# In trapezium ABCD, if BC = 3cm then AD = ?Given...

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$$(\triangle ADO)$$.

• What else is given? I have got $$d^2=\frac{17}{3}BO^2+3-AO^2$$ Nov 9 '19 at 7:42
• Nothing else is given...sorry.But, how did you get the equation? Nov 10 '19 at 8:24

I have used the equations $$6AO\times BO=CO\times D=$$ $$AO\times DO=CO\times BO$$ $$d^2=AO^2+DO^2-2AO\times DO\times \cos(\pi-\delta)$$ $$9=BO^2+CO^2-2BO\times CO\times \cos(\pi-\delta)$$

• Okay thanks. I think that the question was incomplete to start with :-( Nov 11 '19 at 11:53

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 $$AD$$ = $$\sqrt{3^2+(\sqrt{6}-1)^2}\simeq3.332.$$
• By providing two cases where the length of $AD$ differs I have shown that there is insufficent information to determine $AD$ uniquely. Nov 10 '19 at 15:23