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?

Solution:

First, construct a quadrilateral ABCD and join BD.

We know that

C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m

The diagram is:

Now, apply Pythagoras theorem in ΔBCD

BD2 = BC2 +CD2

⇒ BD2 = 122+52

⇒ BD2 = 169

⇒ BD = 13 m

Now, the area of ΔBCD = (½ ×12×5) = 30 m2

The semi perimeter of ΔABD

(s) = (perimeter/2)

= (8+9+13)/2 m

= 30/2 m = 15 m

Using Heron’s formula,

Area of ΔABD

= 6√35 m2 = 35.5 m2 (approximately)

∴ The area of quadrilateral ABCD = Area of ΔBCD+Area of ΔABD

= 30 m2+35.5m2 = 65.5 m2