Question

In a $△ABC$, $\angle A={90}^0$, $AB=5$ cm and $AC=12$ cm. If $AD\perp BC$, then the length of $AD$ is:

• A. $\frac{13}{2}$ cm
• B. $\frac{2\sqrt{15}}{13}$ cm
• C. $\frac{60}{13}$ cm
• D. $\frac{13}{60}$ cm

Answer 1

The correct option is C $\frac{60}{13}$ cm

Given, $\angle A={90}^{\circ }$
By Pythagoras theorem,
$B{C}^2=A{B}^2+A{C}^2$
$B{C}^2=5^2+{12}^2$
$BC=13$ cm
Area of triangle = $\frac{1}{2}\times base\times height$

Thus, $\frac{1}{2}AB\times AC=\frac{1}{2}AD\times BC$

$5\times 12=13\times AD$

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