In the given figure, in $ riangle A B C, A D \perp B C$ and $\angle B A C=90^{\circ}$. If $B C=16 \

Question

In the given figure, in $	riangle A B C, A D \perp B C$ and $\angle B A C=90^{\circ}$. If $B C=16 \mathrm{~cm}$ and $D C=4 \mathrm{~cm}$, then the value of $x$ is :\

Professor PC Thomas Classes · Answer

(c):In $	riangle A B C$ and $	riangle D A C$ $$ \begin{aligned} & \angle B A C=\angle A D C=90^{\circ} \ & \angle C=\angle C \end{aligned} $$ (Common)\ So, $	riangle A B C \sim 	riangle D A C$\ (By AA similarity)\ $	herefore \quad \frac{B C}{A C}=\frac{A C}{D C}$\ $\Rightarrow \frac{16}{x}=\frac{x}{4}$\ $\Rightarrow \quad x^{2}=16 	imes 4=64$\ $\Rightarrow \quad x=8 \mathrm{~cm}$\