Question

In $\triangle A B C, D$ is a point on $A C$ such that $B D \perp A C$. If $\angle A B C=90^{\circ}, A B=20 \mathrm{~cm}$, and $B D=12 \mathrm{~cm}$, then find the length of $B C$ (in $\mathrm{cm}$ ).

• A. 12
• B. 15
• C. 16
• D. 25

Answer 1

Answer: (B)

$ \text { In } \triangle A B D, A D^2=A B^2-B D^2$
$=A D^2=20^2-12^2=400-144=256=16^2 $
$\Rightarrow A D=16 \mathrm{~cm} $
$ \text { We have } B D^2=A D \cdot C D $
$ \Rightarrow 12^2=16 \cdot C D \Rightarrow C D=9 \mathrm{~cm}$
$ \text { In } \triangle B D C, $
$B C^2=B D^2+C D^2=12^2+9^2 $
$=144+81=225=15^2 $
$\Rightarrow B C=15 \mathrm{~cm}$