ABCD is the park as shown in the figure.

Join BD.

In ∆DBC, we have

DB2 = BC^{2} + CD^{2} [Pythagoras theorem]

⇒ DB2 = (12)^{2} + 5^{2}

⇒ DB = 144 + 25 = 169

⇒ DB = 13 m.

Area of ∆DBC = 1/ 2 × base × height

= 1/ 2 × 12 × 5 m^{2} = 30 m^{2}

In ∆ABD, a = 9 m, b = 8 m, c = 13 m

∴ s = (a +b + c) /2 = (9+ 8+ 13) / 2 m = 15 m

∴ Area of the park = area of ∆DBC + area of ∆ABD

= (30 + 35.5) m^{2} = **65.5 m**^{2} Ans.