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 = 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
Answer: (a) r2
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
Answer: (a) 14:11
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
Answer: (b) 16 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
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.
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 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
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 cm2
(b) 220 cm2
(c) 231 cm2
(d) 250 cm2
Answer: (c) 231 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
Answer: (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
Answer: (c) 44 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
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:
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
Answer: (c) 28 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
Answer: (b) 462 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
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 = πr2
⇒ r = 2
Therefore, radius = 2 units
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
Answer: (b) 77/8 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
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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