## Task 3 - 040913

Given is a right-angled triangle $A B C$, whose hypotenuse $\overline{A B} 25 \mathrm{~mm}$ and whose cathetus $\overline{B C}$ $20 \mathrm{~mm}$ long is. On this cathetus, the segment $\overline{B D}$ of length $15 \mathrm{~mm}$ is marked, and from point $D$ the perpendicular $\overline{D E}$ is dropped onto the hypotenuse.

Calculate the perimeter of triangle $B D E$!
It is given that $u=\overline{E B}+\overline{B D}+\overline{D E}$ with $\overline{B D}=15$.

The triangles $\triangle A B C$ and $\triangle D B E$ are similar with 2 equal angles:

$\measuredangle A B C=\measuredangle D B E$ and $\measuredangle A C B=\measuredangle D E B=90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0888.jpg?height=271&width=462&top_left_y=247&top_left_x=797)

Thus, it follows that:

$$
\overline{E B}=\overline{B C} \cdot \overline{B D}: \overline{A B}=20 \cdot 15: 25=12
$$

and by the Pythagorean theorem

$$
\overline{D E}=\sqrt{\overline{B D}^{2}-\overline{E B}^{2}}=\sqrt{15^{2}-12^{2}}=\sqrt{225-144}=\sqrt{81}=9
$$

Therefore, in the above equation: $u=12+15+9=36$. The perimeter of the triangle $\triangle B D E$ is thus $36 \mathrm{~mm}$.