Question

$\text{A park, in the shape of a quadrilateral ABCD, has}\angle {\text{C =90}}^{\circ }\text{, AB = 9m, BC = 12m, CD = 5m}$

$\text{and AD = 8m}$ $\text{.}$ $\text{How much area does it occupy ?}$

Answer 1

Area of quad $ABCD=Ar\left(△BCD\right)+Ar\left(△ABD\right)$

$BD=\sqrt{(12{)}^2+(5{)}^2}$ [By Pythagoras theorem]

$=\sqrt{144-25}$

$Ar\left(△BCD\right)=\frac{1}{2}\times base\times height=\frac{1}{2}\times 5\times 12$

$Ar\left(△BCD\right)=30m^2$

For $ar\left(△ABD\right)$ ; Semi perimeter $=\frac{9+8+13}{2}=\frac{30}{2}=15$

Applying theorem's formula $=\sqrt{5(5-a)(5-b)(5-c)}$

$=\sqrt{15(15-3)(15-13)(15-8)}$

$=\sqrt{1260}$

$ar\left(△ABD\right)=35.4m^2$

Area of quad $ABCD=(30+35.4)m^2$

$=65.4m^2$