Solution of Triangles Long Questions (Question 3 & 4)


Question 3:


The diagram shows a trapezium PQRS. PS is parallel to QR and QRS is obtuse. Find
(a) the length, in cm, of QS,
(b) the length, in cm, of RS,
(c) QRS,
(d) the area, in cm2, of triangle QRS.

Solution:
(a)
Q S sin P = P S sin Q Q S sin 85 = 13.1 sin 28 Q S = 13.1 × sin 85 sin 28 Q S = 27.8  cm

(b)
 RQS = 180o – 85o – 28o
 RQS = 67o
Using cosine rule,
RS2 = QR2 + QS2 – 2 (QR)(QS) ÐRQS
RS2 = 6.42 + 27.82 – 2 (6.4)(27.8) cos 67o
RS2 = 813.8 – 139.04
RS2 = 674.76
RS = 25.98 cm

(c)
Using cosine rule, Q S 2 = Q R 2 + R S 2 2 ( Q R ) ( R S ) cos Q R S 27.8 2 = 6.4 2 + 25.98 2 2 ( 6.4 ) ( 25.98 ) cos Q R S 772.84 = 715.92 332.54 cos Q R S cos Q R S = 715.92 772.84 332.54 cos Q R S = 0.1712 Q R S = 99.86

(d)
Area of triangle QRS
= ½ (QR)(RS) sin R
= ½ (6.4) (25.98) sin 99.86o
= 81.91 cm2


Question 4:
Diagram below shows a quadrilateral PQRS.



(a) Find
(i) the length, in cm, of QS.
(ii) ∠QRS.
(iii) the area, in cm2, of the quadrilateral PQRS.
(b)(i) Sketch a triangle S’Q’R’ which has a different shape from triangle SQR such that S’R’ = SR, S’Q’ = SQ and ∠S’Q’R’ = ∠SQR.
(ii) Hence, state ∠S’R’Q’.

Solution:
(a)(i)
P=1807634=70 QS sin70 = 8 sin34 QS= 8×sin70 sin34  =13.44 cm

(a)(ii)
13.44 2 = 6 2 + 9 2 2( 6 )( 9 )cosQRS 108cosQRS= 6 2 + 9 2 13.44 2 cosQRS= 6 2 + 9 2 13.44 2 108  QRS= cos 1 ( 0.5892 )    = 126 o 6

(a)(iii)
Area of PQRS =Area of PQS+Area of QRS =( 1 2 ×8×13.44×sin76 )+( 1 2 ×6×9×sin 126 o 6 ) =52.16+21.82 =73.98  cm 2

(b)(i)



(b)(ii)
SRQ=SRR    =180 126 o 6    = 53 o 54

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