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21 Jul 2017, 00:44
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C
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Difficulty:
95%
(hard)
Question Stats:
39% (02:56) correct
61%
(02:51)
wrong
based on 143
sessions
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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
A. 2
B. 160
C. 176
D. 178
E. 179
23 Jul 2017, 20:10
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Bunuel
Total angle measure of a polygon = (n-2)*180
Interior angle of a regular polygon = (n-2)*180/n
The maximum value of interior angle has to be less than 180 and since it is specified to be a integer here . Let's equate it with 179 .
(n-2)*180 / n = 179
=> 180n - 360 = 179n
=> n = 360
Polygon X has 360 sides and each angle has a measure of 179 .
Now Polygon Q has 90 sides and angle measure = 88*180 / 90
=176
Answer C
19 Nov 2020, 13:40
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Bunuel
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: \((n-2)(180) = 179n\)
Expand the left side to get: \(180n-360 = 179n\)
Solve: \(n = 360\)
So, 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 = 90
When we plug 90 into our formula above, we get...
Each angle in a 90-sided polygon \(= \frac{(90-2)(180)}{90}= \frac{(88)(2)}{1}=176\)
Answer: C
Cheers,
Brent
thanks Brent 
