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.

Answer 1

We know that, in 60 mins, minute hand revolving = 360°

In 1 min, minute hand revolving = `360^circ/60^circ`

∴ In (6 : 05 am to 6 : 40 am) = 35 mins,

minute hand revolving  = `360^circ/60 xx 35` = (6 × 35)°

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

∴ Area of sector AOBA with angle ∠O

= `(π"r"^2)/360^circ xx ∠"O"`

= `22/7 xx (5)^2/360^circ xx (6 xx 35)^circ`

= `22/7 xx (5 xx 5)/360^circ xx (6 xx 35)^circ`

= `(22 xx 5 xx 5 xx 5)/60^circ`

= `(22 xx 5 xx 5)/12`

= `(11 xx 5 xx 5)/6`

= `275/6`

= `45 5/6 "cm"^2`

Hence, the required area swept by the minute hand is `45 5/6 "cm"^2`.