Question : ABCD is a trapezium where AD$\parallel$ BC. The diagonal AC and BD intersect each other at the point O. If AO = 3, CO = $x-3$, BO = $3x-19$, and DO = $x-5$, the value of $x$ is:
Option 1: -8, 9
Option 2: 8, -9
Option 3: -8, -9
Option 4: 8, 9
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Correct Answer: 8, 9
Solution :
Given, trapezium ABCD with AD$\parallel$ BC
Given, AO = 3, CO = $x-3$, BO = $3x-19$ and DO = $x-5$
Consider $\triangle$ ADO and $\triangle$ CBO,
$\angle$ AOD = $\angle$ COB (vertically opposite angles)
$\angle$ OAD = $\angle$ OCB (alternate interior angles)
$\angle$ ODA = $\angle$ OBC (alternate interior angles)
So, $\triangle$ ADO$\sim$$\triangle$ CBO.
So, $\frac{\text{AO}}{\text{DO}}=\frac{\text{CO}}{\text{BO}}$
Or, $\frac{3}{x-5}=\frac{x-3}{3x-19}$
Or, $9x-57=x^{2}-8x+15$
Or, $x^{2}-17x+72=0$
Or, $(x-9)(x-8)=0$
Or, $x=8,9$
Hence, the correct answer is 8, 9.
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