CBSE Class 10 Mathematics Areas Related to Circles MCQs Set 14

Download CBSE MCQs for Class 10 Mathematics: Chapter 11 Areas Related to Circles

Explore reliable objective questions for Chapter 11 Areas Related to Circles tailored for Class 10 learners. Utilizing these Mathematics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Chapter-wise Objective Questions: Chapter 11 Areas Related to Circles

Navigate directly to the 50 objective questions for Chapter 11 Areas Related to Circles using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: The area of a circle of radius 4.2 cm (use π = 22/7) is:
(a) 55.44 cm²
(b) 26.40 cm²
(c) 52.80 cm²
(d) 110.88 cm²
Show Answer & Explanation

Answer: (a) 55.44 cm²

Explanation:
1. Area = (22/7)(4.2)² = 55.44 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 2: The circumference of a circle whose area is 301.84 cm² (use π = 22/7) is:
(a) 30.8 cm
(b) 49.3 cm
(c) 98.0 cm
(d) 61.6 cm
Show Answer & Explanation

Answer: (d) 61.6 cm

Explanation:
1. r² = 301.84 x 7/22 = 96.04, so r = 9.8 cm and the circumference = 2(22/7)(9.8) = 61.6 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 area of a circle whose circumference is 44 cm (use π = 22/7) is:
(a) 44 cm²
(b) 77 cm²
(c) 154 cm²
(d) 308 cm²
Show Answer & Explanation

Answer: (c) 154 cm²

Explanation:
1. 2πr = 44 gives r = 7 cm, so the area = (22/7)(49) = 154 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: The circumference of a circle exceeds its diameter by 16.8 cm. Using π = 22/7, the circumference is:
(a) 16.80 cm
(b) 24.64 cm
(c) 21.28 cm
(d) 30.80 cm
Show Answer & Explanation

Answer: (b) 24.64 cm

Explanation:
1. πd - d = 16.8 gives (15/7)d = 16.8, so d = 7.84 cm and the circumference = (22/7)(7.84) = 24.64 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: A horse is tied to a pole with a 28 m long string. The area over which the horse can graze (use π = 22/7) is:
(a) 1232 m²
(b) 2464 m²
(c) 4928 m²
(d) 616 m²
Show Answer & Explanation

Answer: (b) 2464 m²

Explanation:
1. The horse grazes a circle of radius 28 m: (22/7)(784) = 2464 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.

Question 6: A steel wire bent into a square encloses an area of 121 cm². The same wire bent into a circle encloses an area (use π = 22/7) of:
(a) 154 cm²
(b) 121 cm²
(c) 77 cm²
(d) 308 cm²
Show Answer & Explanation

Answer: (a) 154 cm²

Explanation:
1. The side of the square is 11 cm, so the wire is 44 cm long. For the circle 2πr = 44, r = 7 cm and the area = 154 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 7: A sector of a circle of radius 4 cm contains an angle of 30 degrees. The area of the sector is:
(a) \( \frac{8\pi}{3} \) cm²
(b) \( \frac{2\pi}{3} \) cm²
(c) \( \frac{4\pi}{9} \) cm²
(d) \( \frac{4\pi}{3} \) cm²
Show Answer & Explanation

Answer: (d) \( \frac{4\pi}{3} \) cm²

Explanation:
1. Area = (30/360)π(4)² = 16π/12 = 4π/3 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 8: A sector of a circle of radius 8 cm contains an angle of 135 degrees. The area of the sector is:
(a) \( 48\pi \) cm²
(b) \( 12\pi \) cm²
(c) \( 24\pi \) cm²
(d) \( 32\pi \) cm²
Show Answer & Explanation

Answer: (c) \( 24\pi \) cm²

Explanation:
1. Area = (135/360)π(64) = (3/8)(64)π = 24π 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 9: The area of a sector of a circle of radius 2 cm is π cm². The angle contained by the sector is:
(a) 60 degrees
(b) 120 degrees
(c) 90 degrees
(d) 45 degrees
Show Answer & Explanation

Answer: (c) 90 degrees

Explanation:
1. (θ/360)π(4) = π gives θ = 90 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 10: A plot is a rectangle ABCD with a semicircle on side BC, where AB = 60 m and BC = 28 m (the semicircle has BC as its diameter). The area of the plot (use π = 22/7) is:
(a) 1680 m²
(b) 1988 m²
(c) 2296 m²
(d) 2142 m²
Show Answer & Explanation

Answer: (b) 1988 m²

Explanation:
1. Rectangle = 60 x 28 = 1680 m². Semicircle of radius 14: (1/2)(22/7)(196) = 308 m². Total = 1988 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.

Question 11: A playground is a rectangle 36 m by 24.5 m with two semicircles, each having a shorter side as diameter, added outside. Its area (use π = 22/7) is:
(a) 1353.625 m²
(b) 882.000 m²
(c) 1235.800 m²
(d) 1471.600 m²
Show Answer & Explanation

Answer: (a) 1353.625 m²

Explanation:
1. Rectangle = 36 x 24.5 = 882 m². The two semicircles together form a circle of radius 12.25 m: (22/7)(150.0625) = 471.625 m². Total = 1353.625 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.

Question 12: The outer circumference of a circular race-track is 528 m and the track is everywhere 14 m wide. The cost of levelling the track at 50 paise per m² (use π = 22/7), in rupees, is:
(a) 6776
(b) 1694
(c) 4235
(d) 3388
Show Answer & Explanation

Answer: (d) 3388

Explanation:
1. Outer radius R = 528 x 7/44 = 84 m and inner radius r = 70 m. Area = (22/7)(84² - 70²) = (22/7)(2156) = 6776 m². Cost = 6776 x 0.50 = Rs. 3388.

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: From the four corners of a rectangular piece 20 m by 15 m, quadrants of radius 3.5 m are cut. The area of the remaining part (use π = 22/7) is:
(a) 338.5 m²
(b) 269.5 m²
(c) 247.0 m²
(d) 261.5 m²
Show Answer & Explanation

Answer: (d) 261.5 m²

Explanation:
1. The four quadrants make one full circle of radius 3.5 m: (22/7)(12.25) = 38.5 m². Remaining = 300 - 38.5 = 261.5 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.

Question 14: Four equal circles, each of radius 5 cm, touch each other (their centres form a square of side 10 cm). The area enclosed between them (use π = 3.14) is:
(a) 78.5 cm²
(b) 43.0 cm²
(c) 21.5 cm²
(d) 31.4 cm²
Show Answer & Explanation

Answer: (c) 21.5 cm²

Explanation:
1. Area of the square of centres = 100 cm². The four quarter-circles inside it total one circle of area 3.14 x 25 = 78.5 cm². Required area = 100 - 78.5 = 21.5 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 15: Four cows are tethered at the corners of a square plot of side 50 m so that each can graze the maximum area without reaching another. The area left ungrazed (use π = 3.14) is:
(a) 1962.5 m²
(b) 537.5 m²
(c) 625.0 m²
(d) 475.0 m²
Show Answer & Explanation

Answer: (b) 537.5 m²

Explanation:
1. Each rope is 25 m. The four quarter-circles make a circle of area 3.14 x 625 = 1962.5 m². Ungrazed = 2500 - 1962.5 = 537.5 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.

Question 16: The circumferences of two circles are in the ratio 2 : 3. The ratio of their areas is:
(a) 4 : 9
(b) 2 : 3
(c) 8 : 27
(d) 3 : 2
Show Answer & Explanation

Answer: (a) 4 : 9

Explanation:
1. Circumference is proportional to the radius, so the radii are in the ratio 2 : 3. Areas are proportional to the squares: 4 : 9.

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: The side of a square is 10 cm. The area of its circumscribed circle (use π = 3.14) is:
(a) 157 cm²
(b) 78 cm²
(c) 314 cm²
(d) 125 cm²
Show Answer & Explanation

Answer: (a) 157 cm²

Explanation:
1. The circle passes through the vertices and its diameter is the diagonal 10√2, so r² = 50 and the area = 3.14 x 50 = 157 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 18: The sum of the radii of two circles is 140 cm and the difference of their circumferences is 88 cm. The diameters of the circles (use π = 22/7) are:
(a) 140 cm and 140 cm
(b) 77 cm and 63 cm
(c) 168 cm and 112 cm
(d) 154 cm and 126 cm
Show Answer & Explanation

Answer: (d) 154 cm and 126 cm

Explanation:
1. 2π(R - r) = 88 gives R - r = 14. With R + r = 140, R = 77 cm and r = 63 cm. The diameters are 154 cm and 126 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 19: The area of a circle inscribed in an equilateral triangle is 154 cm². The perimeter of the triangle (use π = 22/7 and √3 = 1.73) is:
(a) 48.44 cm
(b) 60.55 cm
(c) 72.66 cm
(d) 84.00 cm
Show Answer & Explanation

Answer: (c) 72.66 cm

Explanation:
1. πr² = 154 gives r = 7 cm. For an equilateral triangle r = a/(2√3), so a = 14√3 = 24.22 cm. Perimeter = 3a = 72.66 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 20: A circular field is fenced at Rs. 12 per metre at a total cost of Rs. 2640. The field is then ploughed at Rs. 0.50 per m². Using π = 22/7, the cost of ploughing, in rupees, is:
(a) 1540
(b) 1925
(c) 2310
(d) 3850
Show Answer & Explanation

Answer: (b) 1925

Explanation:
1. Length of fence = 2640/12 = 220 m = 2πr, so r = 35 m. Area = (22/7)(1225) = 3850 m². Cost = 3850 x 0.50 = Rs. 1925.

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: If a square is inscribed in a circle, the ratio of the area of the circle to the area of the square is:
(a) \( 2 : \pi \)
(b) \( \pi : 2 \)
(c) \( \pi : 4 \)
(d) \( \pi : 1 \)
Show Answer & Explanation

Answer: (b) \( \pi : 2 \)

Explanation:
1. If the square has side a, the circle has diameter a√2, so r² = a²/2. Circle : square = πa²/2 : a² = π : 2.

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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FAQs

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