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January 26, 2015

January 26, 2015

Posted by **Blonde Glamour** on Thursday, December 21, 2006 at 10:01pm.

a. what are the other three right triangles in arthimetic series

Actually, I think there are an infinite numbr of such triangles.

You want to have sides a, b and c such that a < b < c and b - a = c - b.

This means that:

c = 2 b - a

You also know that:

c^2 = a^2 + b^2 --->

c^2 - b^2 = a^2

(c-b)(c+b) = a^2

(b-a)(3b-a) = a^2 --->

3 b^2 - 4ba = 0 --->

3b - 4a = 0 --->

b = 4/3 a

E.g.

take a = 27, then b = 36 and c should be 36 plus the difference of 36 and 27, which is 45. Pythagoras gives

c = sqrt[27^2 + 36^2] which is indeed 45.

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