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Regular polygon X has r sides, and each vertex has...

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Regular polygon X has r sides, and each vertex has...

by BTGmoderatorLU » Sun Nov 05, 2017 4:53 pm
Regular polygon X has r sides, and each vertex has an angle measure of s, an integer. If regular polygon Q has r/4 sides, what is the greatest possible value of t, the angle measure of each vertex of polygon Q?

A. 2
B. 160
C. 176
D. 178
E. 179

The OA is C.

Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.
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by DavidG@VeritasPrep » Tue Nov 07, 2017 5:44 pm
LUANDATO wrote:Regular polygon X has r sides, and each vertex has an angle measure of s, an integer. If regular polygon Q has r/4 sides, what is the greatest possible value of t, the angle measure of each vertex of polygon Q?

A. 2
B. 160
C. 176
D. 178
E. 179

The OA is C.

Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.
If we want to find the maximum possible angle measure for the interior angles of Q, we first need to find the max possible angle measure for X, as Q will have 1/4 as many sides as X.

Equation for finding the sum of all interior angles of a polygon with 'n' sides: (n-2) * 180
Equation for finding the measure of each interior angle of a regular polygon with 'n' sides: (n-2)*180/n

If the interior angles of X are integer values, then the greatest possible measure of each angle would be 179 degrees. (180 would be a straight line, and thus not a polygon.)
To find the number of sides of a polygon with interior angles of 179, simply set the above equation to 179:
(n-2)*180/n = 179
(n-2) *180 = 179n
180n - 360 = 179n
-360 = -n
n = 360

So X can have at most 360 sides.

If Q has 1/4 the number of sides as X, then Q, at most, can have 360/4= 90 sides.
If n = 90, each interior angle would be (90-2)*180/90 = 88*180/90 = 88*2 = 176. The answer is C
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