Question

A park, in the shape of a quadrilateral $ABCD$, has $\angle C={90}^o,AB=9m,BC=12m,CD=5m$ and $AD=8m$. How much area does it occupy?

Answer 1

Area of $△BCD=\frac{1}{2}\times 12\times 5$

$=30m^2$

Area of $△ABD=\sqrt{s(s-a)(s-b)(s-c)}$

Here, $a=8m,b=9m,c=BD$

Since $\angle C={90}^o$

So,

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

$B{D}^2=(12{)}^2+(5{)}^2$

$B{D}^2=144+25$

$BD=13m$

then,

$s=\frac{a+b+c}{2}=\frac{8+9+13}{2}=\frac{30}{2}=15m$

Area of $△ABD=\sqrt{15(15-8)(15-9)(15-13)}$

$=\sqrt{15\times 7\times 6\times 2}$

$=\sqrt{1260}$

$=35.46m^2$

Area of the park $=35.46+30$

$=65.46m^2$