In ∆ABC, we have
AB2 + BC2 = 9 + 16 = 25
= AC2
Hence, ABC is a right triangle, right angled at B
[By converse of Pythagoras theorem]
∴ Area of ∆ABC = 1/ 2 × base × height
= 1/ 2 × 3 × 4 cm2 = 6 cm2.
In ∆ACD, a = 5 cm, b = 4 cm, c = 5 cm.
∴ s = a +b + c+2 = 5+ 4+5 / 2 cm = 7cm
Area of the quadrilateral = area of ∆ABC + area of ∆ACD
= (6 + 9.2) cm2 = 15.2 cm2 Ans.
