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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.

- ${183.3\ cm^{2}}$
- ${366.6\ cm^{2}}$
- ${244.4\ cm^{2}}$
- ${188.39\ cm^{2}}$

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Ans is option (D)

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$^{2}$

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$^{2}$