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In trapezium $ABCD$, $AB || CD$. Diagonals $AC$ and $BD$ meet at $O$. Area$(\triangle ABO)$ : area$(\triangle CDO) = 1:6$. If $BC = 3$cm, find $AD$.

I know that $\triangle ABO$ and $\triangle CDO$ are similar and the ratio of their corresponding sides is $1:\sqrt6$. Also,
area$(\triangle BCO) =$ area$(\triangle ADO)$.

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    $\begingroup$ What else is given? I have got $$d^2=\frac{17}{3}BO^2+3-AO^2$$ $\endgroup$ Nov 9 '19 at 7:42
  • $\begingroup$ Nothing else is given...sorry.But, how did you get the equation? $\endgroup$
    – Sapula1
    Nov 10 '19 at 8:24
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I have used the equations $$6AO\times BO=CO\times D=$$ $$AO\times DO=CO\times BO$$ $$d^2=AO^2+DO^2-2AO\times DO\times \cos(\pi-\delta)$$ $$9=BO^2+CO^2-2BO\times CO\times \cos(\pi-\delta)$$

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  • $\begingroup$ Okay thanks. I think that the question was incomplete to start with :-( $\endgroup$
    – Sapula1
    Nov 11 '19 at 11:53
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There isn't a single answer.

For example: assume $AB = 1$, so that $CD=\sqrt6$.

Then, if we assume $ABCD$ is symmetrical this gives $AD=3$, but if instead we assume, say, that $\angle BCD$ is a right angle, then $AD$ = $\sqrt{3^2+(\sqrt{6}-1)^2}\simeq3.332.$

enter image description here

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  • $\begingroup$ Nice, but how can that be an answer? $\endgroup$ Nov 10 '19 at 11:14
  • $\begingroup$ These are only special cases. $\endgroup$ Nov 10 '19 at 11:14
  • $\begingroup$ By providing two cases where the length of $AD$ differs I have shown that there is insufficent information to determine $AD$ uniquely. $\endgroup$
    – nickgard
    Nov 10 '19 at 15:23

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