Carcass 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
Sum of all angles in an n-sided polygon
= (n-2)(180°)If the n-sided polygon is REGULAR, each interior angle in the n-sided polygon
= \frac{(n-2)(180°)}{n}The question tells us that, in polygon X, each angle is an INTEGER.
Since each angle in a regular polygon must be
LESS THAN 180°, 179° must be the biggest possible angle in polygon X
So we can write:
\frac{(n-2)(180)}{n} = 179 [we'll solve this equation for n]Multiply both sides by
n to get:
(n-2)(180) = 179nExpand the left side to get:
180n-360 = 179nSolve:
n = 360So, when polygon X has
360 sides, each interior angle is 179° (the greatest integer value for an interior angle in a regular polygon)
Now that we've maximized the measurement of each angle in polygon X, we can focus our attention on polygon Q
We're told that the number of sides in polygon Q is 1/4 the number of sides in polygon X
So, the number of sides in polygon Q =
360/4 =
90When we plug
90 into our
formula above, we get...
Each angle in a
90-sided polygon
= \frac{(90-2)(180)}{90}= \frac{(88)(2)}{1}=176Answer: C
Cheers,
Brent