10.081. The area of a rectangle is 9 cm $^{2}$, and the measure of one of the angles formed by the diagonals is $120^{\circ}$ (Fig. 10.79). Find the sides of the rectangle.
## Solution.

$C D=x \cdot \triangle O C D$ - equilateral; $A O=O C=C D=x ; A C=2 x$. The area of the rectangle $S=\frac{1}{2} d_{1} d_{2} \sin 60=\frac{1}{2} 2 x \cdot 2 x \cdot \frac{\sqrt{3}}{2}, 9=x^{2} \sqrt{3}$. Therefore, $x=\frac{3}{\sqrt[4]{3}}=\frac{3^{4 / 4}}{3^{1 / 4}}=3^{3 / 4}=\sqrt[4]{27}$ (cm). $A C=2 \sqrt[4]{27}$ (cm).

From $\triangle A C D:$

$$
\begin{aligned}
& A D^{2}=A C^{2}-C D^{2}=4 \sqrt{27}-\sqrt{27}=3 \sqrt{27} \\
& A D=\sqrt{3 \sqrt{27}}=3^{5 / 4}=3 \sqrt[4]{3} \text { (cm). }
\end{aligned}
$$

Answer: $\sqrt[4]{27}$ and $3 \sqrt[4]{3}$ cm.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-595.jpg?height=370&width=534&top_left_y=95&top_left_x=109)

Fig. 10.80