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 cm and sweeps through an angle of 120°. Find the total area cleaned at each sweep of the blades.

Solution

We need to find the area of a sector and then multiply it by two.

Step 1 — Area by one wiper

Let's find the area cleaned by one wiper. The blade length is the radius of the sector, $r = \textbf{28 cm}$ . The angle swept is $\theta = \textbf{120°}$ . The area of a sector is given by the formula: Area $= (\theta / 360°) \times \pi r^2$ .

$\text{Area of one wiper} = \frac{120}{360} \times \frac{22}{7} \times (28)^2$

$= \frac{1}{3} \times \frac{22}{7} \times 784$

$= \frac{1}{3} \times 22 \times 112$

$= \frac{2464}{3}$

$\boxed{\frac{2464}{3} \text{ cm}^2}$

Diagram 1

Step 2 — Total area cleaned

There are two wipers. They do not overlap. So, we multiply the area cleaned by one wiper by 2.

$\text{Total area cleaned} = 2 \times \frac{2464}{3}$

$= \frac{4928}{3}$

Answer

(i) The total area cleaned at each sweep of the blades is $\textbf{4928/3 cm}^2$ .

More questions in Exercise 6.3

Unless stated otherwise, use the approximation $\frac{22}{7}$ for $\pi$ .

Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.

Find the area of a quadrant of a circle whose circumference is 44 cm.

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

A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the corresponding:

(i) minor sector (that subtends 90° at the centre), and (ii) major sector (that subtends 270° at the centre). (Use $\pi \approx 3.14$ .)

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 $\pi \approx 3.14$ and $\sqrt{3} \approx 1.73$ .)

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 $\pi r^2 \left( \frac{1}{6} - \frac{\sqrt{3}}{4} \right)$ .

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 $\frac{3\sqrt{3}}{4\pi} \approx 0.413$ .

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 $\frac{2}{\pi} \approx 0.637$ .

Q10

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 $\frac{3\sqrt{3}}{2\pi} \approx 0.827$ . Can you see why the answer is exactly twice the answer to Question 8?