Question

In a right triangle $\Delta$ABC, right angled at B, BD is a perpendicular dropped onto the hypotenuse AC.

If AC = 2AB, find the area of $\Delta$ABD, given, area of $\Delta$ABC = 5 sq units.

Answer 1

Given: $\frac{AB}{AC}=\frac{1}{2}$

and Ar$\left(\Delta ABC\right)=5sq.units$

In $\Delta ABD$ and $\Delta ABC,$

$\angle BAD=\angle BAC$
(common angle in both the triangles)

$\angle BDA=\angle ABC={90}^{\circ }$

So, $\Delta ABD\sim \Delta ACB$ (by AA similarity criterion) [1 Mark]

$\Rightarrow \frac{Ar\left(\Delta ABD\right)}{Ar\left(\Delta ACB\right)}={\left(\frac{AB}{AC}\right)}^2={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$ [1 Mark]

$\Rightarrow Ar\left(\Delta ABD\right)=\frac{Ar\left(\Delta ACB\right)}{4}$

$\Rightarrow Ar\left(\Delta ABD\right)=\frac{5}{4}=1.25sq.units$

Hence, the area of $\Delta ABD$ is 1.25 sq units. [1 Mark]