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Class 10 Maths MCQ Questions of Areas Related to Circles with Answers?

Answer 1

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. Class 10 Maths MCQ Questions of Areas related to circles are made available online here for college students to attain better marks in exams. 

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Practice MCQ Questions for class 10 Maths 

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 R₂ is 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

Answer 2

1. Answer: (d) 28 cm
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

2. Answer: (c) 9π cm²
Explanation: Size of square = 6 cm, radius = 62 = 3 cm;
Area of the circle = πr² = π × 3 × 3 = 9π cm² 

3.  Answer: (d) 17.5 cm
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

4.  Answer: (a) 346.5 cm²

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

=346.5cm²  

5. 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

6. Answer: (c) 78.5 sq.cm

Explanation: Radius = 5cm

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

7. Answer: (b) 16π 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 

= 16π cm²  

8. 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.

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 + R₂²  

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

11. Answer: (c) 8α units

12. Answer: (a) R₁ + R₂ = R

13. Answer: (a) r²  

14. Answer: (b) 16 π

15. Answer: (b) 14 : 11

16. Answer: (c) 8 units

Explaination: πr² = 2πr × 2

⇒ r = 4

⇒ 2r = 8 units

17. Answer: (c) 18.84 cm²
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 cm²  

18. Answer: (c) 51.3 cm²
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 cm² 

19. Answer: (d) 50 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.

20. Answer: (c) 288 cm²
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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