# Class 10 Maths Chapter 12 Areas Related to Circles MCQs

Class 10 Maths MCQs for Chapter 12 (Areas related to circles) are made available online here for students to score better marks in exams. These multiple-choice questions are provided here, chapter-wise, with answers and detailed explanations as per CBSE syllabus and NCERT guidelines.

## Class 10 Maths MCQs for Areas Related to Circles

Practice 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.

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

Explanation: Radius = 5cm

Area = πr2 = 3.14x5x5 = 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 = 8cm

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