The area of a sector of a circle given the radius (r) and the angle (a) is $\frac{a}{360} \cdot r^2 \cdot \pi$
Here, the angle is $60$ and the radius is $6$ so we have $\frac{60}{360} \cdot 6^2 \pi= \frac{1}{6} \cdot 36 \pi = 6 \pi.$
The area of the equilateral triangle: We first must know the height and width, which is 6, but we do not know the height(I am assuming we do not know the formula for the area of an equilateral triangle.) Let the height be $h.$ We have $(\frac{6}{2})^2 + h^2 = 6^2 \Rightarrow 9 + h^2 = 36 \Rightarrow h^2 = 27 = 3 \sqrt{3}.$
The height is $3\sqrt{3},$ so the area is $\frac{1}{2} \cdot 3\sqrt{3} \cdot 6 = 9 \sqrt{3}.$
The answer is the sector - equilateral triangle = $6 \pi - 9 \sqrt{3}.$
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