CBSE Class 10 Mathematics Areas Related to Circles MCQs Set 12

Multiple Choice Questions (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.

Practice Chapter 11 Areas Related to Circles MCQs for Class 10 Mathematics

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Multiple Choice Questions

Question 1: A drain cover is made from a square metal plate of side 40 cm with 441 circular holes of diameter 1 cm drilled in it. Use π = 22/7. The area of the remaining plate is
(a) 1250.5 cm²
(b) 1253.5 cm²
(c) 1240.2 cm²
(d) 1260.2 cm²
Show Answer & Explanation

Answer: (b) 1253.5 cm²

Explanation:
1. Area of the plate = 40 × 40 = 1600 cm².
2. Area of one hole = \( \frac{22}{7}\times\frac{1}{2}\times\frac{1}{2}=\frac{11}{14} \) cm², so 441 holes = 441 × 11/14 = 346.5 cm².
3. Remaining area = 1600 − 346.5 = 1253.5 cm².

Teacher's Note:
1. The radius is half the diameter.
2. Subtract the total hole area from the plate area.

Question 2: The perimeter of a sector of a circle with central angle 90° is 25 cm (use π = 22/7). The area of the minor segment of the circle is
(a) 14 cm²
(b) 16 cm²
(c) 18 cm²
(d) 24 cm²
Show Answer & Explanation

Answer: (a) 14 cm²

Explanation:
1. Perimeter = 2r + \( \frac{\pi r}{2} \) = 25, so r(2 + 11/7) = 25 and r = 7 cm.
2. Segment = sector − triangle = \( \frac{1}{4} \) × \( \frac{22}{7} \) × 49 − \( \frac{1}{2} \) × 7 × 7 = 38.5 − 24.5 = 14 cm².

Teacher's Note:
1. Perimeter of a sector = 2r + arc length.
2. Segment area = sector area − triangle area.

Question 3: The area of a circular ring formed by two concentric circles of radii 5.7 cm and 4.3 cm is (take π = 3.1416)
(a) 43.98 cm²
(b) 53.67 cm²
(c) 47.24 cm²
(d) 38.54 cm²
Show Answer & Explanation

Answer: (a) 43.98 cm²

Explanation:
1. Area = π(R² − r²) = π(5.7 − 4.3)(5.7 + 4.3) = π × 1.4 × 10 = 14π.
2. 14 × 3.1416 = 43.98 cm².

Teacher's Note:
1. Use R² − r² = (R − r)(R + r).
2. Multiply by π at the end.

Question 4: (Case study) The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (use π = 3.14 and √3 = 1.73205). The length of the side of △ABC is
(a) 200 cm
(b) 105.5 cm
(c) 210.3 cm
(d) 200.5 cm
Show Answer & Explanation

Answer: (a) 200 cm

Explanation:
1. Area = (√3/4)a² = 17320.5, so a² = 17320.5 × 4/1.73205 = 40000.
2. a = 200 cm.

Teacher's Note:
1. Use the area formula for an equilateral triangle.
2. 40000 is a perfect square.

Question 5: (Case study) The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (use π = 3.14 and √3 = 1.73205). The radius of each circle is
(a) 200 cm
(b) 20 cm
(c) 10 cm
(d) 100 cm
Show Answer & Explanation

Answer: (d) 100 cm

Explanation:
1. Radius = half of the side = 200/2 = 100 cm.

Teacher's Note:
1. Read the radius from the given condition.
2. Half of 200 is 100.

Question 6: (Case study) The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (use π = 3.14 and √3 = 1.73205). The area of each sector (inside the triangle at each vertex) is
(a) 5233.3 cm²
(b) 5223.3 cm²
(c) 4233.3 cm²
(d) 522.2 cm²
Show Answer & Explanation

Answer: (a) 5233.3 cm²

Explanation:
1. Each angle of the triangle is 60°, so the sector area = \( \frac{60}{360} \) × 3.14 × 100² = 5233.3 cm².

Teacher's Note:
1. Sector area = (θ/360°) × πr².
2. The angle at each vertex is 60°.

Question 7: (Case study) The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (use π = 3.14 and √3 = 1.73205). The area of the shaded region (inside the triangle but outside the three sectors) is
(a) 17320.5 cm²
(b) 1620.5 cm²
(c) 15700 cm²
(d) 31400 cm²
Show Answer & Explanation

Answer: (b) 1620.5 cm²

Explanation:
1. Three sectors: 3 × 5233.33 = 15700 cm².
2. Shaded area = 17320.5 − 15700 = 1620.5 cm².

Teacher's Note:
1. Subtract the sector areas from the triangle area.
2. The three sectors together make half a circle.

Question 8: (Case study) The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (use π = 3.14 and √3 = 1.73205). The perimeter of △ABC is
(a) 60 cm
(b) 400 cm
(c) 600 cm
(d) 300 cm
Show Answer & Explanation

Answer: (c) 600 cm

Explanation:
1. Perimeter = 3 × 200 = 600 cm.

Teacher's Note:
1. Perimeter = 3 × side.
2. The side is 200 cm.

Question 9: (Case study) A racing track has semicircular ends. The distance between the two inner parallel straight segments is 60 m and each is 106 m long. The track is 10 m wide. (Use π = 22/7.) The radius of the inner semicircular end is
(a) 30 m
(b) 60 m
(c) 10 m
(d) 40 m
Show Answer & Explanation

Answer: (a) 30 m

Explanation:
1. Inner radius = 60 ÷ 2 = 30 m.

Teacher's Note:
1. The distance between the straight parts is the inner diameter.
2. Halve it.

Question 10: (Case study) A racing track has semicircular ends. The distance between the two inner parallel straight segments is 60 m and each is 106 m long. The track is 10 m wide. (Use π = 22/7.) The radius of the outer semicircular end is
(a) 30 m
(b) 50 m
(c) 40 m
(d) 70 m
Show Answer & Explanation

Answer: (c) 40 m

Explanation:
1. Outer radius = 30 + 10 = 40 m.

Teacher's Note:
1. The track is 10 m wide.
2. Add the width to the inner radius.

Question 11: (Case study) A racing track has semicircular ends. The distance between the two inner parallel straight segments is 60 m and each is 106 m long. The track is 10 m wide. (Use π = 22/7.) The distance around the track along its inner edge is
(a) 423.57 m
(b) 400.57 m
(c) 400.32 m
(d) 400 m
Show Answer & Explanation

Answer: (b) 400.57 m

Explanation:
1. Inner edge = 2 × 106 + 2 × \( \frac{22}{7} \) × 30 = 212 + 188.57 = 400.57 m.

Teacher's Note:
1. Two straight parts plus a full circle of the inner radius.
2. The two semicircles make one circle.

Question 12: (Case study) A racing track has semicircular ends. The distance between the two inner parallel straight segments is 60 m and each is 106 m long. The track is 10 m wide. (Use π = 22/7.) The distance around the track along its outer edge is
(a) 462.43 m
(b) 461.43 m
(c) 463 m
(d) 463.43 m
Show Answer & Explanation

Answer: (d) 463.43 m

Explanation:
1. Outer edge = 2 × 106 + 2 × \( \frac{22}{7} \) × 40 = 212 + 251.43 = 463.43 m.

Teacher's Note:
1. Two straight parts plus a full circle of the outer radius.
2. Use r = 40 m.

Question 13: (Case study) A racing track has semicircular ends. The distance between the two inner parallel straight segments is 60 m and each is 106 m long. The track is 10 m wide. (Use π = 22/7.) The area of the track is
(a) 4320 m²
(b) 4230 m²
(c) 2340 m²
(d) 4120 m²
Show Answer & Explanation

Answer: (a) 4320 m²

Explanation:
1. Outer region = 106 × 80 + \( \frac{22}{7} \) × 40² = 8480 + 5028.57 = 13508.57 m²; inner region = 106 × 60 + \( \frac{22}{7} \) × 30² = 6360 + 2828.57 = 9188.57 m².
2. Area of track = 13508.57 − 9188.57 = 4320 m².

Teacher's Note:
1. Track area = outer area − inner area.
2. Each region = rectangle + circle.

Free study material for Mathematics

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

Class 10 Mathematics Chapter 11 Areas Related to Circles Objective Test Questions

Test your conceptual understanding of Chapter 11 Areas Related to Circles with these targeted multiple-choice questions. Designed in alignment with the latest CBSE curriculum for Class 10 Mathematics, these problem sets build accuracy and prepare students for objective exams.

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FAQs

Where can I access latest CBSE Class 10 Mathematics Areas Related to Circles MCQs Set 12?

You can get most exhaustive CBSE Class 10 Mathematics Areas Related to Circles MCQs Set 12 for free on StudiesToday.com. These MCQs for Class 10 Mathematics are updated for the 2026-27 academic session as per CBSE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Mathematics Class 10 material?

Yes, our CBSE Class 10 Mathematics Areas Related to Circles MCQs Set 12 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the CBSE paper is now competency-based.

How do practicing Mathematics MCQs help in scoring full marks in Class 10 exams?

By solving our CBSE Class 10 Mathematics Areas Related to Circles MCQs Set 12, Class 10 students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Mathematics.

Do you provide answers and explanations for CBSE Class 10 Mathematics Areas Related to Circles MCQs Set 12?

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