Question

In $△ABC$, $\angle B={90}^o$, $AB=3,AC=5$ and $BD$ is perpendicular to $AC$, then find $BD$.

Answer 1

Given that:

$AB=3,AC=5,\angle B={90}^{\circ }$

By Pythagoras theorem,

$BC=\sqrt{5^2-3^2}=4$

We know that, Area of a triangle $=\frac{1}{2}\times$ base $\times$height

$\Rightarrow \frac{1}{2}\times AB\times BC=\frac{1}{2}\times BD\times AC$

$\Rightarrow 4\times 3=BD\times 5$

$\Rightarrow BD=\frac{12}{5}$