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If the area of $1000$ sided polygon is $314$ sq. cms. How to find the maximum length (in cm) between any two vertices of this polygon?

$ a)\space 15.7 \quad \quad b)\space 18.1 \quad \quad c)\space 20 \quad\quad d) \space 21.6$

I am not sure how to solve this for a general polygon.Any ideas?

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    For the circle the answer is $20$.however I couldn't follow the rest,if the question means regular polygon and as this have even number of edge then the longest diagonal is the same as the diameter of the circumscribed circle which I think is same,what exactly I am missing?2011-10-08

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Just for the reason,that this question is still unanswered,I am posting this official solution from here,which is very close to Matt Bennet's hint in the comments above.

It can be seen that, the higher the number of sides of a regular polygon, the more closely does its area approach to that of its circum-circle.

In this case, we have a polygon of $1000$ sides and its area will be very close to that of the circle of radius $r$.

To find $r$, we put,

$πr² = 314$ cm$^²$

$\Rightarrow r ≈ 10$ cm

Now, vertices $1$ and $501$ of our $1000$ sided polygon will correspond to the opposite ends of the diameter of the circum-circle of this polygon.

∴ The distance between them will be approximately = $2 × r = 20$ cm

Hence, option $3$.