Question

Draw quadrilateral ABCD with the following measurements. Also find its area.
AB = 4 cm, AC = 8 cm, $\angle$ ABC = ${100}^o$ , $\angle$ ABD = ${50}^o$ and $\angle$ CAD = ${40}^o$ .

Answer 1

Area ABCD = ar $△$ACD + ar $△$ACB

ar $△$ACD = $\sqrt{s(s-a)(s-b)(s-c)},s=\frac{5.8+5.1+8}{2}=\frac{18.9}{2}=9.45cm$

= $\sqrt{9.45(9.45-5.8)(9.45-5.1)(9.45-8)}=\sqrt{9.45\times 3.65\times 4.35\times 1.45}$

= $14.75{cm}^2$

ar $△$ACB = $\sqrt{s(s-a)(s-b)(s-c)},s=\frac{4+6.2+8}{2}=\frac{18.2}{2}=9.1cm$

= $\sqrt{9.1(9.1-4)(9.1-6.2)(9.1-8)}=\sqrt{9.1\times 5.1\times 2.9\times 1.1}$

= $12.20{cm}^2$

$\therefore$ ar ABCD = 14.75 + 12.20 = 26.95${cm}^2$