Class 10 Maths Chapter 11 Areas Related to Circles Extra Questions (HOTS)

Genuinely harder, HOTS-level practice for Class 10 Maths Chapter 11 (Areas Related to Circles), going beyond Exercise 11.1 with fresh numbers, general proofs, and reverse-engineering problems. These Class 10 Mathematics Chapter 11 important questions are handy for last-minute exam practice.

Last Updated: September 10, 2026

  1. Q1 (General derivation). Derive the general formula for the area of a sector in terms of its arc length l and radius r, starting from the sector-angle formula.
    Solution: Arc length l = (θ/360) × 2πr ⇒ θ/360 = l/(2πr). Substituting into Area = (θ/360)πr² gives Area = [l/(2πr)] × πr² = (1/2) × l × r. This general result works for any sector without knowing θ directly, as long as the arc length is known.
  2. Q2 (Reverse-engineering). The area of a sector of a circle of radius 14 cm is 154 cm². Find the angle of the sector.
    Solution: 154 = (θ/360) × (22/7) × 196 ⇒ θ = 154 × 360 × 7 / (22 × 196) = 388080/4312 = 90°Sector area 154 cm² → θ = 90°.
  3. Q3 (Fresh-number word problem). A pendulum swings through an angle of 45° and describes an arc of length 22 cm. Find the length of the pendulum.
    Solution: 22 = (45/360) × 2 × (22/7) × r = (11/14)r ⇒ r = 22 × 14/11 = 28 cmArc 22cm at 45° → pendulum length r=28cm.
  4. Q4 (Assertion-Reason). Assertion (A): The area of a sector with central angle 180° equals half the area of the circle. Reason (R): A sector with central angle 180° is a semicircle.
    (a) Both A and R true, R is the correct explanation of A   (b) Both true, R is not the correct explanation   (c) A true, R false   (d) A false, R true
    Solution: Sector area at 180° = (180/360)πr² = (1/2)πr², exactly half the circle’s area, and this is precisely because 180° sweeps a semicircle. Answer: (a)180° sector = semicircle = half the circle’s area.
  5. Q5 (Combination of figures, fresh numbers). A circular park of radius 20 m has a 7 m wide path built all around it (outside the park). Find the area of the path, and the cost of gravelling it at ₹50 per m².
    Solution: Outer radius R = 20 + 7 = 27 m. Path area = π(R² − r²) = (22/7)(729 − 400) = (22/7) × 329 = 1034 m². Cost = 1034 × 50 = ₹51,700Path area = π(27²−20²) = 1034 m².
  6. Q6 (Clock/time HOTS, fresh numbers). The minute hand of a clock is 10.5 cm long. Find the area swept by it between 8:00 a.m. and 8:35 a.m.
    Solution: In 35 minutes, angle swept = (35/60) × 360° = 210°. Area = (210/360) × (22/7) × 10.5² = (7/12) × 346.5 = 202.13 cm² (approx.)35 min = 210° swept → area ≈ 202.13 cm².

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Written by Satish

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