In a trapezium ABCD, AB is parallel to DC. The diagonals AC and BD intersect at P. If AP:PC=4:(4x-4) and BP:PD=(2x-1):(2x+4), then what is the value of x?

Explanation

The diagonals of a trapezium intersect proportionally, so \frac{AP}{PC} = \frac{BP}{PD}. Thus, \frac{4}{4x-4} = \frac{2x-1}{2x+4} \implies \frac{1}{x-1} = \frac{2x-1}{2x+4}. Cross-multiplying yields 2x+4 = (x-1)(2x-1) = 2x^2-3x+1, giving 2x^2-5x-3=0. The positive root is x=3.

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