Question

In triangle $ABC$, the altitude from $A$ to $BC$ meets $BC$ at $D$ and the altitude from $B$ to $CA$ meets $AD$ at $H$. If $AD$ $=4$ cm, $BD$ $=3$ cm and $CD$ $=2$ cm and if $\frac{AB}{BD}=\frac{AH}{HD}$, then the length of $HD$ is __________.

• A. $\frac{\sqrt{5}}{2}$ cm
• B. $\frac{3}{2}$ cm
• C. $\sqrt{5}$ cm
• D. $\frac{5}{2}$ cm

Answer 1

Answer: (B)

$\frac{3}{2}$ cm

Given, $AD=4$ cm, $BD=3$ cm and $CD=2$ cm

In Triangle$ABD$,

Using Pythagoras Theorem, we have

$(AB{)}^2=(AD{)}^2+(BD{)}^2$

$\Rightarrow$ $(AB{)}^2=9+16$

$\Rightarrow$ $AB=5$ cm

Given that,$\frac{AB}{BD}=\frac{AH}{HD}$

$\Rightarrow$ $AH=\frac{5}{3}HD$

$\Rightarrow AD=AH+HD=\frac{8}{3}HD$

$\Rightarrow$ $HD=\frac{3}{8}\times 4$

$\Rightarrow$ $HD=\frac{3}{2}$ cm

Hence, option B is correct.