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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Solution

2s = 5+4+3 ⇒ S = 6
Area of ΔABC
$= \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{6 (6 - 5) (6 - 4) (6 - 3)} = \sqrt{6 (1) (2) (3)} = \sqrt{36} = 6 c m^{2} \Rightarrow For area of △ A D C, \Rightarrow 2 s = 5 + 4 + 5 \Rightarrow S = 7 c m \Rightarrow A r e a o f △ A D C = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{7 (7 - 5) (7 - 5) (7 - 4)} = \sqrt{7 \times 2 \times 2 \times 3} \sqrt{221} = 9.16 c m^{2} \Rightarrow Area of quad ABCD=Area of (△ A B C + △ A D C)$
= 6 + 9.16
$= 15.16 c m^{2} \approx 15.2 c m^{2}$

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