Question

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

Answer 1

Given -: In quadrilateral $ABCD$ where $\angle C=90°$, $AB=9m$, $BC=12m$, $CD=5m$ and $AD=8m$. Join BD

Step 1: Find area of $∆BCD$

By applying Pythagoras theorem in $∆BCD$, we get

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

$\Rightarrow B{D}^2={12}^2+5^2$

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

$\Rightarrow B{D}^2=169$

$\Rightarrow BD=\sqrt{169}=13m$

Now, the area of $\Delta BCD$ $=\frac{1}{2}\times 12\times 5=6\times 5=30m^2$ ( area of triangle = $\frac{1}{2}\times base\times height$ )

Step 2: Find area of $\Delta ABD$

The semi perimeter of $\Delta ABD$ is $s=\frac{a+b+c}{2}$

$\Rightarrow s=\frac{9+13+8}{2}=\frac{30}{2}=15m$

Using Heron’s formula,

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

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

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

$=\sqrt{3\times 5\times 2\times 3\times 2\times 7}$

$=2\times 3\sqrt{5\times 7}$

$=6\sqrt{35}{m}^2=35.5m^2\left(approx\right)$

Step 3: Find area of quadrilateral $ABCD$

∴ The area of quadrilateral $ABCD$ $=$ Area of $∆BCD$+Area of $\Delta ABD$

$=30m^2+35.5m^2=65.5m^2$

Hence, the area of quadrilateral $ABCD$$=65.5m^2$