Question 135203
the sum of the measures of the interior angles of a polygon is between 2100 and 2400. how many sides does the polygon have? thanks.
<pre>
The expression for the sum of the interior angles of
an {{{n}}}-sided polygon is:

{{{(n-2)180}}}°

{{{2100 <= (n-2)180<=2400}}}

Divide all three sides by {{{60}}}

{{{2100/60 <= ((n-2)180)/60<=2400/60}}}

{{{35 <= (n-2)3<=40}}}

{{{35 <= 3(n-2)<=40}}}

{{{35 <= 3n-6<=40}}}

Add {{{6}}} to all three sides:

{{{41 <= 3n<=46}}}

Divide all three sides by {{{3}}}

{{{41/3 <= (3n)/3<=46/3}}}

{{{13}}}{{{2/3 <= n <= 15}}}{{{1/3}}}

{{{n}}} must be an integer and the only
two integers between {{{13}}}{{{2/3}}} and {{{15}}}{{{1/3}}}

are {{{14}}} and {{{15}}}.

So the polygon could either

1. have {{{14}}} sides, with the sum of the interior angles being

{{{(n-2)180}}}°

{{{(14-2)180}}}°

{{{(12)180}}}°

{{{2160}}}°

OR

2. have {{{15}}} sides, with the sum of the interior angles being

{{{(n-2)180}}}°

{{{(15-2)180}}}°

{{{(13)180}}}°

{{{2340}}}°

So there are two solutions.  The number of sides is {{{14}}} or {{{15}}}.

Edwin</pre>