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

Explanation

we know min, minute hand revolving = `360^(@)`
In 1 min, minute hand revolving = `(360^(@))/(60^(@))`
`:.` In (`6 : 05` am to `6 : 40` am)= 35 min,
minute hand revolving = `(360^(@))/(60^(@))xx35 = 6 xx 35`
Given that, length of minute hand ( r) = 5 cm.
`:.` Area of sector AOBA with angle `angleO = (pir^(2))/(360)xxangleO`
= ` (22)/(7)((5)^(2))/(360^(@))xx 6 xx 35`
= ` (22)/(7)xx(5xx5)/(360^(@))xx6xx35`
= `(22 xx 5xx 5 xx 5xx)/(60^(@))=(22xx5xx5)/(12)`
= ` (11xx5xx5)/(6) = (275)/(6) = 45(5)/(6) cm^(2)`
Hence, tha raquired area swept by the minute land is `45(5)/(6) cm^(2)` .