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Question
In the given figure, ABCD is a trapezium in which AB║DC and its diagonals intersect at O. If AO = (5x – 7), OC = (2x + 1) , BO = (7x – 5) and OD = (7x + 1), find the value of x.
Solution
In trapezium ABCD, AB ‖ CD and the diagonals AC and BD intersect at O. Therefore,
`(AO)/(OC)=(BO)/(OD)`
⇒ `(5x-7)/(2x+1)=(7x-5)/(7x+1)`
⟹ (5x-7) (7x+1) = (7x-5) (2x+1)
⟹`35x^2+5x-49x-7=14x^2-10x+7x-5`
⟹ `21x^2-41x-2=0`
⟹`21x^2-42x+x-2=0`
⟹`21x(x-2)+1(x-2)=0`
⟹`(x-2) (2x+1)=0`
⟹ `x=2,1/21`
∵ `x ≠-1/21`
∴ x=2
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