4B. In a right-angled triangle, the length of the height drawn to the hypotenuse is 4, and the length of the median from the right angle is 5. Calculate the sum of the lengths of the legs of this triangle.
Solution. Let the right triangle be $ABC$ with the right angle at vertex $C$. Let $CD$ be the altitude and $CF$ be the median dropped from $C$. Since triangle $ABC$ is a right triangle, it follows that

$$
\overline{AF}=\overline{FB}=\overline{CF}=5 \text{.}
$$

Triangle $CDF$ is a right triangle, so

![](https://cdn.mathpix.com/cropped/2024_06_05_e56627a2e774dea33268g-03.jpg?height=315&width=551&top_left_y=1847&top_left_x=914)

The Pythagorean theorem implies that $\overline{DF}=\sqrt{5^{2}-4^{2}}=3$.

The Pythagorean theorem again implies

$$
\overline{AC}=\sqrt{\overline{AD}^{2}+\overline{CD}^{2}}=\sqrt{4+16}=2 \sqrt{5} \text{ and } \overline{BC}=\sqrt{\overline{DB}^{2}+\overline{CD}^{2}}=\sqrt{64+16}=4 \sqrt{5} \text{.}
$$

The sum of the legs is $6 \sqrt{5}$.

## 2nd year