Systems of equations finding right triangles I need help setting up the equation for the question, "Find all right triangles for which the perimeter is $24$ units and the area is $24$ square units."
I know that the area is $A = \frac12 b  h$ and perimeter is $P = a + b + h$.
Using this, would the system look something like?
$a + b + c = 24$
$\frac12b c = 24$
(I'm using $c$ for consistency)
Then continue on using substitution?

Let A=24. I am assuming c as hypotenuse and a, b the other two sides of the triangle $\implies c^2=a^2+b^2$.
We now have $1/2*a*b= A$ and $a+b+c=A$
$(a+b)^2=(A-c)^2$
$\implies a^2+b^2+2ab= A^2-2Ac+c^2$ 
$\implies c=\dfrac{A^2-2ab}{2A}=10$
There we have ab=48 and c=10. So the possible right triangle is 6,8,10.
