Question

A park in the shape of a quadrilateral $ABCD$ has $C={90}^0$. $AB=18m$, $BC=24m$, $CD=10m$ and $AD=16m$. how much area does it occupy?

Answer 1

Area of $△BCD=\frac{1}{2}\times 24\times 10=120m^2$

Use pythagoras theorem in $\Delta CBD$

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

${10}^2+{24}^2=D{B}^2$

$676=D{B}^2$

$⟹DB=26$m

Area of $△ABD\Rightarrow$ (By heron's)

$S=\frac{26+18+16}{2}=30$

$A=\sqrt{30(30-26)(30-18)(30-16)}$

$=141.98\approx 142m^2$

$\therefore$ Total area $=142+120=262m^2$