Question

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

Answer 1

The quadrilateral ABCD is divided into two triangles ABD and BCD.

ABD is a right-angle triangle in which $\angle A$ is ${90}^{\circ }$, AD is the altitude and AB is the base.

Area of triangle ABD $=\frac{1}{2}\times AB\times AD$

$=\frac{1}{2}\times 18\times 24=216m^2$

In triangle ABD, we have

$B{D}^2=A{B}^2+A{D}^2$ [Using Pythagoras theorem]

$\Rightarrow B{D}^2={18}^2+{24}^2$

$\Rightarrow B{D}^2=900$

$\therefore BD=30m$

In triangle BCD triangle, we have

Three sides are

$BD=a=30m,BC=b=40m,CD=c=50m$

s = semi perimeter $=\frac{a+b+c}{2}=\frac{30+40+50}{2}=60m$
Using Heron's Formula, Area of triangle BCD $=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{60(60-30)(60-40)(60-50)}$

$=\sqrt{60\left(30\right)\left(20\right)\left(10\right)}$

$=\sqrt{600\times 600}$

$=600m^2$

Total area of the quadrilateral ABCD = Area of triangle ACD + Area of triangle BCD

$=216+600=816m^2$