## Task 5 - 200835

Two rays $s_{1}$ and $s_{2}$, which originate from a point $S$ and form a right angle with each other, are intersected by two parallel lines $g$ and $h$. The line $g$ intersects $s_{1}$ at $A$ and $s_{2}$ at $C$, the line $h$ intersects $s_{1}$ at $B$ and $s_{2}$ at $D$. Furthermore, $\overline{S B}=5 \mathrm{~cm}$, and the area of triangle $S A C$ is exactly $36 \%$ of the area of triangle $S B D$.

Determine the length of the segment $S A$ from these conditions!
The area of triangle $S B D$ is $\frac{S B \cdot S D}{2}=\frac{5 \mathrm{~cm} \cdot S D}{2}$.
The area of triangle $S A C$ is $\frac{S A \cdot S C}{2}$.

By assumption, this is $36 \%$ of the area of triangle $S B D$. Therefore,

$$
S A \cdot S C=\frac{36}{100} \cdot S D \cdot 5 \mathrm{~cm}
$$

According to a part of the ray theorem, $S C: S D=S A: 5 \mathrm{~cm}$, so

$$
S C=\frac{S A \cdot S D}{5 \mathrm{~cm}}
$$

Substituting $S C$ from (2) into (1), we get

$$
\frac{S A^{2} \cdot S D}{5 \mathrm{~cm}}=\frac{36}{100} \cdot S D \cdot 5 \mathrm{~cm}
$$

and from this $S A^{2}=9 \mathrm{~cm}^{2}$ and thus $S A=3 \mathrm{~cm}$.