Answer:
Q1
[tex]by \ pythagoras \ theorem : \\(3x +1 )^2 = (3x - 1)^2 +x^2\\\\9x^2 + 6x +1 = 9x^2 + 1 -6x +x^2\\\\9x^2 + 6x +1 = 10x^2 + 1 -6x \\\\ 10x^2 + 1 -6x - 9x^2 -6x -1 = 0 \\\\x^2 -12x = 0\\x(x - 12) = 0\\x = 0, 12\\since \ length \ cannot \be \ 0, x = 12\\OJ = 3x + 1 = 3(12) +1 = 37[/tex]
Q2
[tex]by \ pythagoras \ thm, \\(7x -12) ^2 = (5x +10)^2 +(2x)^2\\\\49x^ 2 +144 - 168x = 25x^2 +100+ 100x +2x^2\\\\49x^ 2 +144 - 168x - 25x^2 -100- 100x -2x^2 = 0\\\\22x^2 -268x -44 = 0\\11x^2 - 134x - 22 = 0\\x=12.0154[/tex]
