Let the vertices $K, L, M$ and $N$ of the rectangle $K L M N$ be located on the sides $A B, B C, C D$ and $A D$ of the square $A B C D$, respectively, such that $K N=2 K L$. Let $K L=x, \angle A K N=\alpha$. Then

$M N=x, L M=K N=2 x, \angle C M L=\angle B L K=\angle A K N=\alpha, A K=2 x \cos \alpha, B K=x \sin \alpha, B L=x \cos \alpha, C L=L M \sin \alpha=$ $2 x \sin \alpha$.

Since $A B=A K+B K$ and $B C=B L+C L$, we have $2 x \cos \alpha+x \sin \alpha=x \cos \alpha+2 x \sin \alpha$, from which $\sin \alpha=\cos \alpha$, and since $\alpha<90^{\circ}$, then $\alpha=45^{\circ}$.

Since $2 x \cos 45^{\circ}+x \sin 45^{\circ}=A B=\sqrt{18}=3 \sqrt{2}$, then $x=2$, and $S_{K L M N}=2 x^{2}=8$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9617f685748a8dfcd82ag-15.jpg?height=429&width=418&top_left_y=1809&top_left_x=823)

## Answer

8.
In a square with an area of 18, a rectangle is inscribed such that one vertex of the rectangle lies on each side of the square. The sides of the rectangle are in the ratio $1: 2$.

Find the area of the rectangle.