Posted by Achillies on Saturday, March 5, 2011 at 3:30pm.
b = a + 3
c = 2a -3
c^2 = a^2 + b^2 (Pythagorean theorem)
Solve those three equations simultaneously.
c^2 = 4a^2 -12a +9
c^2 = a^2 + (a+3)^2 = 2a^2 +6a +9
2a^2 -18a =0
a = 9
b = 12
c = 15
s^2 + l^2 = h^2
s^2 + (s+3)^2 = (2s-3)^2
s^2 + s^2 + 6 s + 9 = 4 s^2 - 12 s + 9
2 s^2 - 18 s = 0 = 2s(s-9)
s = 0 or s = 9
the s = 0 root is irrelevant since that results in a negative hypotenuse
s = 9
l = 9+3 = 12
h = 15
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