An equilateral triangle is inscribed in a circle of radius $r$. Show that the ratio of the area of the triangle to the area of the circle is equal to $\frac{3\sqrt{3}}{4\pi} \approx 0.413$.                                                                                                                  [Page No. 148]

Let radius of circle = $r$.

For an equilateral triangle inscribed in a circle:

Side, $a = √3r$

Area of triangle:

$$\begin{aligned} &= \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (\sqrt{3} r)^2 \\ &= \frac{3\sqrt{3}}{4} r^2 \end{aligned}$$

$$\text{Area of circle} = \pi r^2$$

Required ratio:

$$\frac{\frac{3\sqrt{3}}{4} r^2}{\pi r^2} = \frac{3\sqrt{3}}{4\pi} = 0.413$$

Hence, proved.