Question

Draw quadrilateral ABCD with the following measurements. Find also its area.
AB = 5.5 cm, BC = 6.5 cm, BD = 7 cm, AD = 5 cm and $\angle$ BAC = ${50}^o$ .

Answer 1

Area ABCD = ar $△$ABD + ar $△$BCD

ar $△$ABD = $\sqrt{s(s-a)(s-b)(s-c)},s=\frac{5.7+5+7}{2}=8.75$

= $\sqrt{8.75(8.75-5.5)(8.75-5)(8.75-7)}=\sqrt{8.75\times 3.25\times 3.75\times 1.75}$

= $13.66{cm}^2$

ar $△$BCD = $\sqrt{s(s-a)(s-b)(s-c)},s=\frac{7+5+6.5}{2}=9.25cm$

= $\sqrt{9.25(9.25-7)(9.25-5)(9.25-6.5)}=\sqrt{9.25\times 2.25\times 4.25\times 2.75}$

= ${15.60cm}^2$

$\therefore$ ar ABCD = 13.66 + 15.60 = 29.20${cm}^2$