10.108. Find the area of a square inscribed in an equilateral triangle with side $a$.
## Solution.

$\triangle B K C$ and $\triangle E M C$ are similar $\Rightarrow \frac{B C}{E C}=\frac{B K}{E M}, B K=\sqrt{B C^{2}-K C^{2}}=$. $=\sqrt{a^{2}-\frac{a^{2}}{4}}=\frac{a \sqrt{3}}{2} ; E M=x$ and $x=\frac{E C \cdot B K}{B C}=\frac{(a-x) a \sqrt{3}}{2 a}$. Therefore, $2 a x=$ $=a \sqrt{3}(a-x)$. Solving the equation, we get $x=\frac{a \sqrt{3}}{2+\sqrt{3}}$. The area of the desired square:

$$
S=x^{2}=\frac{3 a^{2}}{(2+\sqrt{3})^{2}}=3 a^{2}(7-4 \sqrt{3})
$$

Answer: $\quad 3 a^{2}(7-4 \sqrt{3})$

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-608.jpg?height=478&width=550&top_left_y=93&top_left_x=157)

Fig. 10.105

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-608.jpg?height=430&width=410&top_left_y=125&top_left_x=779)

Fig. 10.106