1. The area of a circle with radius r is given by:
A. 2πr
B. πr²
C. πr
D. 2πr²
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Answer: B. πr²
Explanation: Standard formula for the area of a circle of radius r is Area = πr².
2. The circumference of a circle of radius r is given by:
A. πr²
B. 2πr
C. πr
D. 2πr²
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Answer: B. 2πr
Explanation: Standard formula for the perimeter/circumference of a circle is 2πr.
3. Area of a sector of angle θ of a circle with radius r is:
A. (θ / 360°) × 2πr
B. (θ / 180°) × πr²
C. (θ / 360°) × πr²
D. (θ / 180°) × 2πr
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Answer: C. (θ / 360°) × πr²
Explanation: Formula for area of sector = (θ / 360°) × πr².
4. The length of an arc of a sector of angle θ of a circle with radius r is:
A. (θ / 360°) × πr²
B. (θ / 360°) × 2πr
C. (θ / 180°) × πr²
D. (θ / 720°) × 2πr
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Answer: B. (θ / 360°) × 2πr
Explanation: Arc length = (θ / 360°) × 2πr.
5. If the perimeter and 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
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Answer: A. 2 units
Explanation: 2πr = πr² ⇒ r = 2 units.
6. If the radius of a circle is 7 cm, its area is (take π = 22/7):
A. 44 cm²
B. 154 cm²
C. 308 cm²
D. 77 cm²
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Answer: B. 154 cm²
Explanation: Area = (22/7) × 7 × 7 = 154 cm².
7. The perimeter of a quadrant of a circle of radius r is:
A. (πr / 2) + 2r
B. (πr / 2)
C. πr + 2r
D. 2πr + r
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Answer: A. (πr / 2) + 2r
Explanation: Arc length of quadrant = 2πr / 4 = πr/2. Adding the two straight radial edges gives (πr/2) + 2r.
8. Area of a sector of central angle 60° in a circle of radius 6 cm is:
A. 132/7 cm²
B. 66/7 cm²
C. 11/7 cm²
D. 154/7 cm²
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Answer: A. 132/7 cm²
Explanation: Area = (60/360) × (22/7) × 6 × 6 = (1/6) × (22/7) × 36 = 132/7 cm².
9. If the area of a circle is 154 cm², then its perimeter is:
A. 22 cm
B. 44 cm
C. 88 cm
D. 66 cm
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Answer: B. 44 cm
Explanation: πr² = 154 ⇒ r² = 49 ⇒ r = 7 cm. Perimeter = 2 × (22/7) × 7 = 44 cm.
10. The area of a quadrant of a circle whose circumference is 22 cm is:
A. 77/8 cm²
B. 77/4 cm²
C. 77/2 cm²
D. 154 cm²
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Answer: A. 77/8 cm²
Explanation: 2πr = 22 ⇒ r = 7/2 cm. Quadrant Area = (1/4) × πr² = (1/4) × (22/7) × (7/2) × (7/2) = 77/8 cm².
11. Angle subtended by minute hand of a clock in 5 minutes is:
A. 5°
B. 15°
C. 30°
D. 60°
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Answer: C. 30°
Explanation: Minute hand completes 360° in 60 minutes. In 5 minutes, angle = (360° / 60) × 5 = 30°.
12. Area swept by 14 cm long minute hand of a clock in 5 minutes is:
A. 154/3 cm²
B. 154/6 cm²
C. 308/3 cm²
D. 77/3 cm²
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Answer: A. 154/3 cm²
Explanation: Angle = 30°. Area = (30/360) × (22/7) × 14 × 14 = (1/12) × 22 × 28 = 154/3 cm².
13. Area of segment of a circle is calculated by subtracting area of corresponding triangle from area of corresponding:
A. Circle
B. Sector
C. Quadrant
D. Diameter
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Answer: B. Sector
Explanation: Area of segment = Area of sector - Area of triangle.
14. If ratio of areas of two circles is 4 : 9, then ratio of their radii is:
A. 2 : 3
B. 16 : 81
C. 4 : 9
D. 1 : 2
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Answer: A. 2 : 3
Explanation: Ratio of areas = r1² / r2² = 4/9 ⇒ r1 / r2 = √(4/9) = 2/3.
15. If ratio of circumferences of two circles is 3 : 4, then ratio of their areas is:
A. 3 : 4
B. 9 : 16
C. 16 : 9
D. √3 : 2
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Answer: B. 9 : 16
Explanation: Ratio of circumferences = r1/r2 = 3/4. Ratio of areas = (r1/r2)² = (3/4)² = 9/16.
16. If diameter of a circle is increased by 100%, its area increases by:
A. 100%
B. 200%
C. 300%
D. 400%
Show Answer
Answer: C. 300%
Explanation: New radius r' = 2r. New area A' = π(2r)² = 4πr² = 4A. Increase = 4A - A = 3A, which is 300%.
17. A chord of a circle of radius 10 cm subtends a right angle at centre. Area of minor sector is (use π = 3.14):
A. 78.5 cm²
B. 314 cm²
C. 157 cm²
D. 39.25 cm²
Show Answer
Answer: A. 78.5 cm²
Explanation: Area = (90/360) × 3.14 × 10 × 10 = (1/4) × 314 = 78.5 cm².
18. Perimeter of a semicircle with radius r including its diameter is:
A. πr
B. πr + r
C. πr + 2r
D. 2πr + 2r
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Answer: C. πr + 2r
Explanation: Curved arc length = πr. Adding diameter (2r) gives πr + 2r.
19. The area of the largest triangle that can be inscribed in a semicircle of radius r is:
A. r²
B. 2r²
C. r²/2
D. 1/2 r²
Show Answer
Answer: A. r²
Explanation: Base of triangle = diameter = 2r. Maximum height = radius = r. Area = (1/2) × 2r × r = r².
20. Area of a square inscribed in a circle of radius r is:
A. r²
B. 2r²
C. 4r²
D. πr²
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Answer: B. 2r²
Explanation: Diagonal of square = diameter = 2r. Area of square = (1/2) × d² = (1/2) × (2r)² = 2r².
21. The wheel of a car makes 1000 revolutions in moving 88 km. Diameter of the wheel is:
A. 14 m
B. 28 m
C. 7 m
D. 28 cm
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Answer: B. 28 m
Explanation: Distance in 1 rev = 88000 m / 1000 = 88 m. 2πr = 88 ⇒ πd = 88 ⇒ d = 88 × (7/22) = 28 m.
22. Area of ring bounded by two concentric circles of radii R and r (R > r) is:
A. π(R - r)
B. π(R² - r²)
C. π(R² + r²)
D. 2π(R - r)
Show Answer
Answer: B. π(R² - r²)
Explanation: Ring area = Outer Area - Inner Area = πR² - πr² = π(R² - r²).
23. Area of a sector of angle p (in degrees) of a circle with radius R is:
A. (p / 180°) × 2πR
B. (p / 180°) × πR²
C. (p / 360°) × 2πR
D. (p / 720°) × 2πR²
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Answer: D. (p / 720°) × 2πR²
Explanation: (p / 360°) × πR² = (p / 720°) × 2πR² on multiplying numerator and denominator by 2.
24. If circumference of a circle is 44 cm, then area of circle is:
A. 154 cm²
B. 308 cm²
C. 44 cm²
D. 88 cm²
Show Answer
Answer: A. 154 cm²
Explanation: 2πr = 44 ⇒ r = 7 cm. Area = (22/7) × 7 × 7 = 154 cm².
25. Number of revolutions made by a wheel of diameter 0.7 m to cover distance of 1.1 km is:
A. 500
B. 1000
C. 250
D. 100
Show Answer
Answer: A. 500
Explanation: Circumference = (22/7) × 0.7 = 2.2 m. Revolutions = 1100 m / 2.2 m = 500.
26. If sum of areas of two circles with radii R1 and R2 is equal to area of circle of radius R, then:
A. R1 + R2 = R
B. R1² + R2² = R²
C. R1 + R2 < R
D. R1² + R2² < R²
Show Answer
Answer: B. R1² + R2² = R²
Explanation: πR1² + πR2² = πR² ⇒ R1² + R2² = R².
27. If sum of circumferences of two circles with radii R1 and R2 is equal to circumference of circle of radius R, then:
A. R1 + R2 = R
B. R1 + R2 > R
C. R1 + R2 < R
D. R1² + R2² = R²
Show Answer
Answer: A. R1 + R2 = R
Explanation: 2πR1 + 2πR2 = 2πR ⇒ R1 + R2 = R.
28. A horse is tied to a peg at one corner of a square grass field of side 15 m by a 5 m long rope. Area of field horse can graze is:
A. 19.625 m²
B. 78.5 m²
C. 154 m²
D. 225 m²
Show Answer
Answer: A. 19.625 m²
Explanation: Grazing area = Quadrant sector = (1/4) × π × 5² = (1/4) × 3.14 × 25 = 19.625 m².
29. Area of an equilateral triangle of side 2a is:
A. √3 a²
B. (√3/4) a²
C. 2√3 a²
D. √3/2 a²
Show Answer
Answer: A. √3 a²
Explanation: Area = (√3/4) × (side)² = (√3/4) × (2a)² = (√3/4) × 4a² = √3 a².
30. Distance travelled by a circular wheel of diameter d in 1 revolution is:
A. πd
B. 2πd
C. πd²
D. πd/2
Show Answer
Answer: A. πd
Explanation: Distance covered in 1 revolution = circumference = 2πr = π(2r) = πd.