Area Given Perimeter?

We are given that $a + b + c = 18$ and $a^2 + b^2 + c^2 = 128$ , and we need to find $\frac{ab}{2}$ . Note that since this is a right triangle, $a^2 + b^2 = c^2.$ Thus we can substitute this into the second equation to get $2 \times c^2 = 128$ , so $c = 8$ . Then from the first equation, we get $a + b = 10$ and from the second we get $a^2 + b^2 = 64$ . If we square $a + b$ , we get

$(a + b)^2 = a^2 + 2ab + b^2 = 100$ .

Since $a^2+b^2=64$ , $2ab = 36$ . We need to find $\frac{ab}{2}$ , so we divide by $4$ to get an answer of $\boxed{9}$ .