Question

A field is in the shape of a quadrilateral ABCD in which side AB =18 m, side AD=24 m, side BC =40 m, DC =50 m and angle $A={90}^{\circ }$. Find the area of the field.

Answer 1

Since $\angle A={90}^{\circ }$

$\therefore$ By Pythagorus Theorem,

In $\Delta ABD,$

$BD=\sqrt{A{B}^2+A{D}^2}=\sqrt{{18}^2+{24}^2}$

$=\sqrt{324+576}=\sqrt{900}=30m.$

Now, area of $\Delta ABD=\frac{1}{2}\left(18\right)\left(24\right)$

$=\left(18\right)\left(12\right)=216m^2$

Again in $\Delta BCD$; sides are 30,40,50

$\Rightarrow$ By Pythagoras Theorem $\angle CBD={90}^{\circ }$

$[\therefore D{C}^2=B{D}^2+B{C}^2$, Since $(50{)}^2=(30{)}^2+\left(40{)}^2\right]$

$\therefore$ Area of $\Delta BCD=\frac{1}{2}\left(40\right)\left(30\right)=600m^2$

Hence, area of quadrilateral ABCD =Area of $\Delta ABD$+area of $\Delta BCD$

$=216+600=816m^2.$