Solution of Triangles, SPM Paper 2 (Long Question)


Question 5:
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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