A park, in the shape of a quadrilateral ABCD,

Solution:

Let us join BD.

In $\triangle B C D$, applying Pythagoras theorem,

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

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

$=144+25$

$B D^{2}=169$

$B D=13 \mathrm{~m}$



Area of $\triangle B C D=\frac{1}{2} \times B C \times C D=\left(\frac{1}{2} \times 12 \times 5\right) \mathrm{m}^{2}=30 \mathrm{~m}^{2}$

For $\triangle \mathrm{ABD}$,

$s=\frac{\text { Perimeter }}{2}=\frac{(9+8+13) \mathrm{m}}{2}=15 \mathrm{~m}$

By Heron's formula,

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

Area of $\triangle \mathrm{ABD}=[\sqrt{15(15-9)(15-8)(15-13)}] \mathrm{m}^{2}$

$=(\sqrt{15 \times 6 \times 7 \times 2}) \mathrm{m}^{2}$

$=6 \sqrt{35} \mathrm{~m}^{2}$

$=(6 \times 5.916) \mathrm{m}^{2}$

$=35.496 \mathrm{~m}^{2}$

Area of the park $=$ Area of $\triangle \mathrm{ABD}+$ Area of $\triangle \mathrm{BCD}$

$=35.496+30 \mathrm{~m}^{2}=65.496 \mathrm{~m}^{2}=65.5 \mathrm{~m}^{2}$ (approximately)