Question

$ABC$ is an isosceles triangle such that $AB=BC=8$ cm and $\angle ABC={90}^{\circ }$. What is the length of the perpendicular drawn from $B$ on $AC$?

• A. $4$ cm
• B. $4\sqrt{2}$ cm
• C. $2\sqrt{2}$ cm
• D. $2$ cm

Answer 1

The correct option is B $4\sqrt{2}$ cm

Given, $AB=BC=8$ cm, $\angle ={90}^0$

By using pythagorean theorem,

$A{C}^2=A{B}^2+B{C}^2$

$=8^2+8^2$

$=8\sqrt{2}$

Therefore, $AC=8\sqrt{2}$ cm
$\Rightarrow \frac{1}{2}\times AC\times h=8\times 8$
$\Rightarrow 8\sqrt{2}\times h=64$
$\Rightarrow h=\frac{64}{8\sqrt{2}}=\frac{8}{\sqrt{2}}=4\sqrt{2}$ cm