CBSE Class 10 Mathematics Areas Related to Circles MCQs Set 11

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.

Download Chapter 11 Areas Related to Circles MCQs with Answers

View or download the dedicated Chapter 11 Areas Related to Circles MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Mathematics evaluations.

Multiple Choice Questions

Question 1: If the difference between the circumference and the radius of a circle is 37 cm, then, using π = 22/7, the radius of the circle (in cm) is
(a) 154
(b) 44
(c) 14
(d) 7
Show Answer & Explanation

Answer: (d) 7

Explanation:
1. Circumference − radius = 2πr − r = r(2π − 1) = 37.
2. With π = 22/7: r(44/7 − 1) = r × 37/7 = 37.
3. So r = 7 cm.

Teacher's Note:
1. Take r outside as a common factor.
2. Convert 2π to 44/7 before subtracting 1.

Question 2: In the figure, AOB is a sector of a circle of radius 10.5 cm and central angle 60°. The perimeter of the sector (in cm) is (Take π = 22/7.)
60°OAB10.5 cm (a) 32
(b) 21
(c) 11
(d) 35
Show Answer & Explanation

Answer: (a) 32

Explanation:
1. Arc length AB = (60° ÷ 360°) × 2 × (22/7) × 10.5 = 11 cm.
2. Perimeter = OA + OB + arc AB = 10.5 + 10.5 + 11 = 32 cm.

Teacher's Note:
1. The perimeter of a sector includes the two radii as well as the arc.
2. Do not forget to add 2r.

Question 3: The area of the sector of a circle of radius 6 cm whose central angle is 30° is (take π = 3.14)
(a) 9.42 cm²
(b) 7.42 cm²
(c) 8.42 cm²
(d) 10.42 cm²
Show Answer & Explanation

Answer: (a) 9.42 cm²

Explanation:
1. Area = (θ ÷ 360°) × πr² = (30 ÷ 360) × 3.14 × 6 × 6.
2. = (1/12) × 3.14 × 36 = 3 × 3.14 = 9.42 cm².

Teacher's Note:
1. 30° is one-twelfth of a full turn.
2. Use the given value of π.

Question 4: If π is taken as 22/7, the distance (in metres) covered by a wheel of diameter 35 cm in one revolution is
(a) 2.2
(b) 1.1
(c) 9.625
(d) 96.25
Show Answer & Explanation

Answer: (b) 1.1

Explanation:
1. Distance in one revolution = circumference = πd = (22/7) × 35 = 110 cm.
2. 110 cm = 1.1 m.

Teacher's Note:
1. One revolution covers one circumference.
2. Convert centimetres to metres at the end.

Question 5: The circumference of a circular field is 528 cm. Then the radius is (take π = 22/7)
(a) 84 cm
(b) 64 cm
(c) 55 cm
(d) 45 cm
Show Answer & Explanation

Answer: (a) 84 cm

Explanation:
1. 2πr = 528, so 2 × (22/7) × r = 528.
2. r = 528 × 7 ÷ 44 = 84 cm.

Teacher's Note:
1. Write 2πr = 528 and make r the subject.
2. 528 ÷ 44 = 12.

Question 6: The length of the arc of a sector of angle θ° of a circle with radius R is
(a) \( \frac{2\pi R\theta}{180^\circ} \)
(b) \( \frac{2\pi R\theta}{360^\circ} \)
(c) \( \frac{\pi R^2\theta}{180^\circ} \)
(d) \( \frac{\pi R^2\theta}{360^\circ} \)
Show Answer & Explanation

Answer: (b) \( \frac{2\pi R\theta}{360^\circ} \)

Explanation:
1. Arc length is the fraction θ ÷ 360° of the circumference 2πR.
2. Length = (θ ÷ 360°) × 2πR = 2πRθ ÷ 360°.

Teacher's Note:
1. Options with R² give areas, not lengths.
2. A length has only one power of R.

Question 7: If the circumference and the area of a circle are numerically equal, then the diameter of the circle is
(a) \( \frac{\pi}{2} \)
(b) \( 2\pi \)
(c) 2
(d) 4
Show Answer & Explanation

Answer: (d) 4

Explanation:
1. 2πr = πr², so r = 2.
2. Diameter = 2r = 4.

Teacher's Note:
1. Cancel πr from both sides.
2. Do not stop at the radius; the question asks for the diameter.

Question 8: The diameter of the driving wheel of a bus is 140 cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 km/h? (Take π = 22/7.)
(a) 200
(b) 240
(c) 250
(d) 260
Show Answer & Explanation

Answer: (c) 250

Explanation:
1. Distance per minute = 66 × 1000 ÷ 60 = 1100 m.
2. Circumference of the wheel = 2 × (22/7) × 0.70 = 4.4 m.
3. Revolutions per minute = 1100 ÷ 4.4 = 250.

Teacher's Note:
1. Convert the speed to metres per minute.
2. Divide the distance by the circumference.

Question 9: A wheel makes 1000 revolutions in covering a distance of 88 km. The radius of the wheel is (take π = 22/7)
(a) 11 m
(b) 14 m
(c) 12 m
(d) 10 m
Show Answer & Explanation

Answer: (b) 14 m

Explanation:
1. Distance in one revolution = 88 km ÷ 1000 = 88 m.
2. 2πr = 88, so r = 88 × 7 ÷ 44 = 14 m.

Teacher's Note:
1. Distance per revolution = circumference.
2. Keep 88 km = 88000 m in mind before dividing.

Question 10: The inner circumference of a circular race track, 14 m wide, is 440 m. The radius of the outer circle is (take π = 22/7)
(a) 85 m
(b) 82 m
(c) 80 m
(d) 84 m
Show Answer & Explanation

Answer: (d) 84 m

Explanation:
1. Inner radius r: 2πr = 440, so r = 440 × 7 ÷ 44 = 70 m.
2. Outer radius = 70 + 14 = 84 m.

Teacher's Note:
1. Find the inner radius from the inner circumference.
2. Add the width of the track to get the outer radius.

Question 11: Two concentric circles form a ring. The inner and outer circumferences of the ring are \( 50\frac{2}{7} \) m and \( 75\frac{3}{7} \) m respectively. The width of the ring is (take π = 22/7)
(a) 1 m
(b) 2 m
(c) 3 m
(d) 4 m
Show Answer & Explanation

Answer: (d) 4 m

Explanation:
1. Inner circumference = 352/7 m, so 2πr = 352/7 and r = 8 m.
2. Outer circumference = 528/7 m, so 2πR = 528/7 and R = 12 m.
3. Width = R − r = 12 − 8 = 4 m.

Teacher's Note:
1. Change the mixed numbers into improper fractions first.
2. Width = outer radius − inner radius.

Question 12: A sector of 120° cut out from a circle has an area of \( 9\frac{3}{7} \) cm². The radius of the circle is (take π = 22/7)
(a) 1 cm
(b) 1 m
(c) 3 cm
(d) 3 m
Show Answer & Explanation

Answer: (c) 3 cm

Explanation:
1. Area = (120 ÷ 360) × (22/7) × r² = 66/7.
2. (22/21) r² = 66/7, so r² = 66/7 × 21/22 = 9.
3. r = 3 cm.

Teacher's Note:
1. 9 3/7 = 66/7.
2. 120° is one-third of the circle.

Question 13: If the radius of a circle is increased by 75%, then its circumference will increase by
(a) 25%
(b) 50%
(c) 75%
(d) 100%
Show Answer & Explanation

Answer: (c) 75%

Explanation:
1. New radius = 1.75R, so the new circumference = 2π × 1.75R = 1.75 × (2πR).
2. The circumference is proportional to the radius, so it increases by 75%.

Teacher's Note:
1. Circumference is directly proportional to radius.
2. Area, not circumference, depends on the square of the radius.

Question 14: A goes around a circular path 8 times in 40 minutes. If the diameter of the circle is increased to 10 times the original diameter, the time required by A to go around the new path once at the same speed is
(a) 20 min
(b) 25 min
(c) 50 min
(d) 100 min
Show Answer & Explanation

Answer: (c) 50 min

Explanation:
1. Original path: 8 rounds in 40 minutes, so one round takes 5 minutes.
2. The new circumference is 10 times the old one, so one round takes 10 × 5 = 50 minutes.

Teacher's Note:
1. Time for one round is proportional to the circumference at constant speed.
2. Find the time for one old round first.

Question 15: The number of rounds that a wheel of diameter \( \frac{7}{11} \) m will make in moving a distance of 4 km is (take π = 22/7)
(a) 1000 rounds
(b) 2000 rounds
(c) 3000 rounds
(d) none of these
Show Answer & Explanation

Answer: (b) 2000 rounds

Explanation:
1. Circumference = πd = (22/7) × (7/11) = 2 m.
2. Number of rounds = 4000 m ÷ 2 m = 2000.

Teacher's Note:
1. Convert 4 km to 4000 m.
2. Rounds = total distance ÷ circumference.

Question 16: A pendulum swings through an angle of 30° and describes an arc 6.6 cm long. The length of the pendulum is (take π = 22/7)
(a) 12.6 cm
(b) 10.6 cm
(c) 5 cm
(d) 10 cm
Show Answer & Explanation

Answer: (a) 12.6 cm

Explanation:
1. The pendulum length is the radius r of the arc.
2. 6.6 = (30 ÷ 360) × 2 × (22/7) × r = (11/21) r.
3. r = 6.6 × 21 ÷ 11 = 12.6 cm.

Teacher's Note:
1. The swinging length is the radius of the circle traced.
2. 30° is one-twelfth of a full circle.

Question 17: If a chord of a circle of radius 14 cm makes an angle of 90° at the centre, then the area of the major segment is (take π = 22/7)
90°OAB14 cm (a) 560 cm²
(b) 300 cm²
(c) 160 cm²
(d) none of these
Show Answer & Explanation

Answer: (a) 560 cm²

Explanation:
1. Major sector (270°) = (270 ÷ 360) × (22/7) × 14² = 462 cm².
2. Triangle OAB = ½ × 14 × 14 = 98 cm².
3. Major segment = major sector + triangle = 462 + 98 = 560 cm².

Teacher's Note:
1. For a major segment add the triangle; for a minor segment subtract it.
2. The angle of the major sector is 360° − 90° = 270°.

Question 18:
Assertion (A): A sector is the region between a chord and its corresponding arc.
Reason (R): To define an arc we need at least three points.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. A sector is bounded by an arc and the two radii to its end points; the region between a chord and an arc is a segment. The Assertion is false.
2. An arc is fixed by its two end points and one more point on it, so three points are needed. The Reason is true.
3. Answer: (d).

Teacher's Note:
1. Do not confuse sector (two radii) with segment (a chord).
2. Check the Assertion and the Reason on their own.

Question 19:
Assertion (A): The area enclosed by a chord and the major arc is the major segment.
Reason (R): If a circle is divided into three equal arcs, then each is a major arc.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (c) A is true, but R is false.

Explanation:
1. The region between a chord and its major arc is the major segment, so the Assertion is true.
2. Three equal arcs are each 120°, which is less than a semicircle, so each is a minor arc. The Reason is false.
3. Answer: (c).

Teacher's Note:
1. A major arc is longer than a semicircle.
2. Divide 360° by 3 to test the Reason.

Question 20: Ramesh visits a stadium that has a circular swimming pool with centre O and radius 7 m. Two points A and B on the wall of the pool are joined by a rope so that ∠AOB = 90°. The rope separates the shallow section (the smaller section) from the deep section. (Take π = 22/7.) The area of △AOB is
90°OABDeepShallow (a) 49 m²
(b) 24.5 m²
(c) 98 m²
(d) 140 m²
Show Answer & Explanation

Answer: (b) 24.5 m²

Explanation:
1. OA = OB = 7 m and ∠AOB = 90°.
2. Area = ½ × 7 × 7 = 24.5 m².

Teacher's Note:
1. The two radii are perpendicular, so they serve as base and height.
2. Use ½ × base × height.

Question 21: Ramesh visits a stadium that has a circular swimming pool with centre O and radius 7 m. Two points A and B on the wall of the pool are joined by a rope so that ∠AOB = 90°. The rope separates the shallow section (the smaller section) from the deep section. (Take π = 22/7.) The area of the minor sector AOB is
90°OABDeepShallow (a) 77 m²
(b) 38.5 m²
(c) 154 m²
(d) 70 m²
Show Answer & Explanation

Answer: (b) 38.5 m²

Explanation:
1. A 90° sector is one quarter of the circle.
2. Area = ¼ × (22/7) × 7 × 7 = 38.5 m².

Teacher's Note:
1. 90° is one-quarter of 360°.
2. The sector is bigger than its triangle.

Question 22: Ramesh visits a stadium that has a circular swimming pool with centre O and radius 7 m. Two points A and B on the wall of the pool are joined by a rope so that ∠AOB = 90°. The rope separates the shallow section (the smaller section) from the deep section. (Take π = 22/7.) The area of the shallow region of the pool is
90°OABDeepShallow (a) 28 m²
(b) 56 m²
(c) 14 m²
(d) 7 m²
Show Answer & Explanation

Answer: (c) 14 m²

Explanation:
1. The shallow region is the minor segment cut off by the rope.
2. Area = sector − triangle = 38.5 − 24.5 = 14 m².

Teacher's Note:
1. Segment = sector − triangle (for a minor segment).
2. Use the two results from the previous parts.

Question 23: Ramesh visits a stadium that has a circular swimming pool with centre O and radius 7 m. Two points A and B on the wall of the pool are joined by a rope so that ∠AOB = 90°. The rope separates the shallow section (the smaller section) from the deep section. (Take π = 22/7.) The area of the whole swimming pool is
90°OABDeepShallow (a) 77 m²
(b) 38.5 m²
(c) 154 m²
(d) 70 m²
Show Answer & Explanation

Answer: (c) 154 m²

Explanation:
1. Area = πr² = (22/7) × 7 × 7 = 154 m².

Teacher's Note:
1. The pool is a full circle of radius 7 m.
2. 7 × 7 cancels the 7 in the denominator of 22/7.

Question 24: Ramesh visits a stadium that has a circular swimming pool with centre O and radius 7 m. Two points A and B on the wall of the pool are joined by a rope so that ∠AOB = 90°. The rope separates the shallow section (the smaller section) from the deep section. (Take π = 22/7.) The area of the deep section of the pool is
90°OABDeepShallow (a) 77 m²
(b) 150 m²
(c) 154 m²
(d) 140 m²
Show Answer & Explanation

Answer: (d) 140 m²

Explanation:
1. Deep section = whole pool − shallow section = 154 − 14 = 140 m².

Teacher's Note:
1. The deep part is everything except the shallow segment.
2. Subtract the previous answer from 154.

Question 25: A circular wooden table top is to be cut into a square table top of side 2 m, with the four vertices of the square on the edge of the circle (see the figure). (Take π = 3.14.) The radius of the circular table top is
ABCDO2 m (a) \( 2\sqrt{2} \) m
(b) 2.62 m
(c) \( \sqrt{2} \) m
(d) 2 m
Show Answer & Explanation

Answer: (c) \( \sqrt{2} \) m

Explanation:
1. The diagonal of the square is a diameter: BD² = 2² + 2² = 8, so BD = 2√2 m.
2. Radius = BD ÷ 2 = √2 m.

Teacher's Note:
1. The diagonal of a square inscribed in a circle is the diameter.
2. The radius is half the diagonal.

Question 26: A circular wooden table top is to be cut into a square table top of side 2 m, with the four vertices of the square on the edge of the circle (see the figure). (Take π = 3.14.) The area of the circular table top is
ABCDO2 m (a) 6 m²
(b) 12.5 m²
(c) 6.28 m²
(d) 25.14 m²
Show Answer & Explanation

Answer: (c) 6.28 m²

Explanation:
1. Area = πr² = 3.14 × (√2)² = 3.14 × 2 = 6.28 m².

Teacher's Note:
1. (√2)² = 2.
2. Use the given π = 3.14.

Question 27: A circular wooden table top is to be cut into a square table top of side 2 m, with the four vertices of the square on the edge of the circle (see the figure). (Take π = 3.14.) The area of each of the four segments cut off when the square is drawn is
ABCDO2 m (a) 6 m²
(b) 1.5 m²
(c) 2.5 m²
(d) 0.57 m²
Show Answer & Explanation

Answer: (d) 0.57 m²

Explanation:
1. Each segment is a quarter circle sector minus a right triangle with legs √2 m.
2. Sector = 6.28 ÷ 4 = 1.57 m²; triangle = ½ × √2 × √2 = 1 m².
3. Segment = 1.57 − 1 = 0.57 m².

Teacher's Note:
1. Each side of the square subtends 90° at the centre.
2. Segment = sector − triangle.

Question 28: A circular wooden table top is to be cut into a square table top of side 2 m, with the four vertices of the square on the edge of the circle (see the figure). (Take π = 3.14.) The total area that has to be cut away from the circular table top is
ABCDO2 m (a) 2.28 m²
(b) 6 m²
(c) 10 m²
(d) 4.24 m²
Show Answer & Explanation

Answer: (a) 2.28 m²

Explanation:
1. Area cut away = area of circle − area of square = 6.28 − 4 = 2.28 m².
2. Check: 4 × 0.57 = 2.28 m².

Teacher's Note:
1. The four segments together make up circle minus square.
2. Cross-check with 4 times one segment.

Question 29: A circular wooden table top is to be cut into a square table top of side 2 m, with the four vertices of the square on the edge of the circle (see the figure). (Take π = 3.14.) The area of the new square table top is
ABCDO2 m (a) 2 m²
(b) 4 m²
(c) 8 m²
(d) 6 m²
Show Answer & Explanation

Answer: (b) 4 m²

Explanation:
1. Area = side × side = 2 × 2 = 4 m².

Teacher's Note:
1. The side of the square is given as 2 m.
2. Compare with 6.28 m² of the original table top.

Question 30: A wheel is made for a school project with silver wire in the form of a circle of diameter 42 mm. The wire is also used to make 6 diameters, which divide the circle into 12 equal sectors. The cost of silver wire is Rs. 30 per mm. (Take π = 22/7.) The length of silver wire required to make the wheel is
diameter 42 mm (a) 132 mm
(b) 252 mm
(c) 384 mm
(d) 396 mm
Show Answer & Explanation

Answer: (c) 384 mm

Explanation:
1. Circumference = 2 × (22/7) × 21 = 132 mm.
2. Six diameters = 6 × 42 = 252 mm.
3. Total length = 132 + 252 = 384 mm.

Teacher's Note:
1. Add the rim and the six diameters.
2. The radius is 21 mm.

Question 31: A wheel is made for a school project with silver wire in the form of a circle of diameter 42 mm. The wire is also used to make 6 diameters, which divide the circle into 12 equal sectors. The cost of silver wire is Rs. 30 per mm. (Take π = 22/7.) The area of each sector of the wheel is
diameter 42 mm (a) 1386 mm²
(b) 115.5 mm²
(c) 231 mm²
(d) 57.75 mm²
Show Answer & Explanation

Answer: (b) 115.5 mm²

Explanation:
1. 12 equal sectors, so each has angle 360° ÷ 12 = 30°.
2. Area = (30 ÷ 360) × (22/7) × 21 × 21 = 115.5 mm².

Teacher's Note:
1. Each sector is one-twelfth of the circle.
2. Total area = (22/7) × 21² = 1386 mm².

Question 32: A wheel is made for a school project with silver wire in the form of a circle of diameter 42 mm. The wire is also used to make 6 diameters, which divide the circle into 12 equal sectors. The cost of silver wire is Rs. 30 per mm. (Take π = 22/7.) The cost of the silver wire used to make the wheel is
diameter 42 mm (a) Rs. 11520
(b) Rs. 3960
(c) Rs. 7560
(d) Rs. 11880
Show Answer & Explanation

Answer: (a) Rs. 11520

Explanation:
1. Cost = 384 × 30 = Rs. 11520.

Teacher's Note:
1. Cost = length × rate.
2. Use the total length from the first part.

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Multiple Choice Questions (MCQs) for Class 10 Mathematics Chapter 11 Areas Related to Circles

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