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 .