Question

In a right $\Delta ABC$ with right angle at B, BD is a perpendicular dropped onto the hypotenuse. If AC = 2AB, what is the area of $\Delta ABD$?
Note: Area of $\Delta ABC$ = 5 sq units

Answer 1

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