Question

Find the areas of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, AD = 5 cm and AC = 5cm and AC = 5 cm.

Answer 1

Area of $\Delta ABC=\sqrt{s(s-a)(s-b)(s-c)}$

$a=3$ $b=4$ $c=5$

$\Rightarrow \sqrt{6(6-3)(6-4)(6-5)}=\sqrt{6\times 6}=6sq.cm.$

Area of $\Delta ADC$

$=\sqrt{7(7-5)(7-5)(7-4)}=\sqrt{7\times 2\times 2\times 3}$

$=2\sqrt{21}$

$=2\times 4.58$

$=9.16sqcm$

Area of quadrilateral $=$ area of $\Delta ABC+\Delta ADC$

$=6+9.16$

$=15.16sqcm$