A park, in the shape of a quadrilateral ABCD, has C = 90, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m How much area does it occupy? in answer how bd=13m comes?

The quadrilateral given is shown below;

Considering right triangle BCD, we have;
$$\begin{gathered} {\mathrm{BD}}^2 = {\mathrm{BC}}^2+{\mathrm{CD}}^2 \\ \Rightarrow \mathrm{BD} = \sqrt{{12}^2+5^2} = \sqrt{169} = 13 \mathrm{m} \end{gathered}$$
And area of triangle BCD = $\frac{1}{2}\times \mathrm{BC}\times \mathrm{CD} = \frac{1}{2}\times 12\times 5 = 30 {\mathrm{m}}^2$
Now perimeter of triangle ABD = AB + BD + AD = 9 m + 13 m + 8 m = 30 m
so, semi perimeter of triangle ABD;s = $\frac{30}{2} = 15 \mathrm{m}$
So using heron's formula we have;
Area of triangle ABD = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} = \sqrt{15\left(15-9\right)\left(15-13\right)\left(15-8\right)} = \sqrt{1260} = 35.496 {\mathrm{m}}^2$
Therefore area of quadrilateral ABCD = area of triangle ABD + area of triangle BCD
$$\begin{gathered} = 30 {\mathrm{m}}^2+35.496 {\mathrm{m}}^2 \\ = 65.496 {\mathrm{m}}^2 \end{gathered}$$