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The minute hand is $10$ cm long. Find the area of the face of the clock described by the minute hand between $9$ a.m and $9:35$ a.m.

  1. ${183.3\ cm^{2}}$
  2. ${366.6\ cm^{2}}$
  3. ${244.4\ cm^{2}}$
  4. ${188.39\ cm^{2}}$
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Ans is option (A)

The minute covers  $360^{\circ}$  in $60$ minutes. Hence, it will cover  $\frac{360^{\circ}\times 35}{60}=210^{\circ}$  in $35$ minutes.

Now, area of  a sector with radius $r$ units and $\theta$ angle subtended at the centre ,is :  $\frac{1}{2}\times r^{2}\times \theta$  units   ($\theta$ in radian) .

So, $210^{\circ}=210^{\circ}\times \frac{\pi}{180^{\circ}}$  radians  $=\frac{11}{3}$ radians

$\therefore$  Area of the sector:  $\frac{1}{2}\times 100\times \frac{11}{3}=183.33$ cm$^{2}$
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Answer is A

60 min = 360 degree, so 1 min = 6 degree.

9:00 am to 9:35 am = 35 min, so degree is (35x6) = 210 𝜃.

Area of a sector is 𝜃/360 * (𝜋r^2) where r is the radius of the circle and 𝜃 is the angle of the sector. Here  𝜃 is 210 degree.

So, area of the face of the clock described by the minute hand between 9 a.m and 9:35 a.m = (210/360) * (22/7) * (10^2) = 183.3 cm2

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