Find the area of the above trapezium.

88 c m^{2}

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82 c m^{2}

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78 c m^{2}

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68 c m^{2}

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Solution

The correct option is A $88 c m^{2}$
$C X$ is $⊥$ drawn on $A B$ .
$C X = A B$
In $△ B X C$
$=> B C^{2} = B X^{2} + C X^{2}$
$=> 10^{2} = 6^{2} + C X^{2}$
$=> C X^{2} = 100 - 36$
$=> C X^{2} = 64$
$=> C X = 8 c m$
Area of trapezium $A B C D =$ Area of $△ B X C +$ Area of $△ X C D$
$= \frac{1}{2} \times B X \times C X + A X^{2}$
$= \frac{1}{2} \times 6 \times 8 + 8^{2}$
$= 24 + 64$
$= 88 c m^{2}$

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