Question 651402
The lengths of the sides of a right triangle are such that the shortest side is 7 inches shorter than the middle side, while the longest side (hypotenuse) is 1 in. longer than the middle side. Find the lengths of the sides.

I know to use A sq. + b sq. = c sq. 
a= x-7 sq
b= x sq 
c= x+1 sq

I think I would also use sq. binomial rule as well. But I just can't get it to work. 


Let the length of the middle side be M
Then length of shortest side = M - 7
Length of longest side (hypotenuse) = M + 1


Using pythagorean theorem formula, {{{a^2 + b^2 = c^2}}}, where a and b are the lengths of the legs, and c is the length of the hypotenuse, we get: 


{{{M^2 + (M - 7)^2 = (M + 1)^2}}}


{{{M^2 + M^2 - 14M + 49 = M^2 + 2M + 1}}}


{{{M^2 + M^2 - M^2 - 14M - 2M + 49 - 1 = 0}}}


{{{M^2 - 16M + 48 = 0}}}


(M - 4)(M - 12) = 0


M, or middle side = 4  in. (ignore as this would make shortest side, - 3 (4 - 7), and length CANNOT be negative


M, or length of middle side = {{{highlight_green(12)}}} in.


Length of shortest side = 12 - 7, or {{{highlight_green(5)}}} in.


Length of longest side = 12 + 1, or {{{highlight_green(13)}}} in.


This is actually one of the special right triangles, or a pythagorean triple.


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