+4622 Let \(n\) equal the number of sides in a regular polygon. For \(3\leq n < 10\) , how many values of \(n\) result in a regular polygon where the common degree measure of the interior angles is not an integer?
+115 Use the formula for solving the sum of degrees of the interior angles, then divide by \(n\) to find out only one of the interior angles.
Use this formula: \(\frac{(n-2)\times180}{n}\)
I'll do the first one:
\(n=3\)
\(\frac{(3-2)*180}{2}=\frac{180}{2}=90\)
90 is an integer, so 3 will not work.
Keep on doing this until you get to 9.
+9479 Let f(n) be the degree measure of an interior angle in the regular polygon.
f(n) = [ 180(n - 2) ] / n
f(3) = [ 180(3 - 2) ] / 3 = 180 / 3 = 60
f(4) = [ 180(4 - 2) ] / 4 = 360 / 4 = 90
f(5) = [ 180(5 - 2) ] / 5 = 540 / 5 = 108
f(6) = [ 180(6 - 2) ] / 6 = 720 / 6 = 120
f(7) = [ 180(7 - 2) ] / 7 = 900 / 7 ≈ 128.57
f(8) = [ 180(8 - 2) ] / 8 = 1080 / 8 = 135
f(9) = [ 180(9 - 2) ] / 9 = 1260 / 9 = 140
So....only f(7) is not an integer. Only one value of n results in a regular polygon where the common degree measure of the interior angles is not an integer.
+129852 Thanks, hectictar....
Note that in the "formula" provided by MIRB, 3, 4, 5 , 6, 9 and 10 all divide 180, while ( n - 2) will always be an integer
So...we only need to check the values n = 7 and n = 8
( 7 - 2) * 180 / 7 = (5/7)*180 ≈ 128.57°
(8 - 2) * 180 / 8 = (3/4)* 180 = 135°
So.....a regular heptagon [ 7 sides ] will not have integer-valued interior angle measures
