Discussion
A circle is inscribed in an equilateral triangle whose height is \(3\sqrt{3}\) cm. Then find the ratio of area of equilateral triangle to area of circle inscribed in the equilateral triangle.
Answer: D
The answer is None of these: the ratio works out to \(3\sqrt{3} : \pi\).
For an equilateral triangle, height \(h = \frac{\sqrt{3}}{2}a\), so \(a = 6\) cm, and the inradius is one third of the height, \(r = \sqrt{3}\) cm.
\[ \begin{aligned} \text{triangle} &= \frac{\sqrt{3}}{4} \times 6^2 = 9\sqrt{3} \\ \text{circle} &= \pi r^2 = 3\pi \\ \text{ratio} &= 9\sqrt{3} : 3\pi = 3\sqrt{3} : \pi \end{aligned} \]
Option A, \(3\sqrt{3} : 4\), has 4 where \(\pi\) should be, and B inverts it. Neither matches, and C is unrelated.
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