5. In a square with side $a$, another square is inscribed such that its vertices lie on the sides of the first square, but in such a way that the sides of the given and inscribed squares form angles of $30^{\circ}$. What fraction of the area of the given square is covered by the area of the inscribed square? Express this ratio as a percentage.
Solution. Let $x$ be the length of the side of the inscribed square (see diagram). Since $\measuredangle L M A=30^{\circ}$, the right triangle is half of an equilateral triangle with side length $L M=x$. Therefore, $A L=\frac{x}{2}$ and $A M=\frac{x \sqrt{3}}{2}$.

Accordingly, the length of the side of the original square is

$$
\begin{aligned}
A B & =A M+M B=A M+A L \\
& =\frac{x \sqrt{3}}{2}+\frac{x}{2}=\frac{x(\sqrt{3}+1)}{2} .
\end{aligned}
$$

Thus,

$$
\begin{aligned}
& P_{K L M N}=x^{2} \text { and } \\
& P_{A B C D}=\frac{x^{2}(\sqrt{3}+1)^{2}}{4}=\frac{x^{2}(2+\sqrt{3})}{2}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_e4a4e8557567895ef107g-3.jpg?height=399&width=396&top_left_y=272&top_left_x=1071)

Accordingly,

$$
P_{K L M N}: P_{A B C D}=x^{2}: \frac{x^{2}(2+\sqrt{3})}{2}=\frac{2}{2+\sqrt{3}}=\frac{2(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}=2(2-\sqrt{3})
$$

Thus,

$$
P_{K L M N}=2(2-\sqrt{3}) P_{A B C D} \approx 0.535898 P_{A B C D}
$$

In other words, the area of the inscribed square is approximately $53.5898\%$ of the area of the original square.

## Eighth Section