CBSE Class 10 Mathematics Areas Related to Circles MCQs Set 15

Mathematics Objective Questions and Answers: Chapter 11 Areas Related to Circles

Access targeted multiple-choice questions for Chapter 11 Areas Related to Circles designed to align with the latest CBSE academic syllabus for Class 10 Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Multiple Choice Questions

Question 1: A park is a rectangle 120 m x 100 m. At its centre is a circular lawn. The area of the park excluding the lawn is 8700 m². The radius of the lawn (use π = 22/7), to two decimal places, is:
(a) 32.40 m
(b) 28.50 m
(c) 36.75 m
(d) 45.00 m
Show Answer & Explanation

Answer: (a) 32.40 m

Explanation:
1. Area of the lawn = 12000 - 8700 = 3300 m². (22/7)r² = 3300 gives r² = 1050 and r = 32.40 m.

Teacher's Note:
1. Circumference of a circle = 2 pi r and area = pi r^2. Area of a sector = (theta/360) pi r^2 and length of an arc = (theta/360) 2 pi r.

Question 2: The radii of two circles are 8 cm and 6 cm. The radius of the circle whose area equals the sum of their areas is:
(a) 14 cm
(b) 7 cm
(c) 12 cm
(d) 10 cm
Show Answer & Explanation

Answer: (d) 10 cm

Explanation:
1. πR² = π(64) + π(36) = 100π, so R = 10 cm.

Teacher's Note:
1. Circumference of a circle = 2 pi r and area = pi r^2. Area of a sector = (theta/360) pi r^2 and length of an arc = (theta/360) 2 pi r.

Question 3: The radii of two circles are 19 cm and 9 cm. The area of the circle whose circumference equals the sum of their circumferences (use π = 22/7) is:
(a) 1232 cm²
(b) 1936 cm²
(c) 2464 cm²
(d) 3136 cm²
Show Answer & Explanation

Answer: (c) 2464 cm²

Explanation:
1. 2πR = 2π(19) + 2π(9), so R = 28 cm. Area = (22/7)(784) = 2464 cm².

Teacher's Note:
1. Circumference of a circle = 2 pi r and area = pi r^2. Area of a sector = (theta/360) pi r^2 and length of an arc = (theta/360) 2 pi r.

Question 4: A car travels 1 kilometre and each of its wheels makes 450 complete revolutions. The radius of a wheel (use π = 22/7) is:
(a) 17.68 cm
(b) 70.71 cm
(c) 35.35 cm
(d) 44.44 cm
Show Answer & Explanation

Answer: (c) 35.35 cm

Explanation:
1. One revolution covers 1000/450 = 20/9 m = 222.22 cm. So 2πr = 222.22 and r = 222.22 x 7/44 = 35.35 cm.

Teacher's Note:
1. Circumference of a circle = 2 pi r and area = pi r^2. Area of a sector = (theta/360) pi r^2 and length of an arc = (theta/360) 2 pi r.

Question 5: The area enclosed between two concentric circles is 770 cm². If the outer radius is 21 cm, the inner radius (use π = 22/7) is:
(a) 7 cm
(b) 14 cm
(c) 12 cm
(d) 16 cm
Show Answer & Explanation

Answer: (b) 14 cm

Explanation:
1. (22/7)(441 - r²) = 770 gives 441 - r² = 245, so r² = 196 and r = 14 cm.

Teacher's Note:
1. Circumference of a circle = 2 pi r and area = pi r^2. Area of a sector = (theta/360) pi r^2 and length of an arc = (theta/360) 2 pi r.

Question 6: The length of the arc that subtends an angle of 30 degrees at the centre of a circle of radius 4 cm is:
(a) \( \frac{2\pi}{3} \) cm
(b) \( \frac{4\pi}{3} \) cm
(c) \( \frac{\pi}{3} \) cm
(d) \( \frac{2\pi}{9} \) cm
Show Answer & Explanation

Answer: (a) \( \frac{2\pi}{3} \) cm

Explanation:
1. Arc = (30/360)(2π)(4) = 8π/12 = 2π/3 cm.

Teacher's Note:
1. Circumference of a circle = 2 pi r and area = pi r^2. Area of a sector = (theta/360) pi r^2 and length of an arc = (theta/360) 2 pi r.

Question 7: The angle subtended at the centre of a circle of radius 5 cm by an arc of length 5π/3 cm is:
(a) 30 degrees
(b) 90 degrees
(c) 120 degrees
(d) 60 degrees
Show Answer & Explanation

Answer: (d) 60 degrees

Explanation:
1. Arc = (θ/360)(2π)(5) = 5π/3 gives θ/360 = 1/6 and θ = 60 degrees.

Teacher's Note:
1. Circumference of a circle = 2 pi r and area = pi r^2. Area of a sector = (theta/360) pi r^2 and length of an arc = (theta/360) 2 pi r.

Question 8: An arc of length 20π cm subtends an angle of 144 degrees at the centre of a circle. The radius of the circle is:
(a) 20 cm
(b) 30 cm
(c) 50 cm
(d) 25 cm
Show Answer & Explanation

Answer: (d) 25 cm

Explanation:
1. (144/360)(2πr) = 20π gives (2/5)(2r) = 20, so r = 25 cm.

Teacher's Note:
1. Circumference of a circle = 2 pi r and area = pi r^2. Area of a sector = (theta/360) pi r^2 and length of an arc = (theta/360) 2 pi r.

Question 9: An arc of length 15 cm subtends an angle of 45 degrees at the centre of a circle. The radius of the circle is:
(a) \( \frac{30}{\pi} \) cm
(b) \( \frac{120}{\pi} \) cm
(c) \( \frac{60}{\pi} \) cm
(d) \( \frac{15}{\pi} \) cm
Show Answer & Explanation

Answer: (c) \( \frac{60}{\pi} \) cm

Explanation:
1. (45/360)(2πr) = 15 gives πr/4 = 15, so r = 60/π cm (about 19.1 cm).

Teacher's Note:
1. Circumference of a circle = 2 pi r and area = pi r^2. Area of a sector = (theta/360) pi r^2 and length of an arc = (theta/360) 2 pi r.

Question 10: The angle subtended at the centre of a circle of radius a by an arc of length aπ/4 is:
(a) \( 90^\circ \)
(b) \( 45^\circ \)
(c) \( 30^\circ \)
(d) \( 60^\circ \)
Show Answer & Explanation

Answer: (b) \( 45^\circ \)

Explanation:
1. The angle in radians is arc/radius = π/4, which is 45 degrees.

Teacher's Note:
1. Circumference of a circle = 2 pi r and area = pi r^2. Area of a sector = (theta/360) pi r^2 and length of an arc = (theta/360) 2 pi r.

Question 11: A chord AB of a circle of radius 15 cm makes an angle of 60 degrees at the centre. Taking π = 3.14 and √3 = 1.73, the area of the minor segment, to one decimal place, is:
(a) 20.4 cm²
(b) 17.6 cm²
(c) 24.9 cm²
(d) 40.8 cm²
Show Answer & Explanation

Answer: (a) 20.4 cm²

Explanation:
1. Sector area = (60/360)(3.14)(225) = 117.75 cm². The triangle OAB is equilateral with area (√3/4)(225) = 97.31 cm². Minor segment = 117.75 - 97.31 = 20.4 cm².

Teacher's Note:
1. Circumference of a circle = 2 pi r and area = pi r^2. Area of a sector = (theta/360) pi r^2 and length of an arc = (theta/360) 2 pi r.

Question 12: In a circle of radius 21 cm, an arc subtends an angle of 60 degrees at the centre. The length of the arc (use π = 22/7) is:
(a) 22 cm
(b) 11 cm
(c) 44 cm
(d) 66 cm
Show Answer & Explanation

Answer: (a) 22 cm

Explanation:
1. Arc = (60/360)(2)(22/7)(21) = 22 cm.

Teacher's Note:
1. Circumference of a circle = 2 pi r and area = pi r^2. Area of a sector = (theta/360) pi r^2 and length of an arc = (theta/360) 2 pi r.

Question 13: In a circle of radius 21 cm, an arc subtends 60 degrees at the centre. The area of the sector formed by the arc (use π = 22/7) is:
(a) 154 cm²
(b) 308 cm²
(c) 462 cm²
(d) 231 cm²
Show Answer & Explanation

Answer: (d) 231 cm²

Explanation:
1. Area = (60/360)(22/7)(441) = 231 cm².

Teacher's Note:
1. Circumference of a circle = 2 pi r and area = pi r^2. Area of a sector = (theta/360) pi r^2 and length of an arc = (theta/360) 2 pi r.

Question 14: In a circle of radius 21 cm, an arc subtends 60 degrees at the centre. Taking π = 22/7 and √3 = 1.732, the area of the segment formed by the corresponding chord is:
(a) 20.03 cm²
(b) 80.10 cm²
(c) 40.05 cm²
(d) 98.55 cm²
Show Answer & Explanation

Answer: (c) 40.05 cm²

Explanation:
1. Sector = 231 cm². The triangle is equilateral with area (√3/4)(441) = 190.95 cm². Segment = 231 - 190.95 = 40.05 cm².

Teacher's Note:
1. Circumference of a circle = 2 pi r and area = pi r^2. Area of a sector = (theta/360) pi r^2 and length of an arc = (theta/360) 2 pi r.

Question 15: A chord of a circle of radius 10 cm subtends a right angle at the centre. The area of the minor segment (use π = 3.14) is:
(a) 50.0 cm²
(b) 28.5 cm²
(c) 57.0 cm²
(d) 78.5 cm²
Show Answer & Explanation

Answer: (b) 28.5 cm²

Explanation:
1. Minor sector = (1/4)(3.14)(100) = 78.5 cm². Triangle = (1/2)(10)(10) = 50 cm². Minor segment = 78.5 - 50 = 28.5 cm².

Teacher's Note:
1. Circumference of a circle = 2 pi r and area = pi r^2. Area of a sector = (theta/360) pi r^2 and length of an arc = (theta/360) 2 pi r.

Question 16: A chord of a circle of radius 10 cm subtends a right angle at the centre. The area of the major segment (use π = 3.14) is:
(a) 235.5 cm²
(b) 285.5 cm²
(c) 264.0 cm²
(d) 314.0 cm²
Show Answer & Explanation

Answer: (b) 285.5 cm²

Explanation:
1. Circle = 314 cm². Minor segment = 28.5 cm² (sector 78.5 minus triangle 50). Major segment = 314 - 28.5 = 285.5 cm².

Teacher's Note:
1. Circumference of a circle = 2 pi r and area = pi r^2. Area of a sector = (theta/360) pi r^2 and length of an arc = (theta/360) 2 pi r.

Question 17: A circular grassy plot 42 m in diameter has a path 3.5 m wide running round it on the outside. The cost of gravelling the path at Rs. 4 per m² (use π = 22/7), in rupees, is:
(a) 2002
(b) 1001
(c) 2502
(d) 4004
Show Answer & Explanation

Answer: (a) 2002

Explanation:
1. Outer radius = 24.5 m, inner radius = 21 m. Area = (22/7)(600.25 - 441) = (22/7)(159.25) = 500.5 m². Cost = 500.5 x 4 = Rs. 2002.

Teacher's Note:
1. Circumference of a circle = 2 pi r and area = pi r^2. Area of a sector = (theta/360) pi r^2 and length of an arc = (theta/360) 2 pi r.

Question 18: Four equal circles are described about the four corners of a square of side 14 cm so that each touches two of the others. The area of the region between the circles (inside the square, outside the circles) (use π = 22/7) is:
(a) 154 cm²
(b) 84 cm²
(c) 28 cm²
(d) 42 cm²
Show Answer & Explanation

Answer: (d) 42 cm²

Explanation:
1. Each circle has radius 7 cm. The four quarter-circles inside the square total a circle of area (22/7)(49) = 154 cm². Region = 196 - 154 = 42 cm².

Teacher's Note:
1. Circumference of a circle = 2 pi r and area = pi r^2. Area of a sector = (theta/360) pi r^2 and length of an arc = (theta/360) 2 pi r.

Question 19: The radius of the circle whose area is the sum of the areas of two triangles with sides 35, 53, 66 cm and 33, 56, 65 cm, to three decimal places (use π = 22/7), is:
(a) 14.000 cm
(b) 21.000 cm
(c) 24.249 cm
(d) 29.698 cm
Show Answer & Explanation

Answer: (c) 24.249 cm

Explanation:
1. For each triangle s = 77. Area 1 = √(77 x 42 x 24 x 11) = 924 cm². Area 2 = √(77 x 44 x 21 x 12) = 924 cm². Sum = 1848 cm², so (22/7)r² = 1848 and r² = 588, r = 24.249 cm.

Teacher's Note:
1. Heron's formula: area = square root of s(s - a)(s - b)(s - c), where s is half the perimeter.

Question 20: In an equilateral triangle of side 24 cm, a circle is inscribed touching its sides. The area of the remaining portion of the triangle (use π = 22/7 and √3 = 1.732) is:
(a) 78.55 cm²
(b) 128.45 cm²
(c) 98.55 cm²
(d) 150.86 cm²
Show Answer & Explanation

Answer: (c) 98.55 cm²

Explanation:
1. Triangle area = (√3/4)(576) = 249.41 cm². Inradius r = a/(2√3) = 4√3, so r² = 48 and circle area = (22/7)(48) = 150.86 cm². Remaining = 98.55 cm².

Teacher's Note:
1. Circumference of a circle = 2 pi r and area = pi r^2. Area of a sector = (theta/360) pi r^2 and length of an arc = (theta/360) 2 pi r.

Question 21: An athletic track 14 m wide consists of two straight sections 120 m long joined by semicircular ends whose inner radius is 35 m. The area of the track (use π = 22/7) is:
(a) 3696 m²
(b) 7056 m²
(c) 6720 m²
(d) 7392 m²
Show Answer & Explanation

Answer: (b) 7056 m²

Explanation:
1. The two straight parts: 2 x 120 x 14 = 3360 m². The two semicircular ends form a ring with radii 49 m and 35 m: (22/7)(2401 - 1225) = 3696 m². Total = 7056 m².

Teacher's Note:
1. Circumference of a circle = 2 pi r and area = pi r^2. Area of a sector = (theta/360) pi r^2 and length of an arc = (theta/360) 2 pi r.

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