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.

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.

A.

96cm2

B.

90cm2

C.

81cm2

D.

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