How do you find the area of a segment of a circle?

Area of a (minor) segment = area of the corresponding sector − area of the triangle formed by the two radii and the chord. For a right-angle sector, the triangle is right-angled with the two radii as legs.

What value of π should I use?

Use π = 22/7 when the radius is a multiple of 7, and π = 3.14 otherwise, unless the question specifies one. Keep the value consistent throughout a problem.

NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles — Exercise Solutions

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Complete NCERT Solutions for CBSE Class 10 Maths Chapter 11 — Areas Related to Circles: circumference and area, sectors, segments, and arc length.

Key Concepts (Quick Recap)

Circumference = 2πr; Area = πr².
Sector (angle θ): area = (θ/360)πr²; arc length = (θ/360)·2πr.
Segment area = sector area − triangle area.

Exercise 11.1 — Area, Circumference, Sectors

Q1. The circumference of a circle is 22 cm. Find its area. 2πr = 22 → r = 22 × 7/(2 × 22) = 3.5 cm. Area = πr² = (22/7)(3.5)² = (22/7)(12.25) = 38.5 cm².

Q2. Find the area of a sector of radius 7 cm with central angle 60°. Area = (60/360) × (22/7) × 7² = (1/6)(22/7)(49) = (1/6)(154) = 25⅔ cm² ≈ 25.67 cm².

Q3. Find the length of the arc of a sector of radius 21 cm and central angle 60°. Arc = (60/360) × 2 × (22/7) × 21 = (1/6)(2 × 22 × 3) = (1/6)(132) = 22 cm.

Q4. Find the area of a circle that can be inscribed in a square of side 10 cm. Diameter = side = 10 cm → r = 5 cm. Area = πr² = 3.14 × 25 = 78.5 cm².

Exercise 11.2 — Segments

Q1. Find the area of the minor segment of a circle of radius 10 cm cut off by a chord that subtends a right angle (90°) at the centre. (Use π = 3.14) Sector area = (90/360) × 3.14 × 10² = (1/4)(314) = 78.5 cm². Triangle area = ½ × 10 × 10 = 50 cm² (right angle between the two radii). Minor segment = 78.5 − 50 = 28.5 cm².

Q2. A chord of a circle of radius 12 cm subtends 120° at the centre. Find the area of the corresponding minor segment. (Use π = 3.14, √3 = 1.73) Sector area = (120/360) × 3.14 × 144 = (1/3)(452.16) = 150.72 cm². Triangle area = (√3/4) × r² × … for the isosceles triangle with apex 120°: area = ½ r² sin120° = ½ × 144 × (√3/2) = 36√3 = 36 × 1.73 = 62.28 cm². Minor segment = 150.72 − 62.28 = 88.44 cm².

Q3. The length of the minute hand of a clock is 14 cm. Find the area swept by it in 5 minutes. 5 minutes = (5/60) × 360° = 30° of sweep. Area = (30/360) × (22/7) × 14² = (1/12)(22/7)(196) = (1/12)(616) = 51.33 cm².

Common Mistakes

Using diameter instead of radius in πr².
Forgetting to subtract the triangle when finding a segment.
Inconsistent π — pick 22/7 or 3.14 and stay with it.

Exam Tips

Choose π = 22/7 when r is a multiple of 7 to keep arithmetic clean.
Minute/hour-hand problems: convert the elapsed time to a central angle first.
Minor segment = sector − triangle; major segment = circle area − minor segment.

Frequently Asked Questions

Area of a sector with central angle θ (in degrees) and radius r is (θ/360) × πr². The length of its arc is (θ/360) × 2πr. A sector is the 'pizza slice' region bounded by two radii and an arc.