In figure below, if $AB\ ∥\ CD$, find the value of $x$.

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Given:

In the given figure, $AB||CD$.

To do:

We have to find the value of $x$.

Solution:

We know that,

Diagonals of a Trapezium divide each other proportionally.

Therefore,

$\frac{AO}{CO} =\frac{BO}{DO}$

$\frac{6x-5}{2x+1} =\frac{5x-3}{3x-1}$

$( 6x-5)( 3x-1) =( 5x-3)( 2x+1)$

$6x( 3x-1) -5( 3x-1) =5x( 2x+1) -3( 2x+1)$

$18x^{2} -6x-15x+5=10x^{2} +5x-6x-3$

$( 18-10) x^{2} +( -6-15-5+6) x+( 5+3) =0$

$8x^{2} -20x+8=0$

$4\left( 2x^{2} -5x+2\right) =0$

$2x^{2} -5x+2=0$

$2x^{2} -4x-x+2=0$

$2x( x-2) -1( x-2) =0$

$( x-2)( 2x-1) =0$

$x-2=0$ or $2x-1=0$

$x=2$ or $2x=1$

$x=2$ or $x=\frac{1}{2}$

”Š

$x=\frac{1}{2}$ is not possible because $OD=6(\frac{1}{2})-5=-2$ is not possible.

(Length cannot be negative)

Therefore,

**
**

**The value of $x$ is $2$.**