2008 CEMC Gauss (Grade 8) Problems/Problem 2

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 $108 \text{cm}$ and each side has length $12 \text{cm}$ . How many sides does this polygon have?

$\text{ (A) }\ 6 \qquad\text{ (B) }\ 7 \qquad\text{ (C) }\ 8 \qquad\text{ (D) }\ 9 \qquad\text{ (E) }\ 10$

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 $s$ is the number of side lengths of the polygon:

$12s = 108$

$s = \frac{108}{12} = \boxed {\textbf {(D) } 9}$

~anabel.disher

Solution 2 (answer choices)

We can multiply the answer choices by $12$ to see if we get $108$ or not, and see if the result is too high or too low.

$8 \times 12 = 96$ , which is too low (it is smaller than $108$ ), meaning answer choices A, B, and C cannot be the answer.

$10 \times 12 = 120$ , 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 $\boxed {\textbf {(D) } 9}$ .

~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)

Page

Toolbox

Search

2008 CEMC Gauss (Grade 8) Problems/Problem 2

Problem

Solution 1

Solution 2 (answer choices)