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Please help me do this I've been stuck for over an hour

Find the missing length indicated. Leave your answer in simplest radical form.



Please help me do this I've been stuck for over an hour Find the missing length indicated-example-1

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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. Check the picture below.

so the longest leg of the larger triangle will be "z" and the shorter leg of it will be "y", so let's find those fellows using proportions


\cfrac{x}{16}=\cfrac{9}{x}\implies x^2=144\implies x=√(144)\implies \boxed{x=12} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{y}{x}=\cfrac{25}{16}\implies \cfrac{y}{12}=\cfrac{25}{16}\implies y=\cfrac{(12)(25)}{16}\implies \boxed{y=\cfrac{75}{4}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{y}{z}=\cfrac{9}{x}\implies \cfrac{~~ (75 )/(4 ) ~~}{z}=\cfrac{9}{12}\implies \cfrac{75}{4z}=\cfrac{3}{4}\implies 300=12z \\\\\\ \cfrac{300}{12}=z\implies \boxed{25=z}

Please help me do this I've been stuck for over an hour Find the missing length indicated-example-1
User Joel Boehland
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