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The length of minute hand of a clock is 14 cm. Find the area swept by the minute hand in 15 minutes.

Mensuration

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Answer

Let minute hand be at A and after 15 minute it reaches B.

In the figure, ABC is a right angled triangle right angled at B. Semicircles are drawn on AB, BC and CA as diameter. Show that the sum of areas of semicircles drawn on AB and BC as diameter is equal to the area of the semicircle drawn on CA as diameter. Mensuration, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

From figure,

Area of sector OAB = πr2×1560πr^2 \times \dfrac{15}{60}

=227×142×1560=227×196×14=22×28×14=22×7=154 cm2.= \dfrac{22}{7} \times 14^2 \times \dfrac{15}{60} \\[1em] = \dfrac{22}{7} \times 196 \times \dfrac{1}{4} \\[1em] = 22 \times 28 \times \dfrac{1}{4} \\[1em] = 22 \times 7 \\[1em] = 154 \text{ cm}^2.

Hence, area swept by minute hand in 15 minutes = 154 cm2.

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