## Task B-3.6.

One leg of the right triangle $ABC$ is twice as long as the other. The triangle $A^{\prime} B^{\prime} C$ is formed by rotating the triangle $ABC$ around the right-angle vertex $C$ by $30^{\circ}$. The points $A, A^{\prime}, B, B^{\prime}$, and $C$ are the vertices of a convex pentagon with an area of $20+8 \sqrt{3}$. Calculate the lengths of the legs of the triangle $ABC$.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_ab4e7c3a83d03c059773g-15.jpg?height=431&width=877&top_left_y=1389&top_left_x=515)

1 point

The area of the pentagon is equal to the sum of the areas of three triangles. With the labels as in the figure, we have

$$
\begin{aligned}
P\left(A A^{\prime} B B^{\prime} C\right) & =P\left(C A A^{\prime}\right)+P\left(C A^{\prime} B\right)+P\left(C B B^{\prime}\right) & & 1 \text { point } \\
& =\frac{1}{2} b \cdot b \sin 30^{\circ}+\frac{1}{2} a \cdot b \sin 60^{\circ}+\frac{1}{2} a \cdot a \sin 30^{\circ} & & 3 \text { points } \\
& =\frac{1}{4} b^{2}+\frac{1}{4} 2 b^{2} \sqrt{3}+\frac{1}{4} b^{2}=\frac{5+2 \sqrt{3}}{4} b^{2} . & & 2 \text { points }
\end{aligned}
$$

From $\frac{5+2 \sqrt{3}}{4} b^{2}=20+8 \sqrt{3}$

it follows that $b^{2}=16$, or $b=4$ and $a=8$.