The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes. Class 9

The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes. Class 9

Question 1.

The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes. Class 9

Solution:

Length of minute hand (r) = 7cm,

Angle traced by minute hand in 60 minutes = 360°

∴ Angle traced by minute hand in 10 minutes

= $\frac{360}{60}$ × 10 = 60°

∴ Area of sector

$\begin{aligned} & =\frac{\theta}{360^{\circ}} \times \pi r^2 \\ & =\frac{60}{360} \times \frac{22}{7} \times(7)^2 \\ & =\frac{1}{6} \times \frac{22}{7} \times 49 \\ & =\frac{1}{6} \times 154 \\ & =25.666 \ldots\end{aligned}$

∴ Area swept by minute hand in 10 minutes = 25.67 cm²

Question 2.

A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the corresponding: Class 9

i. minor sector (that subtends 90° at the centre), and

ii. major sector (that subtends 270° at the centre). (Use π ≈ 3.14.)

Solution:

i. Minor sector

∵ θ = 90°, r = 10 cm.

∴ Area of sector

$\begin{aligned} & =\frac{\theta}{360} \times \pi r^2=\frac{90}{360} \times 3.14 \times(10)^2 \\ & =\frac{1}{4} \times 3.14 \times 100 \\ & =78.5\end{aligned}$

∴ Area of minor sector = 78.50 cm²

ii. Major sector

∵ Total angle of circle = 360°

∴ Angle of major sector = 360° - 90° = 270°

∴ Area of sector

$\begin{aligned}=\frac{\theta}{360} \times \pi r^2 & =\frac{270}{360} \times 3.14 \times(10)^2 \\ & =\frac{3}{4} \times 3.14 \times 100 \\ & =235.5\end{aligned}$

∴ Area of major sector = 235.50 cm²