A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the corresponding: (Use π ≈ 3.14.)
- (i) minor sector (that subtends 90° at the centre), and
- (ii) major sector (that subtends 270° at the centre).
Answer: (i) 78.5 cm2 (ii) 235.5 cm2
Step-by-step solution
Idea: Area of a sector = πr2 × θ360. The minor and major sectors together make the whole circle.
(i) minor sector (that subtends 90° at the centre), and
- Area = πr2 × 90°360° = 3.14 × 100 × 14.1 mark
- = 78.5 cm2.½ mark
78.5 cm2
(ii) major sector (that subtends 270° at the centre).
- Area = 3.14 × 100 × 270°360° = 314 × 34 = 235.5 cm2.1 mark
- Check: 78.5 + 235.5 = 314 = area of the whole circle ✓.½ mark
235.5 cm2
(i) Minor sector = 78.5 cm² (ii) Major sector = 235.5 cm²
Answer to write in the exam
(i)
Minor sector = 3.14 × 102 × 90°360° = 314 × 14
∴ Minor sector = 78.5 cm2
(ii)
Major sector = 3.14 × 102 × 270°360° = 314 × 34
∴ Major sector = 235.5 cm2
Common mistakes that cost marks
- Using 22/7 when the question says π ≈ 3.14.
- Finding the segment (sector minus triangle) instead of the sector.
- Writing the major sector as 360 − 78.5. Subtract the minor sector from the whole circle: 314 − 78.5 = 235.5 cm2.
How this can come in the exam
MCQ (1 mark)
In a circle of radius 20 cm, a chord subtends 90° at the centre. The area of the major sector is (use π = 3.14)
- 314 cm2
- 628 cm2
- 942 cm2
- 1256 cm2
Show answer
(C) 942 cm2
Circle = 3.14 × 400 = 1256; major sector = 34 × 1256 = 942 cm2.
Try one yourself
A chord of a circle of radius 12 cm subtends 120° at the centre. Find the areas of the minor and major sectors. (Use π = 3.14.)
Show answer
Circle = 452.16; minor = 13 of it = 150.72 cm2; major = 301.44 cm2.
More questions like this
- 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.)
- 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.
- A chord of a circle of radius r subtends an angle of 60° at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to πr2(16 − √34).
- 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.