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avatar+3994 

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
 #1
avatar+113 
+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
avatar+3994 
0

sorry, but it's wrong...

 Dec 28, 2017
 #3
avatar+7350 
+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
avatar+98107 
+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

 

 

 

cool cool cool

 Dec 28, 2017
edited by CPhill  Dec 28, 2017

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