$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.

96cm²

90cm²

81cm²

27cm²

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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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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.

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The diagram, MOPQ is a trapezium with QP||MO, MQ||NP, NQ||OP, |QP| = 9cm and the height of Δ\Delta QNP = 6cm, calculate the area of the trapezium.

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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.