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

MCQs

CBSE CLASS XII Related Questions

1.
Find the inverse of each of the matrices,if it exists \(\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}\)

2.
If A=\(\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}\)verify that A³-6A²+9A-4 I=0 and hence find A^-1

3.
Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix} 1 & 3\\ 2 & 7\end{bmatrix}\)

4.
Find the vector and the cartesian equations of the lines that pass through the origin and(5,-2,3).

5.
Evaluate \(\begin{vmatrix} cos\alpha cos\beta &cos\alpha sin\beta &-sin\alpha \\ -sin\beta&cos\beta &0 \\ sin\alpha cos\beta&sin\alpha\sin\beta &cos\alpha \end{vmatrix}\)

6.
Solve system of linear equations, using matrix method.
x-y+2z=7
3x+4y-5z=-5
2x-y+3z=12

Similar Mathematics Concepts

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MCQs on Areas Related to Circles

MCQs

CBSE CLASS XII Related Questions

1. Find the inverse of each of the matrices,if it exists \(\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}\)

2. If A=\(\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}\)verify that A3-6A2+9A-4 I=0 and hence find A-1

3. Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix} 1 & 3\\ 2 & 7\end{bmatrix}\)

4. Find the vector and the cartesian equations of the lines that pass through the origin and(5,-2,3).

5. Evaluate \(\begin{vmatrix} cos\alpha cos\beta &cos\alpha sin\beta &-sin\alpha \\ -sin\beta&cos\beta &0 \\ sin\alpha cos\beta&sin\alpha\sin\beta &cos\alpha \end{vmatrix}\)

6. Solve system of linear equations, using matrix method. x-y+2z=7 3x+4y-5z=-5 2x-y+3z=12

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Find the inverse of each of the matrices,if it exists \(\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}\)

2.
If A=\(\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}\)verify that A³-6A²+9A-4 I=0 and hence find A^-1

3.
Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix} 1 & 3\\ 2 & 7\end{bmatrix}\)

4.
Find the vector and the cartesian equations of the lines that pass through the origin and(5,-2,3).

5.
Evaluate \(\begin{vmatrix} cos\alpha cos\beta &cos\alpha sin\beta &-sin\alpha \\ -sin\beta&cos\beta &0 \\ sin\alpha cos\beta&sin\alpha\sin\beta &cos\alpha \end{vmatrix}\)

6.
Solve system of linear equations, using matrix method.
x-y+2z=7
3x+4y-5z=-5
2x-y+3z=12