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.

For ΔABC,

AC2 = AB2 + BC2

(5)2 = (3)2 + (4)2

Therefore, ΔABC is a right-angled triangle, right-angled at point B.

Area of ΔABC

For ΔADC,

Perimeter = 2s = AC + CD + DA = (5 + 4 + 5) cm = 14 cm

s = 7 cm

By Heron’s formula,

Area of triangle

Area of ABCD = Area of ΔABC + Area of ΔACD

= (6 + 9.166) cm2 = 15.166 cm2 = 15.2 cm2 (approximately)

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