Example 42. The sides of a parallelogram are equal to $a$ and $b(a>b)$. Find the area of the parallelogram, knowing that the acute angle between its diagonals is $\alpha$.

Given (Fig. 46): $ABCD$ is a parallelogram, $|AD|=a$, $|AB|=b, \widehat{COD}=\alpha$.

Find: $S_{ABCD}$.
S o l u t i o n. This is a problem of type $F$. To find $S_{A B C D}$, we will use the direct counting method. We write:

$$
S_{A B C D}=\frac{1}{2}|A C| \cdot|B D| \cdot \sin \alpha
$$

For brevity, let $|A C|=d_{1},|B D|=d_{2}$. We will find $d_{1} d_{2}$ by applying the cosine theorem to $\triangle A O D$ and $\triangle C O D$.

In the process of calculations, we sequentially obtain:

$$
a^{2}=\left(\frac{d_{1}}{2}\right)^{2}+\left(\frac{d_{2}}{2}\right)^{2}-2 \frac{d_{1}}{2} \cdot \frac{d_{2}}{2} \cos \left(180^{\circ}-\alpha\right)
$$

or

$$
\begin{gathered}
a^{2}=\frac{d_{1}^{2}}{4}+\frac{d_{2}^{2}}{4}+\frac{d_{1} d_{2}}{2} \cos \alpha \\
b^{2}=\left(\frac{d_{1}}{2}\right)^{2}+\left(\frac{d_{2}}{2}\right)^{2}-2 \frac{d_{1}}{2} \cdot \frac{d_{2}}{2} \cos \alpha
\end{gathered}
$$

or

$$
b^{2}=\frac{d_{1}^{2}}{4}+\frac{d_{2}^{2}}{4}-\frac{d_{1} d_{2}}{2} \cos \alpha
$$

68

Subtracting equality (42.2) from equality (42.1) term by term, we find:

$$
d_{1} d_{2} \cos \alpha=a^{2}-b^{2}
$$

from which $d_{1} d_{2}=\frac{a^{2}-b^{2}}{\cos \alpha}$.

Thus,

$$
S_{A B C D}=\frac{1}{2}\left(a^{2}-b^{2}\right) \operatorname{tg} \alpha
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-070.jpg?height=323&width=240&top_left_y=244&top_left_x=1459)

Fig. 47

We omit the investigation.