Posted by Blonde Glamour on .
consider the right traingel with sices 3, 4, and 5. the lengths form an arithmetic series.
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
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.