Question : Find the area of the shaded portion of an equilateral triangle with sides 6 units shown in the following figure. A circle of radius 1 unit is centred at the midpoint of a side of the triangle.
Option 1: $\frac{1}{2}\left(9 \sqrt{3}-\frac{11}{7}\right)$ unit$^2$
Option 2: $\frac{1}{4}\left(9 \sqrt{3}-\frac{11}{7}\right)$ unit $^2$
Option 3: $\frac{1}{2}\left(6 \sqrt{3}-\frac{11}{7}\right)$ unit$^2$
Option 4: $\frac{1}{2}\left(9 \sqrt{3}-\frac{22}{7}\right)$ unit$^2$
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Correct Answer: $\frac{1}{2}\left(9 \sqrt{3}-\frac{11}{7}\right)$ unit$^2$
Solution : Area of equilateral triangle = $\frac{\sqrt 3}{4}(side)^2$ $= \frac{\sqrt3}{4}(6)^2 = 9\sqrt{3}\ \mathrm{unit}^2$ Area of circle = $\pi r^2 = \pi\ \mathrm{unit^2}$ where $r$ = radius of circle = 1 unit. Area of shaded region = $\frac{1}{2}$ Area of triangle - $\frac{1}{4}$ Area of circle Area of shaded region = $\frac{9\sqrt{3}}{2} - \frac{\pi}{4}$ $= \frac{9\sqrt{3}}{2} - \frac{22}{7\times 4}$ $= \frac{9\sqrt{3}}{2} - \frac{11}{7\times 2}$ $= \frac{1}{2}\left(9 \sqrt{3}-\frac{11}{7}\right)$ Hence, the correct answer is $\frac{1}{2}(9 \sqrt{3}-\frac{11}{7})$ unit 2 .
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