# 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

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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 = πr^{2} = 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) r^{2}

(b) 1/2r^{2}

(c) 2r^{2}

(d) √2r^{2}

Answer: **(a) r ^{2}**

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

= r^{2}

**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 = a^{2} = (πr/2)^{2}

A_{circle}/A_{square} = πr^{2}/(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 cm^{2}

(b) 16 π cm^{2}

(c) 12 π cm^{2}

(d) 9 π cm^{2}

Answer: **(b) 16 π cm ^{2}**

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^{2}

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

(a) 256 cm^{2}

(b) 128 cm^{2}

(c) 642 cm^{2}

(d) 64 cm^{2}

Answer: **(b) 128 cm ^{2}**

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

16^{2}= a^{2}+a^{2}

256 = 2a^{2}

a^{2}= 256/2

a^{2}= 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 r^{2}

∴ Area of the sector with angle 60° = (60°/360°) × π r^{2} cm^{2}

= (36/6) π cm^{2}

= 6 × (22/7) cm^{2}

= 132/7 cm^{2}

**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 cm^{2}

(b) 220 cm^{2}

(c) 231 cm^{2}

(d) 250 cm^{2}

Answer: **(c) 231 cm ^{2}**

Explanation: The angle subtended by the arc = 60°

So, area of the sector = (60°/360°) × π r^{2} cm^{2}

= (441/6) × (22/7) cm^{2}

= 231 cm^{2}

**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 R^{2}

(c) p/360 × 2πR

(d) p/720 × 2πR^{2}

Answer: **(d) p/720 × 2πR ^{2}**

Explanation: The area of a sector = (θ/360°) × π r^{2}

Given, θ = p

So, area of sector = p/360 × π R^{2}

Multiplying and dividing by 2 simultaneously,

= [(p/360)/(π R^{2})]×[2/2]

= (p/720) × 2πR^{2}

**11. If the area of a circle is 154 cm ^{2}**

**, 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 cm^{2}

πr^{2} = 154

(22/7) × r^{2} = 154

r^{2} = (154 × 7)/22

r^{2} = 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 R**_{1}** and R**_{2}** is equal to the area of a circle of radius R, then**

(a) R_{1} + R_{2} = R

(b) R_{1}^{2} + R_{2}^{2} = R^{2}

(c) R_{1} + R_{2} < R

(d) R_{1}^{2} + R_{2}^{2} < R^{2}

Answer:** (b) R _{1}^{2} + R_{2}^{2} = R^{2}**

Explanation:

According to the given,

πR_{1}^{2} + πR_{2}^{2} = πR^{2}

π(R_{1}^{2} + R_{2}^{2}) = πR^{2}

R_{1}^{2} + R_{2}^{2} = R^{2}

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

(a) (πr^{2}θ)/360

(b) (πr^{2}θ)/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:

R_{1} = 16/2 = 8 m

R_{2} = 12/2 = 6 m

Let R be the radius of the new circular park.

If the areas of two circles with radii R_{1} and R_{2} is equal to the area of circle with radius R, then

R^{2} = R_{1}^{2} + R_{2}^{2}

= (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 R_{1} and R_{2} is equal to the circumference of a circle of radius R, then R_{1} + R_{2} = R.

Here,

R_{1} = 36/2 = 18 cm

R_{2} = 20/2 = 10 cm

R = R_{1} + R_{2} = 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 cm^{2}

(b) 462 cm^{2}

(c) 386 cm^{2}

(d) 512 cm^{2}

Answer: **(b) 462 cm ^{2}**

Explanation:

Given, radius (r) = 21 cm

Central angle = θ = 120

Area of sector = (πr^{2}θ)/360

= (22/7) × (21 × 21) × (120/360)

= 22 × 21

= 462 cm^{2}

**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 = πr^{2}

⇒ r = 2

Therefore, radius = 2 units

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

(a) 77 cm^{2}

(b) 77/8 cm^{2}

(b) 35.5 cm^{2}

(c) 77/2 cm^{2}

Answer:** (b) 77/8 cm ^{2}**

Explanation:

Given, circumference = 22 cm

2πr = 22

2 × (22/7) × r = 22

r = 7/2 cm

Area of quadrant of a circle = (1/4)πr^{2}

= (1/4) × (22/7) × (7/2) × (7/2)

= 77/8 cm^{2}

**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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