SOLUTION: Wesley measured all the interior angles in a polygon.he added them up to 991 degrees but he had missed out one angle. What type of polygon did Wesley measure? What is the size of t

Algebra ->  Polygons -> SOLUTION: Wesley measured all the interior angles in a polygon.he added them up to 991 degrees but he had missed out one angle. What type of polygon did Wesley measure? What is the size of t      Log On


   



Question 925315: Wesley measured all the interior angles in a polygon.he added them up to 991 degrees but he had missed out one angle. What type of polygon did Wesley measure? What is the size of the missing angle?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The n angles of a polygon with n sides add up to %28n-2%29%2A180%5Eo (see NOTE below).
So, for n=7 , the sum of the angle measures is 5%2A180%5Eo=900%5Eo ,
for n=8 , the sum of the angle measures is 6%2A180%5Eo=1080%5Eo , and
for n=9 , the sum of the angle measures is 7%2A180%5Eo=1260%5Eo .
The missing angle could measure
1080%5Eo-991%5Eo=89%5Eo if it was a polygon with 8 sides,
and that is a very reasonable solution.
Wesley measured a polygon with 8 sides, an highlight%28octagon%29 .

The numbers do not work out for less sides or more sides.
For n%3C=7 , the sum of the angle measures would be at most 900%5Eo , which is less than 991%5Eo .
For n%3E=9 , the missing angle would have to measure at least 1260%5Eo-991%5Eo=269%5Eo ,
and that would make it a convex polygon, something your teacher does not expect:

NOTE: A triangle has 3 sides, and 3 interior angles whose measures add up to 180%5Eo : drawing%28300%2C300%2C-1.5%2C1.5%2C-2%2C1%2Ctriangle%28-1%2C0%2C1%2C0%2C0%2C-1.732%29%29 .
Attaching another triangle with a common side you gain 1 extra side, and get a quadrilateral, like the kite below:
, with 3%2B1=4 sides, and 3%2B1=4 angles whose measures add up to 180%5Eo%2B180%5Eo=2%2A180%5Eo .
Repeating the process we get a pentagon,
with 3%2B2=5 sides, and 3%2B2=5 angles whose measures add up to 180%5Eo%2B2%2A180%5Eo=3%2A180%5Eo ,
and repeating the process once more we get a hexagon,
with 3%2B3=6 sides, and 3%2B3=6 angles whose measures add up to 180%5Eo%2B3%2A180%5Eo=4%2A180%5Eo :
and Following the pattern, the n angles of a polygon with n sides add up to %28n-2%29%2A180%5Eo

This is not a proof, but I thought it was a good way to remember the formula.