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Solve for x , y and z using geometric mean altitude theorem and leg theorem

Solve for x , y and z using geometric mean altitude theorem and leg theorem-example-1

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Check the picture below.

when running a line, in a right-triangle, from the 90° angle perpendicular to its opposite side, we will end up with three similar triangles, one Small, one Medium and a containing Large one. So let's use proportions to get x,y and z.


\cfrac{x}{30}~~ = ~~\cfrac{30}{24}\implies \cfrac{x}{30}~~ = ~~\cfrac{5}{4}\implies x=\cfrac{150}{4}\implies \boxed{x=\cfrac{75}{2}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{y}{24}~~ = ~~\cfrac{x-24}{y}\implies \cfrac{y}{24}~~ = ~~\cfrac{(75)/(2)-24}{y}\implies \cfrac{y}{24}~~ = ~~\cfrac{~~ (27)/(2)~~}{y}\implies \cfrac{y^2}{24}=\cfrac{27}{2} \\\\\\ y^2=\cfrac{(24)(27)}{2}\implies y=\sqrt{\cfrac{(24)(27)}{2}}\implies \boxed{y=18} \\\\[-0.35em] ~\dotfill


\cfrac{z}{y}~~ = ~~\cfrac{30}{24}\implies \cfrac{z}{18}~~ = ~~\cfrac{5}{4}\implies z=\cfrac{(18)(5)}{4}\implies \boxed{z=\cfrac{45}{2}}

Solve for x , y and z using geometric mean altitude theorem and leg theorem-example-1
User Trevorp
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