CBSE Class 10 Mathematics Areas Related to Circles MCQs Set 13

Answer: (b) 22 cm

Explanation:
1. Arc length = \( \frac{\theta}{360} \times 2\pi r \).
2. = \( \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 \) = 22 cm.

Teacher's Note:
1. Use the fraction θ/360 of the full circumference.
2. 60° is one sixth of a full turn.

Answer: (b) 10 cm

Explanation:
1. A right angle gives a quarter circle, so \( \frac{1}{4} \times 2\pi r = 15.7 \).
2. \( \frac{\pi r}{2} = 15.7 \), so r = 31.4 ÷ 3.14 = 10 cm.

Teacher's Note:
1. 90° is one quarter of the circle.
2. Solve for r after substituting π = 3.14.

Answer: (b) 169.56 cm²

Explanation:
1. Area of sector = \( \frac{\theta}{360} \pi r^2 \).
2. = \( \frac{60}{360} \times 3.14 \times 18 \times 18 \) = 169.56 cm².

Teacher's Note:
1. Area of a sector is a fraction of the whole circle's area.
2. Keep π = 3.14 as given.

Answer: (a) 32 cm

Explanation:
1. Arc length = \( \frac{60}{360} \times 2 \times \frac{22}{7} \times 10.5 \) = 11 cm.
2. Perimeter = arc + two radii = 11 + 21 = 32 cm.

Teacher's Note:
1. The perimeter of a sector includes both bounding radii.
2. Do not forget to add 2r.

Answer: (c) 22 cm

Explanation:
1. Radius = 21 cm.
2. Arc = \( \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 \) = 22 cm.

Teacher's Note:
1. Convert diameter to radius first.
2. A 60° arc has length one sixth of the circumference.

Answer: (d) 924 cm²

Explanation:
1. Area swept by one blade = \( \frac{120}{360} \times \frac{22}{7} \times 21^2 \) = 462 cm².
2. Two wipers: 2 × 462 = 924 cm².

Teacher's Note:
1. Two blades mean doubling the sector area.
2. 120° is one third of a full circle.

Answer: (c) 228 cm²

Explanation:
1. The diagonal OB of the square is the radius: OB = 20√2 cm, so r² = 800.
2. Quadrant = \( \frac{1}{4} \times 3.14 \times 800 \) = 628 cm²; square = 400 cm².
3. Shaded area = 628 − 400 = 228 cm².

Teacher's Note:
1. The diagonal of the inscribed square is the radius.
2. Square the radius directly: r² = 20² + 20².

Answer: (a) 282 cm

Explanation:
1. Arc of the 270° sector = \( \frac{270}{360} \times 2 \times \frac{22}{7} \times 42 \) = 198 cm.
2. Perimeter = 198 + 42 + 42 = 282 cm.

Teacher's Note:
1. The remaining sector is 360° − 90° = 270°.
2. Add the two straight edges OA and OB.

Answer: (c) 21 cm

Explanation:
1. Arc length = \( \frac{60}{360} \times 2\pi r \) = \( \frac{\pi r}{3} \) = 22.
2. r = 66 ÷ π = 66 × 7 ÷ 22 = 21 cm.

Teacher's Note:
1. The wire length equals the arc length.
2. Rearrange the arc formula for r.

Answer: (c) 3500 cm²

Explanation:
1. Minor segment = sector − triangle = \( \frac{1}{4} \times \frac{22}{7} \times 35^2 \) − \( \frac{1}{2} \times 35 \times 35 \) = 962.5 − 612.5 = 350 cm².
2. Circle = \( \frac{22}{7} \times 35^2 \) = 3850 cm².
3. Major segment = 3850 − 350 = 3500 cm².

Teacher's Note:
1. Major segment = circle − minor segment.
2. For 90° the triangle is right-angled with legs equal to r.

Answer: (c) 9.71 cm²

Explanation:
1. Four quarter circles of radius 1 cm add up to one full circle of area π cm².
2. Shaded area = 16 − π − π = \( 16 - \frac{44}{7} \) = \( 9\frac{5}{7} \) cm² ≈ 9.71 cm².

Teacher's Note:
1. Four quarter circles equal one whole circle.
2. Subtract all circular areas from the square.

Answer: (b) 128 cm²

Explanation:
1. The diagonal of the square is the diameter, 16 cm.
2. Area = (diagonal)² ÷ 2 = 256 ÷ 2 = 128 cm².

Teacher's Note:
1. Area of a square = d²/2.
2. The diagonal of an inscribed square equals the diameter.

Answer: (b) 14 cm

Explanation:
1. \( \frac{1}{4} \times \frac{22}{7} r^2 = 38.5 \), so r² = 49 and r = 7 cm.
2. Diameter = 14 cm.

Teacher's Note:
1. A quadrant is one quarter of a circle.
2. Remember to double the radius.

Answer: (b) 35 m

Explanation:
1. Circumference = 5280 ÷ 24 = 220 m.
2. 2 × \( \frac{22}{7} \) × r = 220, so r = 35 m.

Teacher's Note:
1. Cost of fencing uses the circumference.
2. First find the length from cost ÷ rate.

Answer: (c) 10

Explanation:
1. One round = 2 × \( \frac{22}{7} \) × 49 = 308 m.
2. Rounds = 3080 ÷ 308 = 10.

Teacher's Note:
1. Divide total distance by the circumference.
2. The circumference is 308 m.

Answer: (c) r² sq. units

Explanation:
1. Base = diameter = 2r; the greatest height is the radius r.
2. Area = \( \frac{1}{2} \times 2r \times r \) = r².

Teacher's Note:
1. Take the diameter as base.
2. The height is greatest at the top of the semicircle.

Answer: (a) Rs. 1925

Explanation:
1. Circumference = 220 m, so r = 35 m.
2. Area = \( \frac{22}{7} \times 35^2 \) = 3850 m².
3. Cost = 3850 × 0.50 = Rs. 1925.

Teacher's Note:
1. Find the radius from the fencing cost.
2. Then use area × rate.

Answer: (d) \( \frac{77}{2} \) cm²

Explanation:
1. 2πr = 22 gives r = 3.5 cm.
2. Area = \( \frac{22}{7} \times 3.5^2 \) = 38.5 = 77/2 cm².

Teacher's Note:
1. Find r from the circumference first.
2. 38.5 and 77/2 are the same number.

Answer: (d) 770 m²

Explanation:
1. Inner radius r = 88 ÷ (2π) = 14 m; outer radius R = 21 m.
2. Area of road = π(R² − r²) = \( \frac{22}{7} \)(441 − 196) = 770 m².

Teacher's Note:
1. Area of a ring = π(R² − r²).
2. Outer radius = inner radius + width.

Answer: (b) \( 18\frac{2}{7} \) cm²

Explanation:
1. Diameter = diagonal = 8 cm, so r = 4 cm.
2. Circle = \( \frac{22}{7} \times 16 = \frac{352}{7} \); square = 8² ÷ 2 = 32.
3. Shaded = \( \frac{352}{7} - 32 = \frac{128}{7} \) = \( 18\frac{2}{7} \) cm².

Teacher's Note:
1. The diagonal of the square is the diameter.
2. Area of square = d²/2.

Answer: (c) 56 cm

Explanation:
1. Circumferences are proportional to diameters.
2. d = 36 + 20 = 56 cm.

Teacher's Note:
1. Circumference = πd, so diameters add directly.
2. Do not add areas by mistake.

Answer: (d) 88 mm

Explanation:
1. Circumference = πd = \( \frac{22}{7} \times 28 \) = 88 mm.

Teacher's Note:
1. Circumference = π × diameter.
2. Substitute 22/7 as the diameter is a multiple of 7.

Answer: (b) 200 mm

Explanation:
1. Circle: 88 mm. Four diameters: 4 × 28 = 112 mm.
2. Total = 88 + 112 = 200 mm.

Teacher's Note:
1. Add the circumference and the four diameters.
2. Each diameter is 28 mm.

Answer: (d) 62.86 mm

Explanation:
1. Silver radius = 44 ÷ (2 × \( \frac{22}{7} \)) = 7 mm.
2. Outer radius = 7 + 3 = 10 mm.
3. Outer circumference = 2π × 10 = 62.86 mm (approximately).

Teacher's Note:
1. Find the silver radius from its circumference.
2. Add the width of the ring for the outer radius.

Answer: (c) 51π mm²

Explanation:
1. Silver radius 7 mm, outer radius 10 mm.
2. Gold ring area = π(10² − 7²) = 51π mm².

Teacher's Note:
1. Area of the gold ring is π(R² − r²).
2. Read the question wording carefully.