In the given figure, ABCD is a trapezium and AB!! DC. If OD = 3cm, OC = (x – 5) cm, OB = (3-x) cm and OA = (3x – 19) cm, then which Can the following be the value of ‘x’?
Diagonals of a trapezium divide each other. proportionally. So, (AO/OC) = (BO/OD) Or {(3x – 19)/ (x – 5)} = {(3-x)/3} Or, 3 × (3x – 19) = (x – 5) (3 – x) Or 9x – 57 = 3x-x ² -15+5x Or x ² +x-42=0 x² +7x-6x-42=0 x(x+7)-6(x+7) =0 x=6, -7
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