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  • 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 a m.

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 a m.

Answers (1)

\left [ 45\frac{5}{6}cm^{2} \right ]

Solution

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

In 1 minute it is revolving =\frac{360}{60}=6^{\circ}

Time difference =(6:40am -6:05am) =35 min

In 6:05 am and 6.40 am there is 35 minutes

In 35 minutes angle between min hand and hour hand =\left (6\times 35 \right )^{\circ} =210^{\circ}

Length of minute hand (r)=5cm

Area of sector =\frac{\pi r^{2}\theta}{360}

\\=\frac{22}{7}\times \frac{5 \times 5 \times 210^{\circ}}{360}\;\;\;\left \{\because \theta = 210^{\circ} \right \}\\ =\frac{11\times 5\times 5}{6}\\ =\frac{275}{6}\\ =45\frac{5}{6}cm^{2}                                                        

Hence required area is 45\frac{5}{6}cm^{2}

Posted by

infoexpert24

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