Question

in the given figure ABCD is a trapezium with AB||CD . If AO = x-1 , CO = BO = x - 1 and OD = x + 4 , find the value of x​

Answer 1

Step-by-step explanation:

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Answer 2

Given: In the given figure ABCD is a trapezium with $AB||CD$. If $AO = x-1 , CO = BO = x - 1$ and $OD = x + 4$.

To find: the value of x.

Solution:

Know that, In trapezium ABCD, AB ‖ CD and the diagonals AC and BD intersect at O.  Therefore,

Find the value of x.

$\[\frac{{AO}}{{CO}} = \frac{{BO}}{{OD}}\]$

$\[ \Rightarrow \frac{{x - 1}}{{x - 1}} = \frac{{x - 1}}{{x + 4}}\]$

$\[\begin{array}{l} \Rightarrow \left( {x - 1} \right)\left( {x + 4} \right) = \left( {x - 1} \right)\left( {x - 1} \right)\\ \Rightarrow {x^2} + 4x - x - 4 = {x^2} - x - x + 1\\ \Rightarrow 3x - 4 = - 2x + 1\end{array}\]$

$\[\begin{array}{l} \Rightarrow 3x + 2x = 1 + 4\\ \Rightarrow 5x = 5\\ \Rightarrow x = 1\end{array}\]$

Hence, the value of x is 1.