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- Jan 26, 2012

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Welcome to MHB!

If you have a polygon with $n$ sides then each angle can be expressed as \(\displaystyle \frac{(n-2) \times 180^\circ}{n}\). Can you use this formula and the given information to solve for $n$?

Jameson

- Jan 26, 2012

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The implication of the question is that the polygon is not necessarily regular (possibly not even convex) - though given the nature of the question the average of the interior angles of an n-gon is probably an invariant.

Welcome to MHB!

If you have a polygon with $n$ sides then each angle can be expressed as \(\displaystyle \frac{(n-2) \times 180^\circ}{n}\). Can you use this formula and the given information to solve for $n$?

Jameson

In fact it is trivial to show that your formula is the average interior angle for an arbitrary convex n-gon, so that is all-right then!

CB

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- Jan 26, 2012

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I don't see that but trust that you know this better than I do. From the level of the other thread the OP made here it seems more likely to me that this is a straightforward question, but I should consider irregular polygons in the future.The implication of the question is that the polygon is not necessarily regular (possibly not even convex)

CB

EDIT: Ah I think I see your point now. The word "average" could definite imply that the polygon isn't regular although I think it might just be badly worded.

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However, when I saw the term "average measure" for the interior angles, I assumed then a convex polygon, not necessarily regular, as the same number of sides results either way.