So... hmmm if you check the first picture below, for 2)

we could use the proportions of those small, medium and large similar triangles like $\bf \cfrac{small}{large}\qquad \cfrac{x}{12}=\cfrac{6}{x}\impliedby \textit{solve for "x"}
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\cfrac{small}{large}\qquad \cfrac{z}{18}=\cfrac{6}{z}\impliedby \textit{solve for "z"}
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\cfrac{large}{medium}\qquad \cfrac{y}{12}=\cfrac{18}{y}\impliedby \textit{solve for "y"}$

now.. for 3) will be the second picture below

$\bf \cfrac{large}{medium}\qquad \cfrac{x+10}{2\sqrt{30}}=\cfrac{2\sqrt{30}}{10}\impliedby \textit{solve for "x"}
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\textit{now, because you already know what "x" is, we can use it below}
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\cfrac{large}{small}\qquad \cfrac{z}{x}=\cfrac{x+10}{z}\impliedby \textit{solve for "z"}
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\textit{and let us use "x" again below}
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\cfrac{small}{medium}\qquad \cfrac{y}{10}=\cfrac{x}{y}\impliedby \textit{solve for "y"}$

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3 years ago