5. In the right-angled triangle $ABC$ (with the right angle at vertex $C$), one of the acute angles is three times larger than the other. Calculate the area of this triangle if the length of the hypotenuse is $c=8 \mathrm{~cm}$.
Solution. From the condition of the problem, it follows that the acute angles of the given triangle are $67.5^{\circ}$ and $22.5^{\circ}$. Let, for example, $\measuredangle A B C=22.5^{\circ}$. Let us draw a triangle $B C D$ congruent to triangle $A B C (\overline{A B}=\overline{B D})$, see the diagram. Then

$$
\measuredangle A B D=2 \measuredangle A B C=2 \cdot 22.5^{\circ}=45^{\circ}
$$

Then the area of triangle $A B D$ is

$$
P_{A B D}=2 P_{A B C}=\frac{\overline{A B} \cdot \overline{D E}}{2}=4 \overline{D E}
$$

where $D E$ is the height dropped from vertex $D$ to side $A B$. However, $B D E$ is an isosceles right triangle and $\overline{B E}=\overline{D E}$, so $2 \overline{D E}^{2}=\overline{B D}^{2}=8^{2}$, i.e., $\overline{D E}=4 \sqrt{2}$. Finally,

$$
\begin{gathered}
P_{A B D}=4 \cdot 4 \sqrt{2}=16 \sqrt{2} \text { and } \\
P_{A B C}=8 \sqrt{2} \mathrm{~cm}^{2}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_6d6f67207af7682f8d1dg-6.jpg?height=524&width=507&top_left_y=264&top_left_x=972)

## IX Department