Class 10 Maths MCQ Questions of Areas related to circles are made available online here for college understudies to attain better marks in exams. These MCQ Questions are provided here, chapter-wise, with answers and detailed explanations as per the CBSE syllabus and NCERT guidelines. Practice the questions given here and choose the right answer, Verify your answers with detailed explanations. Also, find sample papers for class 10 here.

MCQ questions for Class 10 Maths Areas related to circles help to clear the concepts associated with circles like area, circumference, segment, sector, angle, and length of a circle, area for the world of a circle is provided here.

Explore various MCQ Questions for Class 10 with answers provided with detailed solutions by looking below:-

1. The radii of two circles are 19 cm and 9 cm respectively. The radius of the circle which has circumference equal to the sum of the circumference of two circles is
(a) 35 cm
(b) 10 cm
(c) 21 cm
(d) 28 cm

2. The area of the circle that can be inscribed in a square of side 6 cm, is
(a) 18π cm2
(b) 12π cm2
(c) 9π cm2
(d) 14π cm2

3. The area of the circle is 154 cm2. The radius of the circle is
(a) 7 cm
(b) 14 cm
(c) 3.5 cm
(d) 17.5 cm

4.  The area of the circle whose diameter is 21 cm is
(a) 346.5 cm2
(b) 37.68 cm2
(c) 18.84 cm2
(d) 19.84 cm2

5. 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

6. 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

7. The area of the circle that can be inscribed in a square of side 8 cm is
(a) 36π cm2
(b) 16π cm2
(c) 12π cm2
(d) 9π cm2

8. 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

9. 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

10. If the circumference of a circle and the perimeter of a square are equal, then
(a) Area of the circle = Area of the square
(b) Area of the circle > Area of the square
(c) Area of the circle < Area of the square
(d) Nothing definite can be said about the relation between the areas of the circle and square.

11. The perimeter of a square circumscribing a circle of radius a unit is
(a) 2 units
(b) 4α units
(c) 8α units
(d) 16α units

12. If the sum of the circumferences of two circles with radii R1 and Ris equal to circumference of a circle of radius R, then
(a) R1 + R2 = R
(b) R1 + R2 > R
(c) R1 + R2 < R
(d) Can’t say;

13. The area of the largest triangle that can be inscribed in a semicircle of radius r is
(a) r2
(b) 2r2
(c) r3
(d) 2r3

14. The area (in cm2) of the circle that can be inscribed in a square of side 8 cm is
(a) 64 π
(b) 16 π
(c) 8 π
(d) 32 π

15. If the perimeter of a circle is equal to that of a square, then the ratio of their areas is
(a) 22 : 7
(b) 14 : 11
(c) 7 : 22
(d) 11 : 14

16. If the area of a circle is numerically equal to twice its circumference, then the diameter of the circle is
(a) 4 units
(b) n units
(c) 8 units
(d) 2 units

17. The area of the sector of a circle with radius 6 cm and of angle 60° is
(a) 9.42 cm2
(b) 37.68 cm2
(c) 18.84 cm2
(d) 19.84 cm2

18.  The length of the minute hand of a clock is 14 cm. The area swept by the minute hand in 5 minutes is
(a) 153.9 cm2
(b) 102.6 cm2
(c) 51.3 cm2
(d) 205.2 cm2

19. The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm is
(a) 31 cm
(b) 25 cm
(c) 62 cm
(d) 50 cm

20. The area of the largest square that can be inscribed in a circle of radius 12 cm is
(a) 24 cm2
(b) 249 cm2
(c) 288 cm2
(d) 196√2 cm2

Explanation: Let the radii of two circles be r1 and r2 and the radius of large circle be r.
∴ r1 = 19 cm, r2 = 9 cm
Circumference of two circles = C1+ C2 …(where C = circle)
= 2πr1 + 2πr2 = 2π × 19 + 2π × 9 = 38π + 18π = 56π
∴ Circumference of large circle = 56π
⇒ 2πr = 56π
⇒ r = 28
∴ Radius of large circle = 28 cm

Explanation: Size of square = 6 cm, radius = 62 = 3 cm;
Area of the circle = πr2 = π × 3 × 3 = 9π cm2

Explaination: Area of circle = 154 cm2

⇒ πr2 = 154 cm2

⇒ 22/7 × r2 = 154

⇒ r2 = (154 × 7)/22

⇒ r2 = 7 × 7 = 49

∴ r = $$\sqrt{49}$$ = 7

Explaination: Here diameter = 21 cm

∴ Radius r = 21/2 cm

Area of the circle, A = πr2

∴ A=22/7×21/2×21/2=11×3×21/2=693/2

=346.5cm

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

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

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 cm

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.

9. Answer: (b) R12 + R22 = R2

Explanation: According to given condition,

Area of circle = Area of first circle + Area of second circle

πR2 = πR12 + πR22

R2 = R12 + R2

10. Answer: (b) Area of the circle > Area of the square

12. Answer: (a) R1 + R2 = R

15. Answer: (b) 14 : 11

Explaination: πr2 = 2πr × 2

⇒ r = 4

⇒ 2r = 8 units

Explaination: Reason: Here r = 6 cm, θ = 60°
Area of the sector = θ/360
∴ Area = 60/360 × 3.14 × 6 × 6 = 1/6 × 3.14 × 6 × 6
= 3.14 × 6 = 18.84 cm

Explaination: Angle swept by the minute hand in 1 minute = (360° ÷ 60) = 6°
∴ θ = 30°
∴ Angle swept by the minute hand in 5 minutes = 6° × 5 = 30°
Length of minute hand (r) = 14 cm
∴ Area swept = θ/360πr2 = 30/360 × 22/7 × 14 × 14 = 154/3 = 51.3 cm

Explanation: Area of first circle = πr2 = π(24)2 = 576π m2

Area of second circle = πr2 = π(7)2 = 49π m2

Now, we are given that,

Area of the circle = Area of first circle + Area of second circle

∴ πR2 = 576π +49π

(where, R is the radius of the new circle)

⇒ πR2 = 625π

⇒ R2 = 625

⇒ R = 25

∴ Radius of the circle = 25cm

Thus, diameter of the circle = 2R = 50 cm.

Explanation:Let the side of square be a cm and radius of the circle be r cm
Give, r=12 cm

Area of the square =a2cm2
ΔABC is a right angled triangle.
Thus, by Pythagoras theorem, we have
AB2+BC2=AC2
⇒a2+a2 =(2r)2
⇒2a2=4r2
⇒a2=2r2
⇒a2 =2(12)2
⇒a2=2×144
=288cm2

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