🟠 Section A — Multiple Choice 10 × 1 = 10 marks
Q1 · MCQ
The area of a circle of radius 7 cm (π = 22/7) is:
Q2 · MCQ
The circumference of a circle of diameter 14 cm (π = 22/7) is:
Q3 · MCQ
The area of a sector of angle 90° of a circle of radius r is:
Q4 · MCQ
The length of an arc subtending 60° at the centre of a circle of radius 6 cm is:
Q5 · MCQ
If the radius of a circle is doubled, its area becomes:
Q6 · MCQ
The perimeter of a semicircle of radius r is:
Q7 · MCQ
The area of a minor segment equals:
Q8 · MCQ
The value of π is closest to:
Q9 · MCQ
The area of a ring with outer radius R and inner radius r is:
Q10 · MCQ
A sector of angle 120° is what fraction of the circle?
🟣 Section B — Assertion & Reason 3 × 1 = 3 marks
(a) Both A & R true, R explains A · (b) Both true, R doesn't explain A · (c) A true, R false · (d) A false, R true
Q11 · A–R
Assertion: The area of a circle of radius 7 cm is 154 cm².
Reason: Area of a circle = πr².
(a) Both true and R explains A: (22/7)×7² = 22×7 = 154 cm².
Q12 · A–R
Assertion: The perimeter of a semicircle equals half the circumference.
Reason: Perimeter of a semicircle = πr + 2r.
(d) Assertion is false — the perimeter also includes the diameter (2r). The Reason gives the correct formula, so Reason is true.
Q13 · A–R
Assertion: A sector of 90° has area ¼πr².
Reason: The area of a sector = (θ/360) × πr².
(a) Both true and R explains A: (90/360)πr² = ¼πr².
🟢 Section C — Short Answer 12 × 2 = 24 marks
Q14 · Short
Find the area of a circle whose circumference is 44 cm (π = 22/7).
2πr = 44 → r = 7 cm. Area = πr² = (22/7)×49 = 154 cm².
Q15 · Short
Find the area of a sector of angle 60° in a circle of radius 21 cm (π = 22/7).
Area = (60/360)×(22/7)×21² = (1/6)×(22/7)×441 = (1/6)×1386 = 231 cm².
Q16 · Short
Find the length of an arc of a circle of radius 14 cm subtending 90° at the centre (π = 22/7).
ℓ = (90/360)×2×(22/7)×14 = ¼×88 = 22 cm.
Q17 · Short
The area of a circle is 616 cm². Find its radius (π = 22/7).
πr² = 616 → r² = 616×7/22 = 196 → r = 14 cm.
Q18 · Short
A wheel of radius 35 cm makes 20 revolutions. Find the distance covered (π = 22/7).
Distance = 20 × 2πr = 20 × 2 × (22/7) × 35 = 20 × 220 = 4400 cm = 44 m.
Q19 · Short
Find the area of the triangle formed by two radii and a chord for θ = 90°, r = 10 cm.
Area = ½r²sinθ = ½×100×sin90° = ½×100×1 = 50 cm².
Q20 · Short
Find the area of a ring with outer radius 10 cm and inner radius 6 cm (π = 3.14).
A = π(R²−r²) = 3.14×(100−36) = 3.14×64 = 200.96 cm².
Q21 · Short
Find the perimeter of a quadrant of a circle of radius 7 cm (π = 22/7).
Perimeter = (πr/2) + 2r = (22/7×7/2) + 14 = 11 + 14 = 25 cm.
Q22 · Short
The circumference of a circle exceeds its diameter by 30 cm. Find the radius (π = 22/7).
2πr − 2r = 30 → 2r(π−1) = 30 → 2r(22/7−1) = 30 → 2r×(15/7) = 30 → r = 7 cm.
Q23 · Short
Two circles have radii 8 cm and 6 cm. Find the radius of a circle whose area equals their combined area.
πR² = π(8²+6²) = π(64+36) = 100π → R² = 100 → R = 10 cm.
Q24 · Short
A circle is inscribed in a square of side 14 cm. Find the area left in the corners (π = 22/7).
r = 7. Corners = 14² − πr² = 196 − 154 = 42 cm².
Q25 · Short
Find the area of the sector of a clock swept by the minute hand (length 7 cm) in 15 minutes (π = 22/7).
15 min = 90°. Area = (90/360)×(22/7)×49 = ¼×154 = 38.5 cm².
🔴 Section D — Long Answer 5 × 2 = 10 marks
Q26 · Long
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the minor segment (π = 3.14).
Sector = (90/360)πr² = ¼×3.14×100 = 78.5. Triangle = ½r² = 50. Minor segment = 78.5 − 50 = 28.5 cm².
Q27 · Long
A chord of a circle of radius 12 cm subtends 120° at the centre. Find the area of the corresponding minor segment (π = 3.14, √3 = 1.73).
Sector = (120/360)×3.14×144 = 150.72. Triangle = ½r²sin120° = ½×144×(√3/2) = 36×1.73 = 62.28. Segment = 150.72 − 62.28 = 88.44 cm².
Q28 · Long
Four equal circles of radius 7 cm touch each other. Find the area enclosed between them (the central region), taking π = 22/7.
The centres form a square of side 14 cm. Enclosed area = square − 4 quadrants = 14² − 4×(¼πr²) = 196 − πr² = 196 − 154 = 42 cm².
Q29 · Long
The minute hand of a clock is 14 cm long. Find the area swept between 9:00 and 9:35 (π = 22/7).
35 min = (35/60)×360° = 210°. Area = (210/360)×(22/7)×14² = (7/12)×616 = 359.33 cm².
Q30 · Long
A square field of side 14 m has a circular flower bed inscribed in it. Find the area of the field NOT covered by the bed (π = 22/7).
r = 7 m. Uncovered = 14² − πr² = 196 − (22/7×49) = 196 − 154 = 42 m².