle = 8 cm

Diameter of circle = 16 cm = diagonal of the square

Let “a” be the triangle side, and the hypotenuse is 16 cm

Using Pythagoras theorem, we can write

162 = a²+ a²

256 = 2a²

a² = 256 / 2

a² = 128 = area of a square.

Ques. The area of a sector of a circle with radius 6 cm if the angle of the sector is 60°.

142/7
152/7
132/7
122/7

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Ans. c) 132/7

Explanation: Angle of the sector is 60°

Area of sector = (θ/360°) × π r²

∴ Area of the sector with angle 60° = (60°/360°) × π r² cm²

= (36/6) π cm²

= 6 × (22/7) cm²

= 132/7 cm²

Ques. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. The length of the arc is;

20cm
21cm
22cm
25cm

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Ans. c) 22 cm

Explanation: Length of an arc = (θ/360°) × (2πr)

∴ Length of an arc AB = (60°/360°) × 2 × 22/7 × 21

= (1/6) × 2 × (22/7) × 2

or Arc AB Length = 22 cm

Ques. Area of a sector of angle p (in degrees) of a circle with radius R is

p/180 × 2πR
p/180 × π R²
p/360 × 2πR²
p/720 × 2πR²

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Ans. d) p/720 × 2πR²

Explanation: The area of a sector = (θ/360°) × π r2

Given, θ = p

So, area of sector = p/360 × π R²

Multiplying and dividing by 2 simultaneously,

= [(p/360) / (π R²)] × [2/2]

= (p/720) × 2πR²

Ques. If the sum of the areas of two circles with radii R₁ and R₂ is equal to the area of a circle of radius R, then

R₁ + R₂ = R
R₁² + R₂² = R²
R₁ + R₂< R
R₁² + R₂² < R²

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Ans. (b) R₁² + R₂2²= R²

Explanation: According to the given,

πR₁² + πR₂² = πR²

π(R₁² + R₂²) = πR²

R₁² + R₂² = R²

Read Also:

Area of Segment of a Circle	Perimeter and Area of a Circle	Areas Related To Circles Formula
Tangent Circle Formula	Circle Definition	The Sector of a Circle
Types of Quadrilaterals	Tangent Circle Formula	Types of Triangles
Geometry Formula	Circumference of Circle	Geometry Formula

MCQs on Areas Related to Circles

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If $f : \mathbb{N} \rightarrow \mathbb{W}$ is defined as \[ f(n) = \begin{cases} \frac{n}{2}, & \text{if } n \text{ is even} \\ 0, & \text{if } n \text{ is odd} \end{cases} \] then $f$ is :

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MCQs on Areas Related to Circles

MCQs

CBSE CLASS XII Related Questions

1.If $f : \mathbb{N} \rightarrow \mathbb{W}$ is defined as \[ f(n) = \begin{cases} \frac{n}{2}, & \text{if } n \text{ is even} \\ 0, & \text{if } n \text{ is odd} \end{cases} \] then $f$ is :

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3.Let $f(x) = |x|$, $x \in \mathbb{R}$. Then, which of the following statements is incorrect?

5. Evaluate : \[ I = \int_0^{\frac{\pi}{4}} \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} \]

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Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
If $f : \mathbb{N} \rightarrow \mathbb{W}$ is defined as \[ f(n) = \begin{cases} \frac{n}{2}, & \text{if } n \text{ is even} \\ 0, & \text{if } n \text{ is odd} \end{cases} \] then $f$ is :

2.
Let $f'(x) = 3(x^2 + 2x) - \frac{4}{x^3} + 5$, $f(1) = 0$. Then, $f(x)$ is:

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Let $f(x) = |x|$, $x \in \mathbb{R}$. Then, which of the following statements is incorrect?

5.
Evaluate : \[ I = \int_0^{\frac{\pi}{4}} \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} \]