Question

The length of the minute hand of a clock is 5 cm. Find the area swept by the minute hand during the time period 6: 05 am and 6: 40 am.

• A. $35\frac{7}{9}c{m}^2$
• B. $45\frac{3}{8}c{m}^2$
• C. $25\frac{5}{6}c{m}^2$
• D. $45\frac{5}{6}c{m}^2$

Answer 1

The correct option is D

$45\frac{5}{6}c{m}^2$

We know that, in 60 min, minute hand revolving $={360}^{\circ }$

In 1 min, minute hand revolving $=\frac{{360}^{\circ }}{{60}^{\circ }}$

$\therefore$ In ( 6 : 05 am to 6: 40 am) = 35 min.

Minute hand revolving $=\frac{{360}^{\circ }}{{60}^{\circ }}\times 35=6\times 35$

Given that, length of minute hand (r) = 5cm

$\therefore$ Area of sector AOBA with angle $\angle O=\frac{\pi \times r^2}{360}\times \angle O$

$=\frac{22}{7}\times \frac{(5{)}^2}{{360}^{\circ }}\times 6\times 35$

$=\frac{22}{7}\times \frac{5\times 5}{{360}^{\circ }}\times 6\times 35$

$=\frac{22\times 5\times 5\times 5}{{60}^{\circ }}$

$=\frac{275}{6}=45\frac{5}{6}c{m}^2$

Hence, the required area swept by the minute land is $45\frac{5}{6}c{m}^2$.