Measuring Space: Perimeter and Area | Exercise 6.3

Question 6

A car has two wipers which do not overlap. Each wiper has a blade of length 28 cm28\text{ cm} and sweeps through an angle of 120∘120^\circ. Find the total area cleaned at each sweep of the blades.

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Solution
Understand the Question
  • Each wiper blade forms a circular sector with radius r=28 cmr = 28\text{ cm} and central angle θ=120∘\theta = 120^\circ.
  • The area cleaned by one wiper is the area of a sector: Area=θ360∘×πr2\text{Area} = \dfrac{\theta}{360^\circ} \times \pi r^2.
  • Since the two wipers do not overlap, the total area cleaned is 2×Area of one wiper2 \times \text{Area of one wiper}.

Step 1 · Find Area Cleaned by One Wiper

Given radius r=28 cmr = 28\text{ cm} and angle of sector θ=120∘\theta = 120^\circ.Diagram 1

Area cleaned by one wiper=θ360∘×πr2=120∘360∘×227×(28)2=13×227×784=13×22×112=24643 cm2\begin{aligned} \text{Area cleaned by one wiper} &= \dfrac{\theta}{360^\circ} \times \pi r^2 \\[0.6em] &= \dfrac{120^\circ}{360^\circ} \times \dfrac{22}{7} \times (28)^2 \\[0.6em] &= \dfrac{1}{3} \times \dfrac{22}{7} \times 784 \\[0.6em] &= \dfrac{1}{3} \times 22 \times 112 \\[0.6em] &= \dfrac{2464}{3}\text{ cm}^2 \end{aligned}

Step 2 · Calculate Total Area Cleaned

Since there are two identical, non-overlapping wipers:

Total area cleaned=2×24643=49283 cm2\begin{aligned} \text{Total area cleaned} &= 2 \times \dfrac{2464}{3} \\[0.6em] &= \dfrac{4928}{3}\text{ cm}^2 \end{aligned}
Answer

49283 cm2\dfrac{4928}{3}\text{ cm}^2

Common Mistakes
  • Forgetting the Second Wiper: Calculating the area of a single wiper and forgetting to multiply by 22 for both blades.
  • Formula Confusion: Using the arc length formula θ360∘×2πr\dfrac{\theta}{360^\circ} \times 2\pi r instead of the sector area formula θ360∘×πr2\dfrac{\theta}{360^\circ} \times \pi r^2.

More questions in Exercise 6.3

Q1

Unless stated otherwise, use the approximation 227\dfrac{22}{7} for π\pi.

Find the area of a sector of a circle with radius 7 cm7\text{ cm} if the angle of the sector is 60∘60^\circ.

Q2

Find the area of a quadrant of a circle whose circumference is 44 cm44\text{ cm}.

Q3

The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.

Q4

A chord of a circle of radius 10 cm10\text{ cm} subtends 90∘90^\circ at the centre. Find the area of the corresponding:

(i) minor sector (that subtends 90∘90^\circ at the centre), and (ii) major sector (that subtends 270∘270^\circ at the centre). (Use π≈3.14\pi \approx 3.14.)

Q5

A chord of a circle of radius 15 cm15\text{ cm} subtends an angle of 60∘60^\circ at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use π≈3.14\pi \approx 3.14 and 3≈1.73\sqrt{3} \approx 1.73.)

Q6

A car has two wipers which do not overlap. Each wiper has a blade of length 28 cm28\text{ cm} and sweeps through an angle of 120∘120^\circ. Find the total area cleaned at each sweep of the blades.

Q7

A chord of a circle of radius rr subtends an angle of 60∘60^\circ at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to πr2(16−34)\pi r^2 \left( \dfrac{1}{6} - \dfrac{\sqrt{3}}{4} \right).

Q8

An equilateral triangle is inscribed in a circle of radius rr. Show that the ratio of the area of the triangle to the area of the circle is equal to 334π≈0.413\dfrac{3\sqrt{3}}{4\pi} \approx 0.413.

Q9

A square is inscribed in a circle of radius rr. Show that the ratio of the area of the square to the area of the circle is equal to 2π≈0.637\dfrac{2}{\pi} \approx 0.637.

Q10

A hexagon is inscribed in a circle of radius rr. Show that the ratio of the area of the hexagon to the area of the circle is equal to 332π≈0.827\dfrac{3\sqrt{3}}{2\pi} \approx 0.827. Can you see why the answer is exactly twice the answer to Question 8?

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