IMO Maths Olympiad Class 10 Chapter 12 Areas Related to Circles MCQs with Answers

Answer: (b) \( \sqrt{3\pi} \) cm

Explanation:
1. The radius is 2 cm. The shaded part is the quarter of the circle inside the square (angle 90° at O).
2. Shaded area = \( \dfrac{90}{360} \times \pi \times 2^2 = \pi \) cm².
3. Area of square = 3 × shaded area = 3π, so x² = 3π and the side is \( \sqrt{3\pi} \) cm.

Teacher's Note:
1. The corner of the square at the centre cuts out a 90° sector of the circle.
2. Do not confuse √(3π) with 3√π: the square root covers the whole product 3π.

Answer: (b) 7 m

Explanation:
1. The hexagon is made of 6 equilateral triangles, so its side equals the radius r of the circle and its area is \( 6 \times \dfrac{\sqrt{3}}{4}r^2 = \dfrac{3\sqrt{3}}{2}r^2 \).
2. Circle − hexagon: \( r^2\left(3.143 - \dfrac{3}{2} \times 1.732\right) = 0.545\,r^2 = 26.705 \).
3. So r² = 49 and r = 7 m.

Teacher's Note:
1. A regular hexagon has side equal to the radius of its circumscribed circle.
2. The book prints the options in cm although the areas are in m²; the radius is 7 m.

Answer: (a) 3π cm²

Explanation:
1. In quadrilateral ABCD: ∠B = ∠C = 90° and ∠D = 60°, so ∠A = 360° − 90° − 90° − 60° = 120°.
2. The shaded part is a sector of radius PA = 3 cm with angle 120° at A.
3. Area = \( \dfrac{120}{360} \times \pi \times 3^2 \) = 3π cm².

Teacher's Note:
1. The sector angle is the interior angle of the quadrilateral at A.
2. Use the angle sum of 360° of a quadrilateral.

Answer: (b) ₹ 308.91

Explanation:
1. The table is a rectangle 10 m × 6 m with a semicircle of diameter 6 m (radius 3 m) at each end.
2. Area = 10 × 6 + π × 3² = 60 + 3.14 × 9 = 60 + 28.26 = 88.26 m².
3. Cost = 88.26 × 3.50 = ₹ 308.91.

Teacher's Note:
1. The two semicircles together make one full circle of radius 3 m.
2. The 10 m in the figure is the length of the rectangular part only.

Answer: (d) None of these

Explanation:
1. There are two semicircles on the 100 m sides (radius 50 m) and two on the 50 m sides (radius 25 m).
2. Area of beds = π × 50² + π × 25² = 3125π ≈ 9812.5 m² (π = 3.14).
3. Cost at ₹ 0.60 per m² ≈ ₹ 5887.50, which matches none of the options, so the answer is None of these.

Teacher's Note:
1. Convert 60 paise to ₹ 0.60 before multiplying.
2. Two semicircles on opposite sides make one full circle.

Answer: (d) \( \dfrac{2\pi}{3}\left(8\sqrt{3}-1\right) \)

Explanation:
1. The centres A, B, C form an equilateral triangle of side 4 cm, and the centre O is at distance \( \dfrac{4}{\sqrt{3}} \) from each. So \( R = \dfrac{4}{\sqrt{3}} + 2 \) cm.
2. In the figure the whole triangle ABC is shaded. So the shaded region = big circle − 3 small circles + the three 60° sectors lying inside the triangle: \( \pi R^2 - 3\pi \times 2^2 + 3 \times \dfrac{60}{360}\pi \times 2^2 \).
3. \( \pi\left(\dfrac{16}{3} + \dfrac{16}{\sqrt{3}} + 4\right) - 12\pi + 2\pi = \pi\left(\dfrac{16}{\sqrt{3}} - \dfrac{2}{3}\right) = \dfrac{2\pi}{3}\left(8\sqrt{3}-1\right) \) cm² ≈ 26.93 cm².

Teacher's Note:
1. The small circles only touch, so there is no overlap to subtract.
2. None of the book's options equals 26.93 cm² (its key (d) is about 26.28 cm²), so option (d) has been corrected.

Answer: (c) (P) → (ii); (Q) → (iii); (R) → (i)

Explanation:
1. (P) Segment of a circle of radius 21 cm with centre angle 120°: sector − triangle = \( \dfrac{120}{360} \times \dfrac{22}{7} \times 21^2 - \dfrac{1}{2} \times 21^2 \sin 120^\circ \) = 462 − 190.96 = 271.04 cm², which is (ii).
2. (Q) Circle of radius 28 cm minus the regular hexagon inside it = 6 × (sector of 60° − equilateral triangle) = 6 × (410.67 − 339.48) = 427.14 cm², which is (iii).
3. (R) Square of side 35 cm minus two semicircles of radius 17.5 cm = 1225 − 2 × 481.25 = 262.50 cm², which is (i). So (P) → (ii); (Q) → (iii); (R) → (i).

Teacher's Note:
1. Each segment is a sector minus a triangle.
2. Check the three results against the areas given in Column-II.

Answer: (d) \( 4\sqrt{3} - 6 \)

Explanation:
1. ∠BOC = 2 × 60° = 120°, and OM bisects it, so ∠BOM = 60°. Also OM = 2 cm.
2. Let the small radius be r. Then OO₁ = 2 − r and the distance from O₁ to OB is r, so \( r = (2 - r)\sin 60^\circ = (2-r)\dfrac{\sqrt{3}}{2} \).
3. \( r = \dfrac{2\sqrt{3}}{2+\sqrt{3}} = 2\sqrt{3}\left(2-\sqrt{3}\right) = 4\sqrt{3} - 6 \) ≈ 0.93 cm.

Teacher's Note:
1. The angle at the centre is twice the angle at the circumference.
2. The centre of a circle touching two lines lies on the bisector of the angle between them.

Answer: (d) 6 sq. units

Explanation:
1. BC = √(3² + 4²) = 5 units. By Pythagoras the semicircle on BC equals the sum of the semicircles on AB and AC.
2. Shaded region (the two lunes) = (semicircle on AB + semicircle on AC + △ABC) − semicircle on BC.
3. The semicircles cancel, leaving the area of △ABC = ½ × 3 × 4 = 6 sq. units.

Teacher's Note:
1. This is the lunes of Hippocrates: the area of the lunes equals the area of the right triangle.
2. No value of π is needed.

Answer: (b) (i) \( \dfrac{550}{7} \) cm², (ii) \( \dfrac{1100}{7} \) cm², (iii) \( \dfrac{550}{7} \) cm²

Explanation:
1. (i) The inscribed circle has diameter 10 cm, so its area is \( \dfrac{22}{7} \times 5^2 = \dfrac{550}{7} \) cm².
2. (ii) The circumscribed circle has the diagonal as diameter: \( 10\sqrt{2} \), so \( r = 5\sqrt{2} \) and area \( \dfrac{22}{7} \times 50 = \dfrac{1100}{7} \) cm².
3. (iii) Shaded = circumscribed − inscribed = \( \dfrac{1100}{7} - \dfrac{550}{7} = \dfrac{550}{7} \) cm². This is option (b).

Teacher's Note:
1. The diagonal of a square of side a is a√2, and it is the diameter of the circumscribed circle.
2. The circumscribed circle has exactly twice the area of the inscribed circle.