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= 36.16
BD = 6.013 cm

(c) Using sine rule,
AB sin 35 = 6.013 sinA 10 sin 35 = 6.013 sinA sinA= 6.013×sin 35 10 sinA=0.3449 A= 20.18 ABD= 180 35 20.18 ABD= 124.82

(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