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The diagram, MOPQ is a trapezium with QP||MO, MQ||NP, NQ||OP, |QP| = 9cm and the...


Question

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.

Options

A) 96cm2

B) 90cm2

C) 81cm2

D) 27cm2

The correct answer is C.

Explanation:

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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Dicussion (1)

  • 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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