Difference between revisions of "2008 CEMC Gauss (Grade 8) Problems/Problem 2"
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Latest revision as of 13:30, 20 October 2025
- The following problem is from both the 2008 CEMC Gauss (Grade 8) #2 and 2008 CEMC Gauss (Grade 7) #4, so both problems redirect to this page.
Problem
A regular polygon has perimeter 
and each side has length 
. How many sides does this polygon have?
Solution 1
The perimeter of a shape is the shape's side lengths added together. Since a regular polygon's side lengths are all equal to each other, the perimeter of a regular polygon is its side length multiplied by the number of sides that the polygon has.
This means that we can set up an equation, where 
is the number of side lengths of the polygon:
~anabel.disher
Solution 2 (answer choices)
We can multiply the answer choices by 
to see if we get 
or not, and see if the result is too high or too low.
, which is too low (it is smaller than ), meaning answer choices A, B, and C cannot be the answer.
, which is too high, meaning that answer choice E cannot be the answer.
All of the other answer choices have been eliminated, so the answer is 
.
~anabel.disher
| 2008 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| CEMC Gauss (Grade 8) | ||
| 2008 CEMC Gauss (Grade 7) (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| CEMC Gauss (Grade 7) | ||
