In ∆ABC, we have

AB^{2 }+ BC^{2} = 9 + 16 = 25

= AC^{2}

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 cm^{2} = 6 cm^{2}.

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) cm^{2} = 15.2 cm^{2} Ans.