In figure below, if $AB\ ∥\ CD$. If $OA\ =\ 3x\ –\ 19$, $OB\ =\ x\ –\ 4$, $OC\ =\ x\ -\ 3$ and $OD\ =\ 4$, find $x$.
"

Given:

In the given figure $AB\ ∥\ CD$.

 $OA\ =\ 3x\ –\ 19$, $OB\ =\ x\ –\ 4$, $OC\ =\ x\ -\ 3$ and $OD\ =\ 4$.

To do:

We have to find the value of $x$.

Solution:

We know that,

Diagonals of a Trapezium divide each other proportionally.

Therefore,

$ \begin{array}{l}
\frac{AO}{CO} =\frac{BO}{DO}\\
\\
\frac{3x-19}{x-3} =\frac{x-4}{4}\\
\\
4( 3x-19) =( x-4)( x-3)\\
\\
4( 3x-19) =x( x-3) -4( x-3)\\
\\
12x -76=x^{2} -3x-4x+12\\
\\
x^{2} +( -3-4-12) x+( 12+76) =0\\
\\
x^{2} -19x+88=0\\
\\
x^{2} -11x-8x+88=0\\
\\
x( x-11) -8( x-11) =0\\
\\
( x-11)( x-8) =0\\
\\
x-11=0\ \ \ or\ \ x-8=0\\
\\
x=11\ \ \ \ or\ \ \ x=8
\end{array}$

Hence, the value of $x$ is $8$ or $11$.

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Updated on: 10-Oct-2022

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