12.401 The sides of a parallelogram are equal to $a$ and $\boldsymbol{b}(\boldsymbol{a}<\boldsymbol{b}$). From the midpoint of the larger side, the parallel side is seen at an angle $\alpha$. Find the area of the parallelogram.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-288.jpg?height=287&width=1100&top_left_y=389&top_left_x=102)

On the other hand, $B K^{2}+C K^{2}=b^{2}+2 \cdot B K \cdot C K \cdot \cos \alpha$ (cosine theorem for $\triangle B C K) \Rightarrow B K \cdot C K=\frac{4 a^{2}-b^{2}}{4 \cos \alpha} \Rightarrow S_{A B C D}=2 S_{B C K}=B K \cdot C K \cdot \sin \alpha=\frac{4 a^{2}-b^{2}}{4} \tan \alpha$. Answer: $\frac{4 a^{2}-b^{2}}{4} \cdot \tan \alpha$.