## Task B-1.3.

The length of $\overline{A B}$ is the hypotenuse of the right triangle $A B C$. The altitude from vertex $C$, with its foot at point $D$, cuts off a segment $\overline{D B}$ of length 16 on the hypotenuse. Determine the area of triangle $A B C$, if $|A C|=15 ?$
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_ea3ef29009412d1e0d2bg-03.jpg?height=508&width=437&top_left_y=320&top_left_x=818)

Triangles $A D C$, $C D B$, and $A B C$ are similar (by the $K K$ theorem).

From the similarity of triangles $A D C$ and $A B C$, we get:

$$
\frac{|A D|}{|A C|}=\frac{|A C|}{|A B|} \Longrightarrow \frac{|A D|}{15}=\frac{15}{|A B|} \Longrightarrow \frac{|A D|}{15}=\frac{15}{|A D|+16}
$$

If we let $|A D|=x$, then

$$
\begin{aligned}
& x(x+16)=225 \\
& x^{2}+16 x=225
\end{aligned}
$$

We complete the right side of the equation to a perfect square.

$$
\begin{gathered}
x^{2}+16 x+64=225+64 \\
(x+8)^{2}=289 \Longrightarrow x+8=17 \Longrightarrow x=9
\end{gathered}
$$

Then $|A B|=25,|C B|=20$.

The area of the given triangle is $P=\frac{20 \cdot 15}{2}=150$.