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 $B C=21 \mathrm{~cm}$, then find the length of $B D$ (in $\mathrm{cm}$ ).

• A. $\frac{440}{29}$
• B. $\frac{420}{29}$
• C. $\frac{410}{29}$
• D. $\frac{210}{29}$

Answer 1

Answer: (B)

$ \text { In } \triangle A B D, A C^2=A B^2+B C^2$
$=A C^2=20^2+21^2=400+441=841=29^2$
$ \Rightarrow A C=29 \mathrm{~cm} $
$\therefore \frac{1}{2} \times 21 \times 20=\frac{1}{2} \times 29 \times B D$ $\left(\because \text { Area }=\frac{1}{2} b h\right)$
$\Rightarrow B D=\frac{21 \times 20}{29}=\frac{420}{29} \mathrm{~cm}$