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

In quadrilateral ABCD, join BD.
In triangle BDC, by Pythagoras theorem
BD² BC² + CD²
⇒ BD² = 12² + 5² = 144 + 25 = 169
⇒ BD = 13 cm
Area of triangle BDC = 1/2 × BC × DC = 1/2 × 12 × 5 = 30 cm²
Here, the sides of triangle ABD are a = 9 cm, b = 8 cm and c = 13 cm.
So, the semi-perimeter of triangle ABD s = ABD s = a+b+c/2 = 9+8+13/2 = 30/2 = 15 cm
Therefore, using Heron’s formula area of triangle ABD = √s(s -a)(s -b)(s -c)
= √15(15 -9)(15 -8)(15 -13)
= √15(6)(7)(2) = √1260 = 35.5 cm²(approx.)
Total area of park = 30 + 3535 = 65.5 cm²
Hence, total area of park is 65.5 cm².