Question 498590
You must have been given more information about the right triangle.
Look!
{{{a^2+b^2 = c^2}}} In this case, {{{c = sqrt(18)}}} so {{{c^2 = 18}}}.  Now we have:
{{{a^2+b^2 = 18}}} This equation has an infinite number of solutions unless you place some restrictions on a and b.
Let's assume the the numbers a and b must be integers (not an unreasonable assumption), then a and b can only take on the values of {{{highlight(a = 3)}}} and {{{highlight(b = 3)}}}, so you'd get:
{{{3^2+3^2 = 18}}} or 
{{{9+9 = 18}}}
This implies that the triangle is an isosceles right triangle.
No other integers for a and b would work.