Question

Draw quadrilateral ABCD with the following measurements. Also find its area.
AB = 6 cm, BC = 6 cm, $\angle$ BAC = ${50}^o$ , $\angle$ ACD = ${30}^o$ and $\angle$ CAD = ${100}^o$ .

Answer 1

ar ABCD = ar $△$ABC + ar $△$ACD

ar $△$ABC = $\sqrt{s(s-a)(s-b)(s-c)},s=\frac{8+6+6}{2}=10cm$

= $\sqrt{10(10-8)(10-6)(10-6)}=\sqrt{10\times 2\times 4\times 4}=17.90{cm}^2$

ar $△$ACD = $\sqrt{s(s-a)(s-b)(s-c)},s=\frac{8+8.3+4.2}{2}=\frac{20.5}{2}=10.25cm$

= $\sqrt{10.25(10.25-8)(10.25-8.3)(10.25-4.2)}=\sqrt{10.25\times 2.25\times 1.95\times 6.05}$

= $16.50{cm}^2$

$\therefore$ ar ABCD = 17.90 + 16.50 = $34.40{cm}^2$