Question 109127

Please help me answer or solve this question: 
How many sides does a polygon have if the measure of each interior angle is 8 times the measure of each exterior angle? 
It would be very helpful if you could explain the process that you took to solve this problem. 
Thanks in advance, 
Christine 

<pre><font face = "book antiqua" size = 5><b>
The drawing below shows the bottom right-hand corner of the polygon:

{{{drawing(300, 100,-1,2,0,1, line(-1,0,1,0), 
line(0,0,.9396926208,.3429201433), locate(.4,.17,x),
locate(-.2,.17,8x), line(.9396926208,.3429201433, 1.706737064,.984807753)   )}}}  

The small acute angle marked x is an exterior angle of the
polygon and the large obtuse angle marked 8x is an interior 
angle. It measures 8x because we are told that each interior 
angle is 8 times the measure of each exterior angle. 

Since the interior and exterior angle at the same vertex
of the polygon are supplementary, we can equate their sum to
180°.

INTERIOR ANGLE + EXTERIOR ANGLE = 180°

                         8x + x = 180°
                             9x = 180°
                              x = 20°

This means each exterior angle is 20° and each interior
angle is 180°-20° or 160°. This tells us that the polygon
is a regular polygon because all the interior angles are
equal in measure. 

Now we only need to know that the theorem that says: 

The sum of all the exterior angles of any polynomial is 
always 360°.  

Let the number of sides of this polygon be N.

Since we know that this is an N-sided regular polygon, 
the sum of the exterior angles is N times 20° or 20N
degrees. So 

            20N degrees = 360 degrees
                    20N = 360
                      N = 18 sides

So it is an 18-sided regular polygon. 
                    
Edwin</pre>