Class 10 Maths Chapter 12 Areas Related to Circles MCQs

Class 10 Maths Chapter 12 MCQs (Areas Related to Circles) are made available online here for students to score better marks in the exams. These multiple-choice questions are provided here with answers and detailed explanations. These questions are framed as per the CBSE syllabus (2022 – 2023) and NCERT guidelines. Click here to get all class 10 Maths chapter-wise MCQs.

Class 10 Maths MCQs for Areas Related to Circles

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1. The perimeter of a circle having radius 5cm is equal to:

(a) 30 cm

(b) 3.14 cm

(c) 31.4 cm

(d) 40 cm

Answer: (c) 31.4 cm

Explanation: The perimeter of the circle is equal to the circumference of the circle.

Circumference = 2πr

= 2 x 3.14 x 5

= 31.4 cm

2. Area of the circle with radius 5cm is equal to:

(a) 60 sq.cm

(b) 75.5 sq.cm

(c) 78.5 sq.cm

(d) 10.5 sq.cm

Answer: (c) 78.5 sq.cm

Explanation: Radius = 5cm

Area = πr² = 3.14 x 5 x 5 = 78.5 sq.cm

3. The largest triangle inscribed in a semi-circle of radius r, then the area of that triangle is;

(a) r²

(b) 1/2r²

(c) 2r²

(d) √2r²

Answer: (a) r²

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

Area of triangle = ½ x base x height

= ½ x 2r x r

= r²

4. If the perimeter of the circle and square are equal, then the ratio of their areas will be equal to:

(a) 14:11

(b) 22:7

(c) 7:22

(c) 11:14

Answer: (a) 14:11

Explanation: Given,

The perimeter of circle = perimeter of the square

2πr = 4a

a=πr/2

Area of square = a² = (πr/2)²

A_circle/A_square = πr²/(πr/2)²

= 14/11

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

(a) 36 π cm²

(b) 16 π cm²

(c) 12 π cm²

(d) 9 π cm²

Answer: (b) 16 π cm²

Explanation: Given,

Side of square = 8 cm

Diameter of a circle = side of square = 8 cm

Therefore, Radius of circle = 4 cm

Area of circle

= π(4)²

= π (4)²

= 16π cm²

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

(a) 256 cm²

(b) 128 cm²

(c) 642 cm²

(d) 64 cm²

Answer: (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.

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

(a) 142/7

(b) 152/7

(c) 132/7

(d) 122/7

Answer: (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

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

(a) 20cm

(b) 21cm

(c) 22cm

(d) 25cm

Answer: (c) 22cm

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

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

= (1/6) × 2 × (22/7) × 21

Or Arc AB Length = 22cm

9. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. The area of the sector formed by the arc is:

(a) 200 cm²

(b) 220 cm²

(c) 231 cm²

(d) 250 cm²

Answer: (c) 231 cm²

Explanation: The angle subtended by the arc = 60°

So, area of the sector = (60°/360°) × π r² cm²

= (441/6) × (22/7) cm²

= 231 cm²

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

(a) p/180 × 2πR

(b) p/180 × π R²

(c) p/360 × 2πR

(d) p/720 × 2πR²

Answer: (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

11. If the area of a circle is 154 cm², then its perimeter is

(a) 11 cm 

(b) 22 cm 

(c) 44 cm 

(d) 55 cm

Answer: (c) 44 cm

Explanation:

Given,

Area of a circle = 154 cm²

πr² = 154

(22/7) × r² = 154

r² = (154 × 7)/22

r² = 7 × 7

r = 7 cm

Perimeter of circle = 2πr = 2 × (22/7) × 7 = 44 cm

12. 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

(a) R₁ + R₂ = R 

(b) R₁² + R₂² = R²

(c) R₁ + R₂ < R 

(d) R₁² + R₂² < R²

Answer: (b) R₁² + R₂² = R²

Explanation:

According to the given,

πR₁² + πR₂² = πR²

π(R₁² + R₂²) = πR²

R₁² + R₂² = R²

13. If θ is the angle (in degrees) of a sector of a circle of radius r, then the length of arc is

(a) (πr²θ)/360

(b) (πr²θ)/180

(c) (2πrθ)/360

(d) (2πrθ)/180

Answer: (a) (2πrθ)/360

If θ is the angle (in degrees) of a sector of a circle of radius r, then the area of the sector is (2πrθ)/360.

14. It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be

(a) 10 m

(b) 15 m

(c) 20 m

(d) 24 m

Answer: (a) 10 m

Explanation:

Radii of two circular parks will be:

R₁ = 16/2 = 8 m

R₂ = 12/2 = 6 m

Let R be the radius of the new circular park.

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

R² = R₁² + R₂²

= (8)² + (6)²

= 64 + 36

= 100

R = 10 m

15. The radius of a circle whose circumference is equal to the sum of the circumferences of the  two circles of diameters 36 cm and 20 cm is

(a) 56 cm 

(b) 42 cm 

(c) 28 cm 

(d) 16 cm

Answer: (c) 28 cm

Explanation:

If the sum of the circumferences of two circles with radii R₁ and R₂ is equal to the circumference of a circle of radius R, then R₁ + R₂ = R.

Here,

R₁ = 36/2 = 18 cm

R₂ = 20/2 = 10 cm

R = R₁ + R₂ = 18 + 10 = 28 cm

Therefore, the radius of the required circle is 28 cm

16. Find the area of a sector of circle of radius 21 cm and central angle 120°.

(a) 441 cm²

(b) 462 cm²

(c) 386 cm²

(d) 512 cm²

Answer: (b) 462 cm²

Explanation:

Given, radius (r) = 21 cm

Central angle = θ = 120

Area of sector = (πr²θ)/360

= (22/7) × (21 × 21) × (120/360)

= 22 × 21

= 462 cm²

17. The wheel of a motorcycle is of radius 35 cm. The number of revolutions per minute must the wheel make so as to keep a speed of 66 km/hr will be

(a) 50

(b) 100

(c) 500

(d) 1000

Answer: (c) 500

Explanation:

Circumference of the wheel = 2πr = 2 × (22/7) × 35 = 220 cm

Speed of the wheel = 66 km/hr

= (66 × 1000)/60 m/min

= 1100 × 100 cm/min

= 110000 cm/min

Number of revolutions in 1 min = 110000/220 = 500

18. If the perimeter and the area of a circle are numerically equal, then the radius of the circle is

(a) 2 units 

(b) π units 

(c) 4 units 

(d) 7 units

Answer: (a) 2 units

Explanation:

According to the given,

Perimeter of circle = Area of circle

2πr = πr²

⇒ r = 2

Therefore, radius = 2 units

19. The area of a quadrant of a circle with circumference of 22 cm is

(a) 77 cm²

(b) 77/8 cm²

(b) 35.5 cm²

(c) 77/2 cm²

Answer: (b) 77/8 cm²

Explanation:

Given, circumference = 22 cm

2πr = 22 

2 × (22/7) × r = 22

r = 7/2 cm

Area of quadrant of a circle = (1/4)πr²

= (1/4) × (22/7) × (7/2) × (7/2)

= 77/8 cm²

20. In a circle of radius 14 cm, an arc subtends an angle of 30° at the centre, the length of the arc is

(a) 44 cm

(b) 28 cm

(c) 11 cm

(d) 22/3 cm

Answer: (d) 22/3 cm

Explanation:

Given, radius = r = 14 cm

Length of arc = (2πrθ)/360

= 2 × (22/7) × 14 × (30/360)

= 22/3 cm

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