[ Area of a triangle (using two sides and the angle between them)] [ Application of trigonometric formulas (geometry). ]

Three regular polygons are inscribed in a circle, the number of sides of each subsequent one being twice that of the previous one. The areas of the first two are $S_{1}$ and $S_{2}$. Find the area of the third.
Let $S$ be the area of the third polygon, $R$ the radius of the circle, $n$ the number of sides of the first polygon, $\alpha=\pi / 2 n$. Then $S_{1}=1 / 2 n R^{2} \sin 4 \alpha$,

$S_{2}=n R^{2} \sin 2 \alpha, S=n R^{2} \sin \alpha=S_{2 /} \cos \alpha$. From the first two equations, $S_{1 /}=\cos 2 \alpha$ thus,

$\cos \alpha=\sqrt{\frac{1+\cos 2 \alpha}{2}}=\sqrt{\frac{S_{2}+S_{1}}{2 S_{2}}}, S=\sqrt{\frac{2 S_{2}^{3}}{S_{1}+S_{2}}}$

## Answer

$$
S=\sqrt{\frac{2 S_{2}^{3}}{S_{1}+S_{2}}}
$$

## Problem