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

In $△ABD$

$\Rightarrow B{D}^2+4^2=(4\sqrt{5}{)}^2$

$\Rightarrow B{D}^2+16=80$

$\Rightarrow B{D}^2=64$

$\Rightarrow BD=8$

Let $CD=x;BC=y;AC=x+4$

In $△BDC$

$\Rightarrow x^2+8^2=y^2$

$\Rightarrow y^2=x^2+64$ ...$\left(1\right)$

In $△ABC$

$\Rightarrow y^2+80=(x+4{)}^2$

Substitute $\left(1\right)$ in the above equation

$\Rightarrow x^2+64+80=x^2+8x+16$

$\Rightarrow 8x=128$

$\Rightarrow x=16$

$\Rightarrow y=\sqrt{{16}^2+8^2}=8\sqrt{5}$

Therefore, $BC=8\sqrt{5}$

$\Rightarrow$Area of $△ABC=\frac{1}{2}\left(AB\right)\left(BC\right)$

$\Rightarrow$Area of $△ABC=\frac{1}{2}\left(8\sqrt{5}\right)\left(4\sqrt{5}\right)$

$\Rightarrow$Area of $△ABC=80$sq. units