Task B-3.6. The area of a rectangle is $12 \mathrm{~cm}^{2}$, and if the acute angle between its diagonals is halved, the area is $7.5 \mathrm{~cm}^{2}$. What is the length of the diagonal of the rectangle?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_cf2c4989ba380bc8684dg-15.jpg?height=415&width=631&top_left_y=1229&top_left_x=736)

If the angle between the diagonals is $\alpha$, the area is

$$
\begin{array}{r}
P=4 \cdot \frac{\frac{d}{2} \frac{d}{2} \sin \alpha}{2}=\frac{1}{2} d^{2} \sin \alpha=12 \\
d^{2} \sin \alpha=24
\end{array}
$$

If the angle between the diagonals is $\frac{\alpha}{2}$, the area is

$$
\begin{aligned}
P_{1}=\frac{1}{2} d^{2} \sin \frac{\alpha}{2} & =7.5 \\
d^{2} \sin \frac{\alpha}{2} & =15
\end{aligned}
$$

(1 point)

If we divide equations (1) and (2), we get

$$
\begin{aligned}
\frac{\sin \alpha}{\sin \frac{\alpha}{2}} & =\frac{8}{5} \\
\frac{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{\sin \frac{\alpha}{2}} & =\frac{8}{5} \\
\cos \frac{\alpha}{2} & =\frac{4}{5}
\end{aligned}
$$

Then,

$$
\begin{aligned}
\sin \frac{\alpha}{2} & =\frac{3}{5} \\
d^{2}=\frac{15}{\sin \frac{\alpha}{2}}=\frac{15}{\frac{3}{5}} & =25 \\
d & =5 \text{ cm}
\end{aligned}
$$

Note: The area of the rectangle can also be calculated using the lengths of the sides of the rectangle $a, b$ :

$$
\begin{aligned}
& a=d \cos \frac{\alpha}{2}, \quad b=d \sin \frac{\alpha}{2} \\
& P=a \cdot b=d \cos \frac{\alpha}{2} \cdot d \sin \frac{\alpha}{2}=\frac{1}{2} d^{2} \sin \alpha
\end{aligned}
$$