Find the perimeter of right triangle WXY if the radius of the circle is 4 units and WY = 20 units.

Remark
This problem is one of those little gems you come across every so often. It has a couple of things worth knowing even if it is a little late.

Step One
Find a + b [which are the two legs of the triangle.]
r = the inradius
r = 4 in this question.
r = (a + b - c)/2 where c is the hypotenuse. This is easily proved. If you can't, google inradius and perimeter.
4 = (a+b - 20)/2  Multiply by 2
4*2 = a+b - 20
8 = a+b - 20 Add 20 to both sides.
8 + 20 = a + b
28 = a + b

Step two 
Set up the Pythagorean Theorem
a^2 + b^2 = c^2
a^2 + b^2 = 20^2
a^2 + b^2 = 400

Step three
Expand (a + b)^2 to find ab
(a + b)^2 = a^2 + 2ab + b^2
(28)^2 = a^2 + b^2 + 2ab
Remember that a^2 + b^2 = 400
28^2 = 400 + 2ab
784 = 400 + 2ab     Subtract 400 from both sides.
784 - 400 = 2ab   
384 = 2ab              divide by 2
384/2  = ab
ab = 192

Step 4 
Find a and b
ab = 192
a + b = 28
b = 28 - a

ab = 192                    Substitute 28 - a for b
a(28 - a) = 192
-a^2 + 28a = 192
-a^2 + 28a - 192 = 0

You have to use the quadratic formula.
a = -1
b = 28
c = - 192

[tex]\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac } }{2a} [/tex]

When you solve this you get 2 roots.
a = 12 and b = 16 These are the two sides of the triangle or
a = 16 and b = 12
Check
a^2 + b^2 =? 20^2
12^2 + 16^2 = ?400
144 + 256 =? 400
400 = 400

The problem does check
Neat little problem Thanks for posting. 

I would solve 2*r = (a + b - c) for you but I can't get the many diagram programs to work properly for labeling these diagrams. 

There are great discussions on how to make the incircle radius an integer value by choosing the triangle sides with an integer value.