## Task B-1.7.

Two right triangles $P S T$ and $R S T$ have a common hypotenuse $\overline{S T}$. Vertices $P$ and $R$ are on the same side of the hypotenuse $\overline{S T}$. The legs $\overline{P T}$ and $\overline{R S}$ intersect at point $Q$. If $|P S|=6 \mathrm{~cm},|P T|=17 \mathrm{~cm}$ and $|R T|=1 \mathrm{~cm}$, what is the area of triangle $Q S T$?
## Second Solution.

Using the same notation as in the first solution, applying the Pythagorean theorem to triangles $P S T$ and $R S T$ gives:

$$
\begin{gathered}
|S T|^{2}=|P S|^{2}+|P T|^{2}=36+289=325 \\
|R S|=\sqrt{|S T|^{2}-|R T|^{2}}=\sqrt{325-1}=\sqrt{324}=18 \text{ cm}
\end{gathered}
$$

Triangles $P Q S$ and $R Q T$ are right triangles, with right angles at vertex $P$ and vertex $R$, respectively, and $|\angle P Q T|=|\angle R Q T|$.

Therefore, by the AA similarity criterion, triangles $P Q S$ and $R Q T$ are similar, with a similarity ratio $k=\frac{|R T|}{|P S|}=\frac{1}{6}$.

Thus,

$\frac{|T Q|}{18-|R Q|}=\frac{1}{6} \text{ and } \frac{|R Q|}{17-|T Q|}=\frac{1}{6}$.

Solving this system of equations, we get $|R Q|=\frac{12}{5} \text{ cm}$.

Therefore, the area of triangle $R Q T$ is $P_{3}=\frac{|R T| \cdot|R Q|}{2}=\frac{1 \cdot \frac{12}{5}}{2}=\frac{6}{5} \text{ cm}^{2}$.

Since $P_{3}+P_{1}=\frac{|R T| \cdot|R S|}{2}=\frac{1 \cdot 18}{2}=9$, it follows that $P_{1}=9-P_{3}=9-\frac{6}{5}=\frac{39}{5} \text{ cm}^{2}$.