# 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 (2021 – 2022) and NCERT guidelines. Click here to get all class 10 Maths chapter-wise MCQs.

## Class 10 Maths MCQs for Areas Related to Circles

Practise the questions given here and choose the correct answer. Verify your answers with the solution given. Also, find important questions for class 10 Maths here.

Students can also get access to Areas Related to Circles for Class 10 Notes here.

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

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

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

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

Explanation: Given,

The perimeter of circle = perimeter of the square

2πr = 4a

a=πr/2

Area of square = a2 = (πr/2)2

Acircle/Asquare = 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

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

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

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

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

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

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

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

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

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

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

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

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

Explanation:

According to the given,

Perimeter of circle = Area of circle

2πr = πr2

⇒ r = 2

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

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