use the illustration i've made as reference. (see attachment)
taking the left right triangle, we can use Pythagorean Theorem
[tex]x^2 = (\frac{x}{2})^2 + (x-2)^2 [/tex]
[tex]x^2 = \frac{x^2}{4} + (x^2 - 4x +4)[/tex]
[tex]x^2 = \frac{x^2}{4} + x^2 - 4x + 4[/tex]
multiply the whole equation with 4
[tex]4x^2 = x^2 + 4x^2 - 16x + 16[/tex]
transpose all the terms from left to right, equating it to zero
[tex]0=4x^2 - 4x^2 + x^2 - 16x + 16 [/tex]
or
[tex]4x^2-4x^2+x^2-16x+16=0[/tex]
[tex]x^2 - 16x + 16 = 0[/tex]
using quadratic formula
[tex]x = \frac{-b (+-) \sqrt{b^2-4ac} }{2a} [/tex]
a = 1
b = -16
c = 16
[tex]x = \frac{-(-16) (+-) \sqrt{(-16)^2-4(1)(16)} }{2(1)} [/tex]
[tex]x = \frac{16 (+-) \sqrt{256-64} }{2} [/tex]
[tex]x = \frac{16(+-) \sqrt{192} }{2}[/tex]
[tex]x = \frac{16(+-) \sqrt{64(3)} }{2}[/tex]
[tex]x = \frac{16(+-) 8\sqrt{3}}{2}[/tex]
[tex]x = 8 (+-) 4 \sqrt{3}[/tex]
[tex]x_1 = 8 + 4 \sqrt{3}[/tex]
or
[tex]x_1 = 14.9282 units[/tex]
[tex]x_2 = 8 - 4\sqrt {3}[/tex]
or
[tex]x_2 = 1.0718 units[/tex]
therefore you can have the value of the side as 14.9282 units and/or 1.0718 units
