Fewpal
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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.

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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. 

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