Question : The length of a side of an equilateral triangle is 8 cm. The area of the region lying between the circumcircle and the incircle of the triangle is __________ ( Use: $\pi = \frac{22}{7}$)
Option 1: $50\frac{1}{7}\ \text{cm}^2$
Option 2: $50\frac{2}{7}\ \text{cm}^2$
Option 3: $75\frac{1}{7}\ \text{cm}^2$
Option 4: $75\frac{2}{7}\ \text{cm}^2$
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Correct Answer: $50\frac{2}{7}\ \text{cm}^2$
Solution : Given: Side of the equilateral triangle, $a$ = 8 cm Radius of incircle = $\frac{a}{2\sqrt{3}}$ cm Radius of circumcircle = $\frac{a}{\sqrt{3}}$ cm So, the required area = Area of circumcircle – Area of incircle = $\pi (\frac{a}{\sqrt{3}})^2-\pi (\frac{a}{2\sqrt{3}})^2$ = $\pi (\frac{8}{\sqrt{3}})^2-\pi (\frac{8}{2\sqrt{3}})^2$ = $64\pi(\frac{1}{3}-\frac{1}{12})$ = $64×\frac{22}{7}×\frac{1}{4}$ = $\frac{352}{7}$ = $50\frac{2}{7}\ \text{cm}^2$ Hence, the required area is $50\frac{2}{7}\ \text{cm}^2$.
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