A chord of a circle of radius 15 cm subtends an angle of 60° at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use π ≈ 3.14 and √3 ≈ 1.73.)
Step-by-step solution
To find: Areas of the minor and major segments
Idea: Minor segment = sector − triangle OAB. Major segment = whole circle − minor segment.
- Sector OAB = πr2 × 60°360° = 3.14 × 225 × 16 = 117.75 cm2.1 mark
- In △OAB, OA = OB = 15 cm and ∠AOB = 60°, so the other two angles are equal and also 60°: the triangle is equilateral. Area = √34 × 152 = 1.734 × 225 = 97.3125 cm2.1 mark
- Minor segment = 117.75 − 97.3125 = 20.4375 cm2 (≈ 20.44 cm2).1 mark
- Whole circle = 3.14 × 225 = 706.5 cm2. Major segment = 706.5 − 20.4375 = 686.0625 cm2 (≈ 686.06 cm2).1 mark
Check: Minor + major = 20.4375 + 686.0625 = 706.5 = whole circle ✓.
Answer to write in the exam
Sector = 3.14 × 152 × 60°360° = 117.75 cm2
OA = OB, ∠AOB = 60° ⇒ △OAB equilateral; ar(△OAB) = √34 × 152 = 97.3125 cm2
Minor segment = 117.75 − 97.3125 = 20.4375 cm2
Circle = 3.14 × 225 = 706.5 cm2
∴ Major segment = 706.5 − 20.4375 = 686.0625 cm2
Common mistakes that cost marks
- Subtracting the triangle from the whole circle to get the major segment. The major segment = circle − minor segment.
- Using 12 × 15 × 15 for the triangle; it is not right-angled. Use √34a2 for an equilateral triangle.
- Confusing a segment (between chord and arc) with a sector (between two radii and arc).
How this can come in the exam
A circular table top of radius 20 cm has a strip cut off along a chord that subtends 60° at the centre, so that the table can stand against a wall.
(i) What kind of triangle do the centre and the ends of the chord form? (ii) Find the area of the sector cut by the chord. (iii) Find the area of the strip cut off. (iv) Find the area of the table top that remains. (Use π = 3.14, √3 = 1.73.)
Show answer
(i) Equilateral, side 20 cm (1 mark). (ii) 3.14 × 400 × 16 ≈ 209.33 cm2 (1 mark). (iii) Triangle = 1.734 × 400 = 173 cm2; strip = 209.33 − 173 = 36.33 cm2 (1 mark). (iv) 1256 − 36.33 = 1219.67 cm2 (1 mark).Try one yourself
A chord of a circle of radius 10 cm subtends 60° at the centre. Find the area of the minor segment. (Use π = 3.14, √3 = 1.73.)
Show answer
Sector = 314 ÷ 6 ≈ 52.33; triangle = 43.25; segment ≈ 9.08 cm2.
More questions like this
- A car has two wipers which do not overlap. Each wiper has a blade of length 28 cm and sweeps through an angle of 120°. Find the total area cleaned at each sweep of the blades.
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- An equilateral triangle is inscribed in a circle of radius r. Show that the ratio of the area of the triangle to the area of the circle is equal to 3√34π ≈ 0.413.
- A square is inscribed in a circle of radius r. Show that the ratio of the area of the square to the area of the circle is equal to 2π ≈ 0.637.
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