ANSWER
[tex]x = 8 \sqrt{3} [/tex]
EXPLANATION
[tex] {h}^{2} = {y}^{2} + {4}^{2} [/tex]
This implies that,
[tex] {h}^{2} = {y}^{2} + 16...eqn1[/tex]
Also,
[tex] {h}^{2} + {x}^{2} = {16}^{2} ...eqn2[/tex]
And
[tex] {y}^{2} + {12}^{2} = {x}^{2} [/tex]
[tex] {y}^{2}={x}^{2} - 144 ...eqn3[/tex]
Put equation 1 into equation 2.
This will give us,
[tex] {y}^{2} + 16 + {x}^{2} = {16}^{2} [/tex]
[tex] {y}^{2} +{x}^{2} = {16}^{2} - 16[/tex]
[tex] {y}^{2} +{x}^{2} =16 ( 16 - 1)[/tex]
[tex] {y}^{2} +{x}^{2} =240...eqn4[/tex]
We put equation 3 into equation 4,
[tex] {x}^{2} - 144+{x}^{2} =240[/tex]
We simplify to get,
[tex]2 {x}^{2} = 384[/tex]
[tex]{x}^{2} = 192[/tex]
[tex]x = \sqrt{192} [/tex]
[tex]x = 8 \sqrt{3} [/tex]
