Question

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.

Answer 1

In $△$ABC
2s = 5+4+3 ⇒ s = 6 cm
Area of ΔABC
$=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{6(6-5)(6-4)(6-3)}=\sqrt{6\left(1\right)\left(2\right)\left(3\right)}=\sqrt{36}=6c{m}^2\Rightarrow \text{For area of}△ADC,\Rightarrow 2s=5+4+5\Rightarrow s=7cm\Rightarrow Areaof△ADC=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{7(7-5)(7-5)(7-4)}=\sqrt{7\times 2\times 2\times 3}\sqrt{84}=9.16c{m}^2\Rightarrow \text{Area of quad ABCD=Area of}(△ABC+△ADC)$
$=6+9.16$
$=15.16c{m}^2\approx 15.2c{m}^2$