Question

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

• A. $686m^2$
• B. $816m^2$
• C. $716m^2$
• D. none of the above

Answer 1

The correct option is B $816m^2$

$\angle ABD$ is a right angled $△$
Therefore,
Area of $△ABD=\frac{1}{2}\times AB\times AD$
$=\frac{1}{2}\times 18\times 24$
$=216c{m}^2$
Also,
$B{D}^2=A{B}^2+A{D}^2$
$=>B{D}^2={18}^2+{24}^2$
$=>B{D}^2=324+576$
$=>B{D}^2=900$
$=>BD=\sqrt{900}$
$=>BD=30$
Now,
In $△BDC$,
$S=\frac{BD+BC+CD}{2}$
$S=\frac{30+40+50}{2}$
$S=\frac{120}{2}$
$S=60cm$
By Herons formula,
Area of $△BDC=\sqrt{S(S-BD)(S-BC)(S-CD)}$
$=\sqrt{60(60-30)(60-40)(60-50)}$
$=\sqrt{60\left(30\right)\left(20\right)\left(10\right)}$
$=\sqrt{60\times \left(6000\right)}$
$=600c{m}^2$
Therefore,
Area of quadrilateral ABCD=Area of $△ABD+Areaof△BDC$
$=(216+600)c{m}^2$
$=816c{m}^2$