Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

 Solution:

First, construct a diagram with the given parameter.

Now, apply Pythagorean theorem in ΔABC,

AC2 = AB2+BC2

⇒ 52 = 32+42

⇒ 25 = 25

Thus, it can be concluded that ΔABC is a right angled at B.

So, area of ΔBCD = (½ ×3×4) = 6 cm2

The semi perimeter of ΔACD (s) = (perimeter/2) = (5+5+4)/2 cm = 14/2 cm = 7 m

Now, using Heron’s formula,

Area of ΔACD

= 2√21 cm2 = 9.17 cm2 (approximately)

Area of quadrilateral ABCD = Area of ΔABC + Area of ΔACD = 6 cm2 +9.17 cm2 = 15.17 cm2