Solution of Triangles Long Questions (Question 1)


Question 1:

The diagram shows a quadrilateral ABCD. The area of triangle BCD is 12 cm2 and BCD is acute. Calculate
(a) ∠BCD,
(b) the length, in cm, of BD,
(c) ABD,
(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
BCD = 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,
A B sin 35 = 6.013 sin A 10 sin 35 = 6.013 sin A sin A = 6.013 × sin 35 10 sin A = 0.3449 A = 20.18 A B D = 180 35 20.18 A B D = 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

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