Difference between revisions of "1999 CEMC Gauss (Grade 8) Problems/Problem 18"
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<math> \text{ (A) }\ 15 \qquad\text{ (B) }\ 30 \qquad\text{ (C) }\ 90 \qquad\text{ (D) }\ 45 \qquad\text{ (E) }\ 60 </math> | <math> \text{ (A) }\ 15 \qquad\text{ (B) }\ 30 \qquad\text{ (C) }\ 90 \qquad\text{ (D) }\ 45 \qquad\text{ (E) }\ 60 </math> | ||
==Solution== | ==Solution== | ||
Latest revision as of 15:06, 18 June 2025
Problem
The equilateral triangle has sides of 
and 
as shown.
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
The perimeter of the triangle is
Solution
By the definition of an equilateral triangle, all sides of the triangle must be equal to each other. This means that we have:
Subtracting 
from both sides, we get:
The perimeter of an equilateral triangle is 
times its side length, so the perimeter is
~anabel.disher
