11)
when it comes to a right-triangle, having a line cutting the right-angle and perpendicular to the opposite side, like in this case, what happens is, we end up with 3 similar triangles.
a Large one, containing the other two.
a Medium one, one of the small ones.
a Small one.
so then, we can use the ratios from all these triangles, Check the picture below.
[tex]\bf \cfrac{Large}{Small}\qquad \qquad \cfrac{x+10}{20}=\cfrac{20}{10}\implies \cfrac{x+10}{20}=2\implies x+10=40\implies \boxed{x=30} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{Small}{Medium}\qquad \qquad \cfrac{10}{y}=\cfrac{y}{x}\implies \cfrac{10}{y}=\cfrac{y}{30}\implies 300=y^2\implies \sqrt{300}=y[/tex]
[tex]\bf \sqrt{3\cdot 100}=y\implies \sqrt{3\cdot 10^2}=y\implies \boxed{10\sqrt{3}=y} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{Medium}{Large}\qquad \qquad \cfrac{x}{z}=\cfrac{z}{x+10}\implies \cfrac{30}{z}=\cfrac{z}{30+10}\implies \cfrac{30}{z}=\cfrac{z}{40} \\\\\\ 1200=z^2\implies \sqrt{1200}=z\implies \sqrt{3\cdot 20^2}=z\implies \boxed{20\sqrt{3}=z}[/tex]
