# Given the area of $1000$ sided polygon ,how could we find the maximum length between any two of its vertices?

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?

$\lt$Problem Source $\gt$

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Are you sure there is no other constraint? To show that the possible length is unbounded take two parallel lines arbitrarily close together and a cap of 499 very short edges at either end. You can keep all the angles the same if need be, but the parallel lines can always be adjusted in length to give the area you want. –  Mark Bennet Oct 7 '11 at 19:25
What you can say though is that the longest side $s$ is at least ${\pi\over 50}$ cm. This comes from the isoperimetric inequality $A\leq{L^2\over 4\pi}$ and $L\leq 1000 s$. –  Christian Blatter Oct 7 '11 at 19:33
@Christian: strangely, what attracted me to this problem was one that I read yesterday from Henry Dudeney about a farmer who had panels of fence of length 1 unit, and sheep which needed 1 square unit of grazing. How many fence panels does the farmer need to enclose an area sufficient for 10 sheep? And I was left pondering the isoperimetric inequality you cite ... it's small world! –  Mark Bennet Oct 7 '11 at 20:16
@Mark Bennet:I have given all the information now :) –  Quixotic Oct 7 '11 at 20:33
@FoolForMath: Hint - The original does not state regular, but undoubtedly means it. A 1000-sided polygon is nearly a circle. If it were a circle, what would the answer be (back of an envelope calculation). Could the answer to that approximation be so different that another answer would be possible? So the answer is ... [a] my first answer - as large as you like [b] regular, one of the above (you work it out) –  Mark Bennet Oct 7 '11 at 20:55

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 $3$.

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