Class 10 Mathematics Practice Set: Chapter 11 Areas Related to Circles
Explore reliable practice materials for Chapter 11 Areas Related to Circles tailored for Class 10 learners. Utilizing these Mathematics worksheets ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.
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Questions
Part A: Multiple Choice Questions
Q1. If the perimeter of a circle equals that of a square, the ratio of their areas (circle : square) is
(a) 22/7
(b) 14/11
(c) 7/22
(d) 11/14
View Answer & Explanation
Answer: (b) 14/11. Circle 2πr = 4a gives a = πr/2. Ratio = πr²/(π²r²/4) = 4/π = 14/11 (π = 22/7).
Teacher's Note:
1. Equate the perimeters to relate r and a.
2. Then divide the two areas.
Q2. The area of the largest square that can be inscribed in a circle of radius 8 cm is
(a) 256 cm²
(b) 128 cm²
(c) 64√2 cm²
(d) 64 cm²
View Answer & Explanation
Answer: (b) 128 cm². The diagonal of the square is 16 cm, so area = d²/2 = 256/2 = 128 cm².
Teacher's Note:
1. The diagonal of an inscribed square equals the diameter.
2. Area of a square = (diagonal)²/2.
Q3. The area of a sector of a circle of radius 36 cm is 54π cm². The length of the corresponding arc is
(a) 6π cm
(b) 3π cm
(c) 5π cm
(d) 8π cm
View Answer & Explanation
Answer: (b) 3π cm. Area = ½ × l × r, so l = 2 × 54π/36 = 3π cm.
Teacher's Note:
1. Use area of sector = ½ × arc length × radius.
2. Rearrange for the arc length.
Q4. A wheel has diameter 84 cm. The number of complete revolutions it makes to cover 792 m is
(a) 100
(b) 150
(c) 200
(d) 300
View Answer & Explanation
Answer: (d) 300. Circumference = π × 84 = 264 cm. Revolutions = 79200/264 = 300.
Teacher's Note:
1. Convert 792 m to centimetres.
2. Divide distance by circumference.
Q5. The length of an arc of a circle of radius 12 cm is 10π cm. The central angle of this arc is
(a) 120°
(b) 60°
(c) 75°
(d) 150°
View Answer & Explanation
Answer: (d) 150°. Arc = (θ/360) × 2π × 12, so 10π = (θ/360) × 24π and θ = 150°.
Teacher's Note:
1. Use arc length = (θ/360) × 2πr.
2. Solve for θ.
Part B: Short Answer Questions
Q6. If the diameter of a semicircular protractor is 14 cm, find its perimeter.
View Answer & Explanation
Answer: Perimeter = πr + 2r = (22/7)(7) + 14 = 22 + 14 = 36 cm.
Teacher's Note:
1. Add the curved part and the diameter.
2. The radius is 7 cm.
Q7. The diameter of a cycle wheel is 21 cm. How many revolutions will it make to travel 1.98 km?
View Answer & Explanation
Answer: Circumference = (22/7) × 21 = 66 cm. Distance = 198000 cm, so revolutions = 198000/66 = 3000.
Teacher's Note:
1. Convert 1.98 km to centimetres.
2. Divide by the circumference.
Q8. The minute hand of a clock is 14 cm long. Find the area swept by it in 5 minutes.
View Answer & Explanation
Answer: In 5 minutes it turns 30°. Area = (30/360) × (22/7) × 14² = \( \dfrac{154}{3} \) cm² ≈ 51.33 cm².
Teacher's Note:
1. The minute hand turns 6° every minute.
2. Use the area formula for a sector.
Q9. The inner circumference of a circular track is 440 m and the track is 14 m wide. Find the diameter of the outer circle. (π = 22/7)
View Answer & Explanation
Answer: Inner radius = 440 × 7/(2 × 22) = 70 m. Outer radius = 70 + 14 = 84 m, so the outer diameter is 168 m.
Teacher's Note:
1. Find the inner radius from the circumference.
2. Add the width of the track.
Q10. A copper wire bent into a square encloses an area of 121 cm². If the same wire is bent into a circle, find the area of the circle. (π = 22/7)
View Answer & Explanation
Answer: Side = 11 cm, so wire length = 44 cm. Circle: 2πr = 44, so r = 7 cm and area = (22/7) × 49 = 154 cm².
Teacher's Note:
1. The wire length is the same in both shapes.
2. Find r from the circumference.
Q11. A wire is looped in the form of a circle of radius 28 cm. It is bent into a square. Find the side of the square. (π = 22/7)
View Answer & Explanation
Answer: Length of wire = 2 × (22/7) × 28 = 176 cm. Side = 176/4 = 44 cm.
Teacher's Note:
1. Wire length equals the circumference.
2. Perimeter of the square = 4 × side.
Q12. Three horses are tied with 7 m long ropes at the three corners of a triangular field with sides 20 m, 34 m and 42 m. Find (i) the area grazed by the horses and (ii) the area left ungrazed.
View Answer & Explanation
Answer: Semi-perimeter = 48 m; area = √(48 × 28 × 14 × 6) = 336 m². The three sectors have angles summing to 180°, so grazed area = ½ × (22/7) × 49 = 77 m². Ungrazed = 336 − 77 = 259 m².
Teacher's Note:
1. The three angles of a triangle add up to 180°.
2. Use Heron's formula for the field.
Q13. ABC is a quadrant of a circle of radius 14 cm with right angle at A. A semicircle is drawn on BC as diameter, on the side away from A. Find the area of the region between the semicircle and the arc BC.
View Answer & Explanation
Answer: BC = 14√2, so the semicircle radius is 7√2 and its area = ½ × (22/7) × 98 = 154 cm². Quadrant = ¼ × (22/7) × 196 = 154 cm²; triangle ABC = 98 cm². Segment = 154 − 98 = 56 cm². Required area = 154 − 56 = 98 cm².
Teacher's Note:
1. Find the segment cut off by chord BC from the quadrant.
2. Subtract it from the semicircle.
Q14. The length of a minor arc is 2/9 of the circumference of the circle. Find the angle subtended by the arc at the centre.
View Answer & Explanation
Answer: Angle = (2/9) × 360° = 80°.
Teacher's Note:
1. The arc is proportional to its central angle.
2. Multiply the fraction by 360°.
Part C: Long Answer and Application Questions
Q15. The area of an equilateral triangle is 49√3 cm². With each vertex as centre, a circle of radius equal to half the side is drawn. Find the area of the triangle not included in the circles. (√3 = 1.73)
View Answer & Explanation
Answer: √3a²/4 = 49√3 gives a = 14 cm, radius = 7 cm. Three 60° sectors have total area = ½ × (22/7) × 49 = 77 cm². Triangle area = 49 × 1.73 = 84.77 cm². Remaining area = 84.77 − 77 = 7.77 cm².
Teacher's Note:
1. Find the side from the area.
2. The three sectors together make a semicircle.
Q16. Two circles touch externally. The sum of their areas is 130π cm² and the distance between their centres is 14 cm. Find their radii.
View Answer & Explanation
Answer: r₁ + r₂ = 14 and r₁² + r₂² = 130, so r₁r₂ = (196 − 130)/2 = 33. The radii are the roots of t² − 14t + 33 = 0, i.e. 11 cm and 3 cm.
Teacher's Note:
1. The centre distance equals the sum of the radii.
2. Use (r₁ + r₂)² to find the product.
Q17. Two circles touch internally. The sum of their areas is 116π cm² and the distance between their centres is 6 cm. Find their radii.
View Answer & Explanation
Answer: R − r = 6 and R² + r² = 116, so 2Rr = 116 − 36 = 80 and Rr = 40. Hence R = 10 cm and r = 4 cm.
Teacher's Note:
1. The centre distance is R − r.
2. Use (R − r)² to find Rr.
Q18. A pendulum swings through an angle of 30° and describes an arc 8.8 cm long. Find the length of the pendulum. (π = 22/7)
View Answer & Explanation
Answer: Arc = (30/360) × 2 × (22/7) × r = 8.8, so (11/21) r = 8.8 and r = 8.8 × 21/11 = 16.8 cm.
Teacher's Note:
1. The pendulum length is the radius of the arc.
2. Use arc length = (θ/360) × 2πr.
Q19. The side of a square is 10 cm. Find the areas of the circumscribed and the inscribed circles.
View Answer & Explanation
Answer: Circumscribed radius = half the diagonal = 5√2, so area = 50π cm². Inscribed radius = 5, so area = 25π cm².
Teacher's Note:
1. The circumscribed circle's diameter is the diagonal.
2. The inscribed circle's diameter equals the side.
Q20. An umbrella has 8 equally spaced ribs. Treat it as a flat circle of radius 45 cm. Find the area between two consecutive ribs. (π = 3.14)
View Answer & Explanation
Answer: Area = 3.14 × 45²/8 = 3.14 × 2025/8 = 794.81 cm².
Teacher's Note:
1. Each pair of ribs encloses a sector of 45°.
2. Divide the circle's area by 8.
Q21. A poster is prepared on a square sheet of side 60 cm. A quadrant of radius 17.5 cm is drawn at each corner, and a circle of diameter 21 cm is drawn at the centre. Find the area of the remaining sheet. (π = 3.14)
View Answer & Explanation
Answer: Four quadrants = one full circle of radius 17.5: 3.14 × 306.25 = 961.625 cm². Central circle = 3.14 × 10.5² = 346.185 cm². Remaining = 3600 − 961.625 − 346.185 = 2292.19 cm².
Teacher's Note:
1. The four quadrants together equal one circle.
2. Subtract both from the area of the square.
Q22. A circular birthday cake of radius 60 cm is divided equally among six friends. Find the area of each piece. (π = 3.14)
View Answer & Explanation
Answer: Area of cake = 3.14 × 60² = 11304 cm². Each piece = 11304/6 = 1884 cm².
Teacher's Note:
1. Find the area of the whole cake.
2. Divide it by 6.
Free study material for Mathematics
Exam Preparation Worksheet for Class 10 Mathematics Chapter 11 Areas Related to Circles
Chapter 11 Areas Related to Circles Printable Worksheet for Class 10 Mathematics
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FAQs
You can download the teacher-verified PDF for CBSE Class 10 Mathematics Areas Related to Circles Worksheet Set 08 from StudiesToday.com. These practice sheets for Class 10 Mathematics are designed as per the latest CBSE academic session.
Yes, our CBSE Class 10 Mathematics Areas Related to Circles Worksheet Set 08 includes a variety of questions like Case-based studies, Assertion-Reasoning, and MCQs as per the 50% competency-based weightage in the latest curriculum for Class 10.
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