Task B-1.8. The perimeter of a right triangle is $14 \mathrm{~cm}$. A square is constructed outward on each side. The sum of the areas of all the squares is $72 \mathrm{~cm}^{2}$. What is the area of the given right triangle?
## Solution.

Let $a$ and $b$ be the lengths of the legs, and $c$ be the length of the hypotenuse of a right triangle. It is given that

$$
\begin{aligned}
a+b+c & =14 \\
a^{2}+b^{2}+c^{2} & =72
\end{aligned}
$$

Since, by the Pythagorean theorem, $a^{2}+b^{2}=c^{2}$, the last equation becomes $2 c^{2}=72$ or $c=6 \text{ cm}$.

From the perimeter of the triangle, we get

$$
a+b=14-6=8
$$

We will square this equation and use the fact that the area of a right triangle is $P=\frac{a b}{2}$. Then we have

$$
\begin{aligned}
(a+b)^{2} & =64 \\
a^{2}+2 a b+b^{2} & =64
\end{aligned}
$$

which is equivalent to

$$
\begin{aligned}
& c^{2}+4 P=64 \\
& 36+4 P=64
\end{aligned}
$$

from which we get $P=7 \text{ cm}^{2}$.

(1 point)

## ZUPANIJSKO NATJECANJE IZ MATEMATIKE

## 2nd grade - high school - B variant

February 28, 2011.

IF A STUDENT HAS A DIFFERENT METHOD OF SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO EVALUATE THE METHOD AND SCORE IT APPROPRIATELY.