Question

The minute hand of a clock is $\sqrt{21}$ cm long . find the area described by the minute on the face of the clock between 7.00 AM and 7.05 AM.

Answer 1

Given, minute hand of a clock, say, $r=\sqrt{21}$cm

Angle described by the minute hand in $1$ minute is $6^o$

Since the minute hand covers ${360}^o$ in $60$ minutes.

The time difference between $7:00$AM and $7.05$AM is $5$ minutes.

$\therefore$ Angle described by the minute hand in $5$ minutes$=6\times 5$

$={30}^o$

$\therefore$ Area of the sector described by the minute hand on the face of the clock between $7:00$AM and $7:05$AM

$=\frac{30}{360}\times \pi \times (\sqrt{21}{)}^2$ $\{Areaofsector=\frac{\theta }{{360}^o}\times \pi \times r\}$

$=\frac{1}{12}\times \frac{22}{7}\times 21$ [using $\pi =\frac{22}{7}$]

$=5.5$ sq.cm.