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December 21, 2014

December 21, 2014

Posted by **Achillies** on Saturday, March 5, 2011 at 3:30pm.

- math -
**drwls**, Saturday, March 5, 2011 at 4:30pmb = 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

- math -
**Damon**, Saturday, March 5, 2011 at 4:40pms^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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