Measuring Space: Perimeter and Area | Exercise 6.3

Question 3

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

Solution

Understand the Question

The minute hand of a clock acts as the radius of a circle, where $r = 7\text{ cm}$ .
In $60\text{ minutes}$ , the minute hand completes a full circle of $360^\circ$ .
The region swept in $10\text{ minutes}$ forms a circular sector with central angle $\theta$ .
The area of the sector is given by the formula: $\text{Area} = \dfrac{\theta}{360^\circ} \times \pi r^2$

Step 1 · Find the Angle Swept in 10 Minutes

Diagram 1

Angle swept by the minute hand in $60\text{ minutes} = 360^\circ$

\begin{aligned} \text{Angle in 1 minute} &= \frac{360^\circ}{60} \\[0.6em] &= 6^\circ \end{aligned}

\begin{aligned} \text{Angle swept in 10 minutes (}\theta\text{)} &= 10 \times 6^\circ \\[0.6em] &= 60^\circ \end{aligned}

Step 2 · Calculate the Area Swept

Given radius $r = 7\text{ cm}$ and $\theta = 60^\circ$ .

$\text{Area of a sector} = \frac{\theta}{360^\circ} \times \pi r^2$

Substitute the values:

\begin{aligned} \text{Area swept} &= \frac{60^\circ}{360^\circ} \times \pi \times (7\text{ cm})^2 \\[0.6em] &= \frac{1}{6} \times \frac{22}{7} \times 49\text{ cm}^2 \\[0.6em] &= \frac{1}{6} \times 22 \times 7\text{ cm}^2 \\[0.6em] &= \frac{154}{6}\text{ cm}^2 \\[0.6em] &= \frac{77}{3}\text{ cm}^2 \end{aligned}

Answer

$\dfrac{77}{3}\text{ cm}^2$

Common Mistakes

Confusing Minute Hand and Hour Hand: The minute hand covers $360^\circ$ in $60\text{ minutes}$ ( $6^\circ\text{ per minute}$ ), while the hour hand covers $360^\circ$ in $12\text{ hours}$ ( $0.5^\circ\text{ per minute}$ ).
Using Arc Length Instead of Area: Do not confuse the area swept (sector area: $\dfrac{\theta}{360^\circ} \pi r^2$ ) with the distance moved by the tip of the hand (arc length: $\dfrac{\theta}{360^\circ} \times 2\pi r$ ).

More questions in Exercise 6.3

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

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

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

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 $10\text{ cm}$ subtends $90^\circ$ at the centre. Find the area of the corresponding:

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

A chord of a circle of radius $15\text{ cm}$ subtends an angle of $60^\circ$ 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 $28\text{ cm}$ and sweeps through an angle of $120^\circ$ . Find the total area cleaned at each sweep of the blades.

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

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

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

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