In the given figure, $A B C D$ is trapezium in which the parallel sides AB and CD are perpendicular to BC. If $A B = 16 m, C D = 8 m$ and $A D = 17 m$ , what is the area of the trapezium?

140 m^{2}

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168 m^{2}

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180 m^{2}

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156.4 m^{2}

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Solution

The correct option is C $180 m^{2}$
$D C = E B = 8 c m$
Since $A B = 16 c m, A E = A B - E B = 16 - 8 = 8 m$
In $△ A E D$ ,
${D E}^{2} = {A D}^{2} - {A E}^{2} = 17^{2} - 8^{2} = 289 - 64 = 225$
$∴ D E = 15 m$
Area of the trapezium $A B C D = \frac{1}{2} \times (A B + D C) \times D E$
$= \frac{1}{2} \times (16 m + 8 m) \times 15 = 180 m^{2}$

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In the given figure, ABCD is trapezium in which the parallel sides AB and CD are perpendicular to BC. If AB=16m,CD=8m and AD=17m, what is the area of the trapezium?

The correct option is C 180m2DC=EB=8cm Since AB=16cm,AE=AB−EB=16−8=8m In △AED, DE2=AD2−AE2=172−82=289−64=225∴DE=15m Area of the trapezium ABCD=12×(AB+DC)×DE =12×(16m+8m)×15=180m2

In the given figure, $A B C D$ is trapezium in which the parallel sides AB and CD are perpendicular to BC. If $A B = 16 m, C D = 8 m$ and $A D = 17 m$ , what is the area of the trapezium?