Question

In a right triangle $△$ABC right angled at B, BD is a perpendicular, dropped onto the hypotenuse. If $AC=2AB$ and area of $△$ABC $=5$ sq. units, what is the area of $△$ABD?

• A. $1$ sq. units
• B. $5$ sq. units
• C. $0.5$ sq. units
• D. $1.25$ sq. units

Answer 1

The correct option is D

$1.25$ sq. units

Given, $\frac{AB}{AC}=\frac{1}{2}$ and Ar$△ABC=5$ sq.units
Consider $△ABD$ and $△ABC$
$\angle BAD=\angle BAC$ [commom angle]
$\angle BDA=\angle ABC$ [${90}^{\circ }$]
By AA similarity criterion, $△ABD\sim △ACB$
Hence, $\frac{\text{Ar}\left(△ABD\right)}{\text{Ar}\left(△ACB\right)}={\left(\frac{AB}{AC}\right)}^2={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$
i.e., $\text{Ar}\left(△ABD\right)=\frac{1}{4}\times \text{Ar}\left(△ACB\right)$
$\Rightarrow \text{Ar}\left(△ABD\right)=\frac{5}{4}=1.25$ sq.units
Hence, the area of $△ABD$ is $1.25$ sq. units