Question

$ABCD$ is a quadrilateral in which diagonal $BD$ is perpendicular to the side $AB$. If $AB=4.5cm$, $AD=7.5cm$ and the distance of $C$ from $BD$ is $1.5cm$, then find the area of quadrilateral.

Answer 1

Consider the quadrilateral $ABCD$, such that diagonal $BD$ is perpendicular to side $AB$ and $CM$ is the perpendicular to $BD$ from $C.$

Also, we are given that $AB=4.5$cm;$AD=7.5$cm and $CM=1.5$cm

Now, the triangle $ABD$ is a right angled triangle, by pythagoras theorem:

${AB}^2+{BD}^2={AD}^2$

$\Rightarrow {BD}^2={AD}^2-{AB}^2$

$={\left(7.5\right)}^2-{\left(4.5\right)}^2=56.25-20.25=36=6^2$

Area of $△ABD$

$A_1=\frac{1}{2}\times BD\times AB$ since area$=\frac{1}{2}\times$base$\times$altitude.

$\Rightarrow A_1=\frac{1}{2}\times 6\times 4.5=13.5{cm}^2$

Area of $△BCD$

$A_2=\frac{1}{2}\times BD\times CM$ since area$=\frac{1}{2}\times$base$\times$altitude.

$\Rightarrow A_2=\frac{1}{2}\times 6\times 1.5=4.5{cm}^2$

Now, area of the quadrilateral $ABCD$

$=$Area of $△ABD+$Area of $△BCD$

$=A_1+A_2=(13.5+4.5){cm}^2=18{cm}^2$