The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.
Answer: In 60 min the hand turns 360°, so in 10 min it turns 60°. Area = 227 × 72 × 60360 = 773 ≈ 25.67 cm2.
Step-by-step solution
Given: Minute hand = 7 cm (radius); Time = 10 minutes
To find: Area swept
To find: Area swept
Idea: The hand sweeps a sector. Change the time into an angle: 360° in 60 minutes means 6° per minute.
- Angle in 10 minutes = 10 × 6° = 60°.½ mark
- Area swept = πr2 × θ360° = 227 × 7 × 7 × 60°360° = 154 × 16.1 mark
- = 773 ≈ 25.67 cm2.½ mark
Area swept in 10 minutes = 77/3 cm² ≈ 25.67 cm².
Check: In 60 minutes the hand sweeps the whole 154 cm2; 10 minutes is 16 of that ✓.
Answer to write in the exam
Angle = 1060 × 360° = 60°
Area = 227 × 7 × 7 × 60°360° = 154 × 16
∴ Area = 773 ≈ 25.67 cm2
Common mistakes that cost marks
- Using the hour-hand rate (30° per hour). The minute hand turns 6° per minute.
- Taking the angle as 10° (one degree per minute).
How this can come in the exam
MCQ (1 mark)
The minute hand of a clock is 6 cm long. The area it sweeps in 40 minutes is (use π = 3.14)
- 37.68 cm2
- 56.52 cm2
- 113.04 cm2
- 75.36 cm2
Show answer
(D) 75.36 cm2
40 min = 240° = 23 of a turn: 23 × 3.14 × 36 = 23 × 113.04 = 75.36 cm2.
Try one yourself
The minute hand of a clock is 21 cm long. Find the area it sweeps in 20 minutes. (Use π = 227.)
Show answer
Angle 120°: 227 × 441 × 13 = 462 cm2.
More questions like this
- A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the corresponding: (Use π ≈ 3.14.)
- 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.