Solution of Triangles Long Questions (Question 1 & 2)


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


Question 2:
In the diagram below, ABCis a triangle. AGJB, AHC and BKC are straight lines. The straight line JK is perpendicular to BC.


It is given that BG= 40cm, GA = 33 cm, AH = 30 cm, GAH = 85o and JBK= 45o.
(a) Calculate the length, in cm of
  i.      GH
    ii.   HC
(b) The area of triangle GAH is twice the area of triangle JBK. Calculate the length, in cm, 
  of BK.
(c) Sketch triangle  which has a different shape from triangle ABC such that, A’B’ = AB
 A’C’ = AC and A’B’C’ = ABC.

Solution:
(a)(i)
Using cosine rule,
GH2 = AG2 + AH2 – 2 (AG)(AH) GAH
GH2= 332+ 302 – 2 (33)(30) cos 85o
GH2 = 1089 + 900 – 172.57
GH2 = 1816.43
GH = 42.62 cm

(a)(ii)
A C D = 180 45 85 = 50 Using sine rule, A C sin 45 = 73 sin 50 A C = 73 × sin 45 sin 50 A C = 67.38  cm H C = 67.38 30 = 37.38  cm

(b)
Area of  Δ   G A H = 1 2 ( 33 ) ( 30 ) sin 85 = 493.12  cm 2 Let length of  B K = J K = x ×  Area of  Δ   J B K  = Area of  Δ   G A H 2 × [ 1 2 ( x ) ( x ) ] = 493.12 x 2 = 493.12 x = 22.21  cm B K = 22.21  cm

(c)

Leave a Reply

Your email address will not be published. Required fields are marked *