CBSE Class 10 Mathematics Areas Related to Circles MCQs Set 17

Answer: (c) 300%

Explanation:
1. New radius = 2r, so the new area = 4πr², four times the original.
2. Increase = 300%.

Teacher's Note:
1. Area scales with the square of the radius.
2. Increase = new − original.

Answer: (b) \( \frac{3}{11} \)

Explanation:
1. Side = 2r. Square area = 4r²; the four quarter circles inside it total πr².
2. Uncovered : covered = (4 − π)r² : πr² = \( \frac{6}{7} : \frac{22}{7} \) = 3 : 11, i.e. \( \frac{3}{11} \).

Teacher's Note:
1. Four quarter circles at the corners add up to one full circle.
2. Compute the ratio before simplifying.

Answer: (c) \( \frac{a^2}{2} \)

Explanation:
1. AB = \( a\sqrt{2} \), so the semicircle has area \( \frac{1}{2}\pi \cdot \frac{a^2}{2} = \frac{\pi a^2}{4} \).
2. Segment of the quadrant = \( \frac{\pi a^2}{4} - \frac{a^2}{2} \).
3. Shaded area = \( \frac{\pi a^2}{4} - \left(\frac{\pi a^2}{4} - \frac{a^2}{2}\right) = \frac{a^2}{2} \).

Teacher's Note:
1. This is the lune of Hippocrates; its area equals the area of the right triangle.
2. Subtract the segment from the semicircle.

Answer: (b) \( (4 - \pi) \) cm²

Explanation:
1. The centres form a square of side 2 cm (area 4 cm²).
2. The four circles occupy a quarter each inside this square, i.e. a total of π cm².
3. Gap = (4 − π) cm².

Teacher's Note:
1. Join the four centres to form a square.
2. Four quarter circles add up to a full circle.

Answer: (a) 56 m

Explanation:
1. πr² = 9856, so r² = 9856 × \( \frac{7}{22} \) = 3136.
2. r = 56 m.

Teacher's Note:
1. The grazing region is a full circle when the cow is in an open field.
2. Take the square root carefully.

Answer: (c) 346.5 m²

Explanation:
1. At a corner, the horse grazes a quarter circle of radius 21 m.
2. Area = \( \frac{1}{4} \times \frac{22}{7} \times 21^2 \) = 346.5 m².

Teacher's Note:
1. A corner gives a quarter circle (90° sector).
2. The rope is shorter than both sides, so no overlap issues.

Answer: (a) 4π cm²

Explanation:
1. AO = 8 cm. The small circle lies inside the circle centred at A and touches it internally, so AO = 10 − r and r = 2 cm.
2. Area = π × 2² = 4π cm².

Teacher's Note:
1. Internal contact: distance between centres equals the difference of radii.
2. Here the small circle is inside the circle centred at A.

Answer: (a) \( \frac{2\sqrt{3} - \pi}{2} \) cm²

Explanation:
1. The centres form an equilateral triangle of side 2 cm with area \( \sqrt{3} \) cm².
2. Three 60° sectors of radius 1 cm lie inside it: total area \( \frac{\pi}{2} \) cm².
3. Gap = \( \sqrt{3} - \frac{\pi}{2} = \frac{2\sqrt{3} - \pi}{2} \) cm².

Teacher's Note:
1. The three sectors together make half a circle.
2. Area of an equilateral triangle = (√3/4) × side².

Answer: (b) 64π cm²

Explanation:
1. Radii are in the ratio 3 : 4, so diameters are also 3 : 4; with difference 4 cm the diameters are 12 cm and 16 cm.
2. Outer radius = 8 cm, area = 64π cm².

Teacher's Note:
1. Ratio of areas is the square of the ratio of radii.
2. Ratio of diameters equals ratio of radii.

Answer: (d) 7 m

Explanation:
1. Width = R − r = \( \frac{396 - 352}{2\pi} = \frac{44}{2\pi} \).
2. With π = 22/7 this equals 7 m.

Teacher's Note:
1. Circumference = 2πr, so the difference of circumferences is 2π × width.
2. Use π = 22/7.

Answer: (c) 50π − 100

Explanation:
1. Each quarter circle of radius 10 has area 25π and covers the lens plus the triangle of area 50.
2. Lens = 2 × (25π − 50) = 50π − 100.

Teacher's Note:
1. Two quarter circles overlap in a lens shape inside the square.
2. Lens = sum of the two quarter circles − area of the square.

Answer: (c) 154 cm²

Explanation:
1. Side = 11 cm, so the length of the wire = 44 cm.
2. 2πr = 44 gives r = 7 cm, so area = \( \frac{22}{7} \times 49 \) = 154 cm².

Teacher's Note:
1. The length of the wire does not change when reshaped.
2. For a given perimeter the circle encloses the greater area.

Answer: (c) 204π

Explanation:
1. Outer radius = 52 m and inner radius = 50 m.
2. Area = π(52² − 50²) = π × 102 × 2 = 204π m².

Teacher's Note:
1. Use the difference of squares: R² − r² = (R + r)(R − r).
2. Add the width to the radius, not to the diameter.

Answer: (a) Rs. 2002

Explanation:
1. Inner radius = 21 m, outer radius = 24.5 m.
2. Area = \( \frac{22}{7} \times (24.5^2 - 21^2) \) = \( \frac{22}{7} \times 159.25 \) = 500.5 m².
3. Cost = 500.5 × 4 = Rs. 2002.

Teacher's Note:
1. Path area = outer circle − inner circle.
2. Multiply by the rate at the end.

Answer: (c) 4

Explanation:
1. 2πr = πr² gives r = 2.
2. Diameter = 4.

Teacher's Note:
1. Equate the two formulas and cancel πr.
2. The answer is a pure number because units are ignored.

Answer: (b) 8 m

Explanation:
1. OA = OC = r and all sides of the rhombus equal r, so the diagonal AC = r as well: the rhombus consists of two equilateral triangles of side r.
2. Area = \( 2 \times \frac{\sqrt{3}}{4} r^2 = \frac{\sqrt{3}}{2} r^2 = 32\sqrt{3} \), so r² = 64 and r = 8 m.

Teacher's Note:
1. The sides OA, OC, AB and BC are all radii-equal.
2. The shorter diagonal splits the rhombus into two equilateral triangles.

Answer: (b) 42 cm²

Explanation:
1. Radius = 7 cm. Area of the square = 196 cm².
2. Four quarter circles = one full circle = 154 cm².
3. Area between the circles = 196 − 154 = 42 cm².

Teacher's Note:
1. Each circle contributes a quarter inside the square.
2. Subtract the covered area from the square.

Answer: (c) 55 cm

Explanation:
1. Length of the wire = 2π × 35 = 220 cm.
2. The perimeter of the rhombus is 220 cm, so its side = 55 cm.

Teacher's Note:
1. The length of the wire stays the same.
2. A rhombus has four equal sides.

Answer: (c) 625

Explanation:
1. Circumference = \( \frac{22}{7} \times 56 \) = 176 cm.
2. Distance = 1.1 km = 110000 cm, so revolutions = 110000 ÷ 176 = 625.

Teacher's Note:
1. One revolution covers one circumference.
2. Convert kilometres to centimetres first.

Answer: (a) 3818.5 m²

Explanation:
1. Rectangle = 42 × 35 = 1470 m².
2. Semicircles on the two 42 m sides (r = 21): 2 × ½ × \( \frac{22}{7} \) × 441 = 1386 m². On the two 35 m sides (r = 17.5): 2 × ½ × \( \frac{22}{7} \) × 306.25 = 962.5 m².
3. Total = 1470 + 1386 + 962.5 = 3818.5 m².

Teacher's Note:
1. Two semicircles on opposite edges make one full circle.
2. Add all pieces to the rectangle.