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In the adjoining figure $\mathrm{AB}=12 \mathrm{~cm}, \mathrm{CD}=8 \mathrm{~cm}, \mathrm{BD}=20 \mathrm{~cm} ; \angle \mathrm{ABD}=\angle \mathrm{AEC}=\angle \mathrm{EDC}=90^{\circ}$. If $\mathrm{BE}=\mathrm{x}$, then
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The correct answer is:
$x$ has two possible values whose difference is 4
$\frac{12}{x}=\frac{20-x}{8} \Rightarrow x^{2}-20 x+96=0$
$\Rightarrow x=8,12$
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