We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

# HELP

0
508
4

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?

Dec 27, 2017

### 4+0 Answers

#1
+3

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.

Dec 27, 2017
#2
0

sorry, but it's wrong...

Dec 28, 2017
#3
+1

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.

Dec 28, 2017
#4
+3

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   Dec 28, 2017
edited by CPhill  Dec 28, 2017