A right triangle has legs of length 6 and b, and a hypotenuse of length c. The perimeter of the triangle is \(24\). Compute c.
+2666 We have the following system:
\(6+b+c=24\) (1)
\(36+b^2=c^2\) (2)
Solve for \(c\) in the first equation: \(c=18-b\)
Squaring this gives \(b^2-36b+324\)
Substituting this in (2) gives: \(36+b^2=b^2-36b+324\)
Solving, we find that \(b= 8\)
Subsituting this in to (1), we find that \(\color{brown}\boxed{c = 10}\)
+2666 We have the following system:
\(6+b+c=24\) (1)
\(36+b^2=c^2\) (2)
Solve for \(c\) in the first equation: \(c=18-b\)
Squaring this gives \(b^2-36b+324\)
Substituting this in (2) gives: \(36+b^2=b^2-36b+324\)
Solving, we find that \(b= 8\)
Subsituting this in to (1), we find that \(\color{brown}\boxed{c = 10}\)
