[tex]\triangle RQT\ \text{and}\ \triangle SRT\ \text{are similar. Therefore the sides are in proportion.}\\\\\dfrac{RT}{TQ}=\dfrac{TS}{RT}\\\\\text{We have}\ RT=x,\ TQ=16,\ TS=9.\ \text{Substitute:}\\\\\dfrac{x}{16}=\dfrac{9}{x}\qquad\text{cross multiply}\\\\x^2=(9)(16)\\\\x^2=144\to x=\sqrt{144}\\\\\boxed{x=12}[/tex]
