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?

Join BD in ΔBCD, BC and DC are given. So, we can calculate BD by applying Pythagoras theorem ⇒ BD = √BC² + CD² = √12² + 5² = √144 + 25 = 13 m = BD ⇒ Area of ▱ABCD = Area of ΔABD + Area of ΔBCD ⇒ Area of ΔBCD = 1/2 × b × h = 1/2 × 12 × 5 = 30 m² ⇒ Area of ΔABD = √s(s-a)(s-b)(s-c) (Heron's formula) ⇒ 2S = 9 + 8 + 13, S = 30/2 ⇒ S = 15 m ⇒ Area of ΔABD = √15(15-9)(15-8)(15-13) = √15 × 6 × 7 × 2 = √1260 = 6√35 ≈ 35.49 m² ⇒ Area of Park = Quad ABCD = 30 + 35.49 = 65.49 m² ≈ 65.5 m²