Question Number 124522 by bemath last updated on 03/Dec/20 | ||
Answered by mr W last updated on 04/Dec/20 | ||
$${CD}=\mathrm{25}−\mathrm{16}=\mathrm{9} \\ $$$${AC}^{\mathrm{2}} ={BC}×{CD}=\mathrm{16}×\mathrm{9} \\ $$$${AC}=\mathrm{4}×\mathrm{3}=\mathrm{12} \\ $$$${AB}^{\mathrm{2}} ={BC}^{\mathrm{2}} +{AC}^{\mathrm{2}} =\mathrm{16}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} \\ $$$${AB}=\sqrt{\mathrm{16}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }=\mathrm{20} \\ $$$$\frac{{AE}}{{AC}}=\frac{{AC}}{{AB}} \\ $$$$\Rightarrow{AE}=\frac{\mathrm{12}^{\mathrm{2}} }{\mathrm{20}}=\mathrm{7}.\mathrm{2} \\ $$$$\Rightarrow{answer}\:{D} \\ $$ | ||