Question

The sides of a quadrilateral taken in order are $9m,40m,28m$ and $15m$. If the angle between the first two sides is a right angle, find its area.

Answer 1

Let, $AB=9m,BC=40m,CD=28m$ and $DA=15m$

Given that, $△ABC$ is a right angled triangle,

So, applying pythagoras theorem in $△ABC$

$A{C}^2=A{B}^2+B{C}^2$

$\Rightarrow A{C}^2=9^2+{40}^2$

$\Rightarrow A{C}^2=81+1600$

$\Rightarrow A{C}^2=1681$

$\Rightarrow AC=41$

Length of $AC=41m$

Area of $△ABC=\frac{1}{2}\times AB\times BC$

$=\frac{1}{2}\times 9\times 40m^2$

$=180m^2$

In $△ADC$,

Semi-perimeter $s=\frac{AD+CD+AC}{2}$

$=\frac{15+28+41}{2}m=42m$

Area of $△ADC=\sqrt{s(s-15)(s-28)(s-41)}$

$=\sqrt{42(42-15)(42-28)(42-41)}{m}^2$

$=\sqrt{42\times 27\times 14\times 1}{m}^2$

$=126m^2$

Area of $ABCD=Areaof△ABC+Areaof△ADC=(180+126)m^2=306m^2$

Hence, the required area is $306m^2$