In the adjoining figure, ABCD is a trapezium in which CD || AB and its diagonals intersect at O. If AO = (5x – 7) cm, OC = (2x + 1) cm, DO = (7x – 5) cm and OB = (7x + 1) cm, find the value of x.

In the trapezium ABCD, AB || DC.

Also, AC and BD intersect at O.


Thus,




(5x – 7)(7x + 1) = (7x – 5)(2x + 1)


35x2 + 5x – 49x – 7 = 14x2 – 10x + 7x – 5


35x2 – 44x – 7 = 14x2 – 3x – 5


35x2 – 14x2 – 44x + 3x – 7 + 5 = 0


21x2 – 41x – 2 = 0


21x2 – 42x + x – 2 = 0


21x(x – 2) + (x – 2) = 0


(21x + 1)(x – 2) = 0


(21x + 1) = 0 or (x – 2) = 0


x = -1/21 or x = 2


But x = -1/21 doesn’t satisfy the length of intersected lines.


So x ≠ -1/21


And thus, x = 2.


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