## Task A-1.1.

The perimeter of a right triangle is 18, and the area is 9. What is the length of the hypotenuse of this triangle?
## First Solution.

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

Since $a+b+c=18$, we have

$$
(a+b)^{2}=(18-c)^{2}=18^{2}-36 c+c^{2}
$$

The given triangle is a right triangle, so by the Pythagorean theorem, $a^{2}+b^{2}=c^{2}$. 1 point It follows that

$$
(a+b)^{2}=a^{2}+2 a b+b^{2}=c^{2}+2 a b
$$

Since we have expressed $(a+b)^{2}$ in two ways, equating them gives

$$
2 a b=18^{2}-36 c
$$

Since the area of the triangle is 9, we have $a b=18$. 1 point

We conclude that $36=2 a b=18^{2}-36 c$. From this, $c=8$. 2 points

## Second Solution.

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

The given triangle is a right triangle, so by the Pythagorean theorem, $a^{2}+b^{2}=c^{2}$. 1 point

Since $a+b+c=18$, we have

$$
a^{2}+b^{2}=c^{2}=(18-a-b)^{2}
$$

After squaring, we get

$$
a^{2}+b^{2}=18^{2}+a^{2}+b^{2}-36 a-36 b+2 a b
$$

or $18^{2}+2 a b=36(a+b)=36 \cdot(18-c)$. 2 points

Since the area of the triangle is 9, we have $a b=18$. 1 point

We conclude that $18^{2}+36=36 \cdot(18-c)$. From this, $c=8$. 2 points