Question

Find the area of a quadrilateral ABCD in which AB = 3cm, BC = 4cm, CD = 4cm, DA = 5cm and AC = 5cm

Answer 1

In the quadrilateral, AC is the diagonal which divides the figure into two triangles
Now in $\Delta$ ABC, AB = 3 cm, BC = 4cm, AC = 5cm

$\therefore s=\frac{a+b+c}{2}=\frac{3+4+5}{2}=\frac{12}{2}=6$

$\therefore Areaof\Delta ABC=\sqrt{s(s-a)(s-b)(s-c)}$

$v=\sqrt{6(6-3)(6-4)(6-5)}$

$=\sqrt{6\times 3\times 2\times 1}=\sqrt{36}=6c{m}^2$

In $\Delta$ ADC, AC= 5 cm, AD= 5cm, CD=4cm

$\therefore s=\frac{a+b+c}{2}=\frac{5+5+4}{2}=\frac{14}{2}=7$

$\therefore Areaof\Delta ADC=\sqrt{s(s-a)(s-b)(s-c)}$

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

$=2\sqrt{21}=2\times 4.58=9.16c{m}^2$

$\therefore$ Total area of quadrilateral ABCD

= $6+9.16c{m}^2=15.16c{m}^2$