Question

Find the area of a quadrilateral ABCD in which $AB=3cm,BC=4cm,CD=4cm,DA=5cm$ and $AC=5cm$

Answer 1

For $\Delta ABC$, we have
$A{C}^2=A{B}^2+B{C}^2$

$(5{)}^2=(3{)}^2+(4{)}^2$

Therefore,$\Delta ABC$ is a right triangle, right-angled at point B

Area of $\Delta ABC=\frac{1}{2}\times AB\times BC=\frac{1}{2}\times 3\times 4=6c{m}^2$

For $\Delta ADC,$ Perimeter $=2s=AC+CD+DA=(5+4+5)cm=14cm$

$s=7cm$

By Heron's formula, Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

Area of $\Delta ADC=\sqrt{7(7-5)(7-5)(7-4)}c{m}^2$

$=\left(\sqrt{7\times 2\times 2\times 3}\right)c{m}^2$

$=2\sqrt{21}c{m}^2$

$=9.166c{m}^2$

Area of $ABCD=$ Area of $\Delta ABC$ + Area of $\Delta ACD$

$=(6+9.166)c{m}^2$

$=15.166c{m}^2$

$=15.2c{m}^2$ (approx)