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.
Answer: One wiper: 227 × 28 × 28 × 120360 = 24643 ≈ 821.33 cm2. Two wipers (no overlap): 49283 ≈ 1642.67 cm2.
Step-by-step solution
Given: Blade length (radius) = 28 cm; Angle = 120°; 2 wipers, no overlap
To find: Total area cleaned in one sweep
To find: Total area cleaned in one sweep
Idea: Each blade sweeps a sector. They do not overlap, so add the two sector areas.
- Area cleaned by one blade = πr2 × 120°360° = 227 × 28 × 28 × 13 = 2464 × 13.1 mark
- = 24643 ≈ 821.33 cm2.1 mark
- Two blades, no overlap: total = 2 × 24643 = 49283 ≈ 1642.67 cm2.1 mark
Total area cleaned at each sweep = 4928/3 cm² ≈ 1642.67 cm².
Check: 227 × 784 = 2464 is the whole circle; one-third of it is 821.33, and two blades give 1642.67 ✓.
Answer to write in the exam
One blade: πr2 × θ360° = 227 × 28 × 28 × 120°360°
= 24643 ≈ 821.33 cm2
Two blades = 2 × 24643
∴ Total ≈ 1642.67 cm2
Common mistakes that cost marks
- Finding the area for one wiper only.
- Using arc length 2πr × θ360 instead of area.
- Using 120° as 120180 of the circle.
How this can come in the exam
MCQ (1 mark)
A sprinkler sprays water up to 9 m and turns through 120°. The area watered is (use π = 3.14)
- 42.39 m2
- 84.78 m2
- 127.17 m2
- 254.34 m2
Show answer
(B) 84.78 m2
120° is 13 of a turn: 13 × 3.14 × 81 = 13 × 254.34 = 84.78 m2.
Try one yourself
A bus has one wiper of length 35 cm sweeping 108°. Find the area it cleans in one sweep. (Use π = 227.)
Show answer
227 × 1225 × 108360 = 3850 × 0.3 = 1155 cm2.
More questions like this
- 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.
- A hexagon is inscribed in a circle of radius r. Show that the ratio of the area of the hexagon to the area of the circle is equal to 3√32π ≈ 0.827. Can you see why the answer is exactly twice the answer for the inscribed equilateral triangle?
- Identities in algebra can sometimes be shown as area relationships. For example: The figure shown corresponds to the identity (a + b)2 = a2 + 2ab + b2. Do you see how? Draw figures corresponding to the identities (a + b)(a − b) = a2 − b2 and (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca.