Question

Draw quadrilateral ABCD with the following measurements. Also find its area.
AB = 7 cm, BC = 5 cm, AC = 6 cm, CD= 4 cm, and $\angle$ ACD = ${45}^o$ .

Answer 1

Area of ABCD = ar $△$ABC + ar $△$ACD

ar $△$ABC = $\sqrt{s(s-a)(s-b)(s-c)},s=\frac{7+6+5}{2}=9cm$

= $\sqrt{9(9-7)(9-6)(9-5)}=\sqrt{9\times 2\times 3\times 4}=14.70{cm}^2$

ar $△$ACD = $\sqrt{s(s-a)(s-b)(s-b)},s=\frac{4+6+4.2}{2}=7.1cm$

= $\sqrt{7.1(7.1-4)(7.1-6)(7.1-4.2)}=\sqrt{7.1\times 3.1\times 1.1\times 2.9}$

= $840{cm}^2$

$\therefore$ ar ABCD = 14.70 + 8.40 = 23.1${cm}^2$