## Task B-4.2.

The base $\overline{A B}$ of the isosceles triangle $A B C$ lies on the line $x-2 y+20=0$, and the third vertex is at point $C(1,8)$. If the area of the triangle is 15, determine the lengths of its sides.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_8cadc17e7912dc7108beg-17.jpg?height=520&width=676&top_left_y=2093&top_left_x=610)

The height of the triangle is the distance from point $C$ to the line $A B$,

$$
c=\frac{|1-16+20|}{\sqrt{1+4}}=\sqrt{5}
$$

Let $|A B|=a$. We determine the length of the base from the area,

$$
\begin{aligned}
& P=\frac{a v}{2} \\
& a=\frac{2 P}{v}=\frac{30}{\sqrt{5}}=6 \sqrt{5}
\end{aligned}
$$

We can calculate the length of the legs using the Pythagorean theorem:

$$
b=\sqrt{\left(\frac{a}{2}\right)^{2}+v^{2}}=\sqrt{45+5}=5 \sqrt{2}
$$