Question

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

Answer 1

Given: $AB=9m,BC=12m,CD=5m,DA=8m$

In $\Delta BCD,$

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

$B{D}^2=(12{)}^2+(5{)}^2=144+25=169$

$BD=13m$

Area of $\Delta BCD=\frac{1}{2}\times BC\times CD$

$=\frac{1}{2}\times 12\times 5$

$=30m^2$

Now, perimeter of $\Delta ABD=2s=9m+8m+13m$

$2s=30m$

$s=15m$

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

$=\sqrt{15(15-9)(15-8)(15-13)}$

$=\sqrt{15\times 6\times 7\times 2}=6\sqrt{35}=35.496m^2$

Area of quadrilateral $ABCD$

$=$ Area of $\Delta ABD+$ Area of $\Delta BCD$

$=(35.496+30)m^2$

$=65.5m^2$

Hence, the required area is $65.5m^2$.