# Solution of Triangles Long Questions (Question 1)

Question 1:

The diagram shows a quadrilateral ABCD. The area of triangle BCD is 12 cm2 and ÐBCD is acute. Calculate
(a) ÐBCD,
(b) the length, in cm, of BD,
(c) ÐABD,
(d) the area, in cm2, quadrilateral ABCD.

Solution:
(a) Given area of triangle BCD = 12 cm2
½ (BC)(CD) sin C = 12
½ (7) (4) sin C = 12
14 sin C = 12
sin C = 12/14 = 0.8571
C = 59o
ÐBCD = 59o

(b) Using cosine rule,
BD2 = BC2 + CD2 – 2 (7)(4) cos 59o
BD2 = 72 + 42 – 2 (7)(4) cos 59o
BD2 = 65 – 28.84
BD2 = 36.16
$BD=\sqrt{36.16}$
BD = 6.013 cm

(c) Using sine rule,
$\begin{array}{l}\frac{AB}{\mathrm{sin}{35}^{\circ }}=\frac{6.013}{\mathrm{sin}A}\\ \frac{10}{\mathrm{sin}{35}^{\circ }}=\frac{6.013}{\mathrm{sin}A}\\ \mathrm{sin}A=\frac{6.013×\mathrm{sin}{35}^{\circ }}{10}\\ \mathrm{sin}A=0.3449\\ A={20.18}^{\circ }\\ \angle ABD={180}^{\circ }-{35}^{\circ }-{20.18}^{\circ }\\ \angle ABD={124.82}^{\circ }\end{array}$

(d) Area of quadrilateral ABCD
= Area of triangle ABD + Area of triangle BCD
= ½ (AB)(BD) sin B + 12 cm
= ½ (10) (6.013) sin 124.82 + 12
= 24.68 + 12
= 36.68 cm²