Question

In a $△ABC$, if $AB=5cm$, $BC=13cm$ and $CA=12cm$, then the distance of vertex $A$ from the side $BC$ is (in $cm$)

• A. $\frac{25}{13}$
• B. $\frac{60}{13}$
• C. $\frac{65}{12}$
• D. $\frac{144}{13}$

Answer 1

The correct option is B $\frac{60}{13}$

Area of triangle if $a$ is perpendicular to $b$ is, $\frac{1}{2}\times a\times b$

In this triangle there are two perpendiculars $(AD$ to $BC$ and $CA$ to $AB)$,

$\therefore$ By equating area of triangle for both we get,

$\frac{1}{2}\times CA\times AB=\frac{1}{2}\times AD\times CB$

$12\times 5=AD\times 13$

$AD=\frac{60}{13}$