Question

In $△ABC,m\angle B={90}^o,AB=4\sqrt{5}$. $BD\perp AC,AD=4$, then $A\left(△ABC\right)=$?

• A. $96$ sq. units
• B. $80$ sq. units
• C. $120$ sq. units
• D. $160$ sq. units

Answer 1

Answer: (B)

$80$ sq. units

Given : $m\angle B={90}^o$, $BD\perp AC$. So, $△ADB={90}^o$

$\therefore$ $BD=\sqrt{{\left(4\sqrt{5}\right)}^2-{\left(4\right)}^2}$
$=\sqrt{80-16}=\sqrt{64}$$=8$

$\therefore$ $BD=8$.
Also $\frac{1}{{BD}^2}=\frac{1}{{AB}^2}+\frac{1}{{BC}^2}$
$\frac{1}{{BC}^2}=\frac{1}{{BD}^2}-\frac{1}{{AB}^2}$
$\frac{1}{{BC}^2}=\frac{{AB}^2-{BD}^2}{{BD}^2.{AB}^2}$
$BC=\sqrt{\frac{{BD}^2.{AB}^2}{{AB}^2-{BD}^2}}$
$BC=\frac{BD.AB}{\sqrt{{AB}^2-{BD}^2}}=\frac{8\times 4\sqrt{5}}{4}=8\sqrt{5}$
$\therefore$ Area of $△ABC=\frac{1}{2}\times 4\sqrt{5}\times 8\sqrt{5}=80$ sq.units.