Question

A park is in the shape of quadrilateral $ABCD$ as shown. Given that angle $C={90}^o$ Also, $AB=9m,BC=12m,CD=5m$ and $AD=8m$. Using Pythagoras Theorem find the length of diagonal $BD$. Also, find the total area of the given quadrilateral by using Heron's Formula.

Answer 1

In $△BCD$, using Pythagoras theorem,

${BD}^2={BC}^2+{CD}^2$

$\Rightarrow {BD}^2={\left(12\right)}^2+{\left(5\right)}^2$

$\Rightarrow BD=\sqrt{144+25}=\sqrt{169}=13m$

Therefore,

Area of $△BCD=\frac{1}{2}\times 12\times 5=30m^2$

Now,

In $△ABD$,

Semi-perimeter $\left(s\right)=\frac{8+9+13}{2}=15m$

Using Heron's formula, Area of $△ABD=\sqrt{s(s-a)(s-b)(s-c)}$

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

$\Rightarrow$ Area of $△ABD=\sqrt{15\times 7\times 6\times 2}=6\sqrt{35}{m}^2$

Thereforoe,

Area of quadrilateral $ABCD=$ Area of $△ABD+$ Area of $△BCD$

Therefore,

Area of quadrilateral $ABCD=30+6\sqrt{35}=6(5+\sqrt{35})m^2$