The diagram, MOPQ is a trapezium with QP||MO, MQ||NP, NQ||OP, |QP| = 9cm and the height of \(\Delta\) QNP = 6cm, calculate the area of the trapezium.
96cm2
90cm2
81cm2
27cm2
Correct answer is C
Area of \(\Delta\) QNP = \(\frac{1}{2} \times 9 \times 6 \) = 27cm2
Area of \(\Delta\) QMN = Area of \(\Delta\) QNP
= Area of \(\Delta\) PNO (triangles between the same parallels)
Hence, area of the trapezium
3 x area of \(\Delta\) QNP
= 3 x 27
= 81cm2
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