For what value of a is there a right triangle with sides a+1 ,6a , and 6a+1?
We have that
(a + 1)^2 + (6a)^2 = (6a + 1)^2 simplify
a^2 + 2a+ 1 + 36a^2 = 36a^2 + 12a + 1
a^2 + 2a = 12a
a^2 - 10a = 0
a ( a - 10) = 0
So....a = 0 (reject ) or a = 10 which is correct
+100 
