Question

In the given figure, ABCD is trapezium whow diagonals AC and BD intersect at O such that `OA=(3x-1) cm, (OB=(2x+1) cm, OC=(5x-3) cm and OD =(6x-5)cm`. Then, x=?

Answer 1

We know that `AB||DC` in trapezium ABCD and its diagonal intersectst at O. Then, we have
`(AO)/(OC)=(BO)/(OD)=(3x-1)/(5x-1)=(2x+1)/(6x-5)`
`rArr (3x-1)(6x-5)=10x^(2)-x-3`
`rArr 18x^(2)-20x+8=0 rArr 2x^(2)-5x+2=0`
`rArr (x-2)(2x-1)=0`
`rArr x=2 or x=(1)/(2)`
But, `x=(1)/(2)`will make `OC=(5x-3) cm=(5xx(1)/(2)-3) cm =-(1)/(2) cm`
And the distance cannot be negative
`:. x ne(1)/(2)`
Hence, x=2