A chord of a circle of radius 14cm subtends an angle of $30 °$ at the center. Find the areas of both, minor sector and the major sector of the circle. (Take π=22/7).

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\frac{154}{3} c m^{2}, \frac{1694}{3} c m^{2}

\frac{155}{3} c m^{2}, \frac{1695}{3} c m^{2}

\frac{156}{3} c m^{2}, \frac{1696}{3} c m^{2}

\frac{157}{3} c m^{2}, \frac{1697}{3} c m^{2}

answer is A.

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Detailed Solution

Area of major sector = \frac{(360 - θ)}{360} \times π r^{2}

= \frac{360 - 30}{360} \times \frac{22}{7} \times 14^{2}

= \frac{1694}{3} c m^{2}

Area of minor sector = \frac{θ}{360} \times π r^{2}

= \frac{30}{360} \times \frac{22}{7} \times 14^{2}

= \frac{154}{3} c m^{2}

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