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In trapezium ABCD,AB=14cm,AD=10cm,DC=xcm &distance between AB &DC is 8cm. find the value of x & area of the trapezium ABCD

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  1. Drop a perpendicular from D to AB, meeting it at P.

    Then AP = 6 (Pythagorus)

    Since we know nothing about BC, x can be anything from 0 to 8.

    The area is (x+14)/2 * 8 = 4(x+14)

    If the trapezium is isosceles, then x=2, but without some further knowledge, There's no way to pin down x.

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