Question

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

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

Answer 1

The correct option is C 1.25 sq. units


$Given,\frac{AB}{AC}=\frac{1}{2}$
$Ar\left(\Delta ABC\right)=5sq.units$

Consider $\Delta ADBand\Delta ABC$
$\angle BAD=\angle BAC$ (common angle)
$\angle BDA=\angle ABC={90}^{\circ }$

By AA similarity criterion,
$\Delta ADB\sim \Delta ABC$

We know, in two similar triangles, the ratio of their areas is the square of the ratio of their corresponding sides.

Hence,
$\frac{Ar\left(\Delta ADB\right)}{Ar\left(\Delta ABC\right)}={\left(\frac{AB}{AC}\right)}^2={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$

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

Hence, the area of $\Delta ABD$ is
1.25 sq. units.