Question 451434
using the Pythagorean theorem, we get {{{(x+4)^2 + (x+7)^2 = (3x)^2}}}

<==> {{{x^2 + 8x + 16 + x^2 + 14x + 49 = 9x^2}}}
<==> {{{7x^2 -22x - 65 = 0}}}
<==> (7x + 13)(x-5) = 0
==> x = 5 
The perimeter is x+4 +x+7 + 3x = 5x + 11
Hence the perimeter is 5*5 + 11 = 25 + 11 = 36.

(We cannot accept the other x-value -13/7 for an apparent reason.)