Solving Three Equations

Problem Level :3 (to me)
Problem Source: American Math Competitions 10 #18


A right triangle has perimeter 32 and area 20. What is the length of its hypotenuse?


Solution:


First we need to set up the equations. Just Kidding. We need to first identify the variables: a ,b, c for the side lengths.  



Identifying variables makes hard problems simpler than NORMAL.
We know the perimeter is 32 so a +b +c =32.  Assuming a and b are the legs and c is the hypotenuse and the area is 20, we know that 1/2 a b = 20, also ab=40.
Because of the Pythagorean Theorem a²+b²=c²


We also know that a + b = 32 -c, because we rearrange a +b +c =32.


Taking the square of  a²+b²=c², we get (a+b)²-2ab=c² 
Substituting a+b, we get  (32-c)²-2ab=c².
Substituting ab = 40 we get (32-c)²-2(40)=c².
(32-c)²-80=c².

1024 -64c +c ²-80= c²

Subtracting c² from both sides we get

944 – 64c= 0

64c= 944

Answer : c=59/4

Now that used many manipulations.