General Mathematics Paper 2, May/June. 2010
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 Main
Weakness/Remedies
Strength

Question 11

In the diagram, L.PTQ = L.PSR = 900, /PQ/ = 10 ern, /PS/ = 14.4 cm and /TQ/ = 6 cm.
Calculate the area of quadrilateral QRST.
(b) Two opposite sides of a square are each decreased by 10% while the other two are each increased by 15% to form a rectangle. Find the ratio of the area of the rectangle to that of the square.

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Observation

This question was reported not to be very popular and candidates' performance was described as not satisfactory. Majority of the candidates were reported not to apply the concept of similar triangles correctly. Others did not recognize the quadrilateral as a trapezium and so failed to use the correct formula when finding its area. Part (b) of the question was reported not to have been done satisfactorily either. Candidates were expected to show that:
/PT/ = ..../102 - 62 = 8 cm. m: =!J2L i.e ~ = 14.4.Hence, /SR/ = 10.8 cm.
/TO/ /SR/                6 /SR/
Area of quadrilateral QRST = Yz (6 + 10.8) x 6.4 = 53.76 cm2. Some candidates were reported to have subtracted the area of triangle PQT from triangle PRS. This was also in order. In part (b) if the side of the square was y, then new breadth = 90 x y = 0.9y.
100
New length = 115 x Y = 1.15y. New area = 1.15y x 0.9y = 1.035/.
100

Hence, ratio = 1.035y² : y² = 1.035 : 1 or 207:200 .