Since we are dealing with a right triangle, we can use the Pythagorean theorem, shown below
[tex]H^2=L^2_1+L^2_2[/tex]
In our case, H=4, L_1=2, L_2=x; then,
[tex]4^2=2^2+x^2[/tex]
Solving for x,
[tex]\begin{gathered} \Rightarrow x^2=16-4 \\ \Rightarrow x^2=12 \\ \Rightarrow x=\sqrt[]{12}=\sqrt[]{4\cdot3} \\ \Rightarrow x=2\sqrt[]{3} \end{gathered}[/tex]
The answer is x=2sqrt(3)
