The Total solution for NCERT class 6-12
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-angledat point B.
For ΔADC,
Perimeter = 2s =AC + CD + DA = (5 + 4 + 5) cm = 14 cm
s = 7 cm
By Heron’s formula,
Area ofABCD = Area of ΔABC + Area of ΔACD
= (6 + 9.166) cm2 =15.166 cm2 = 15.2 cm2 (approximately)