2012 AMC 8 Problems/Problem 23
Contents
Problem
An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is 4, what is the area of the hexagon?
Solution 1
Let the perimeter of the equilateral triangle be . The side length of the equilateral triangle would then be
and the sidelength of the hexagon would be
.
A hexagon contains six equilateral triangles. One of these triangles would be similar to the large equilateral triangle in the ratio , since the sidelength of the small equilateral triangle is half the sidelength of the large one. Thus, the area of one of the small equilateral triangles is
. The area of the hexagon is then
.
Solution 2
Let the side length of the equilateral triangle be and the side length of the hexagon be
. Since the perimeters are equal, we must have
which reduces to
. Substitute this value in to the area of an equilateral triangle to yield
.
Setting this equal to gives us
.
Substitue into the area of a regular hexagon to yield
.
Therefore, our answer is .
Notes
The area of an equilateral triangle with side length is
.
The area of a regular hexagon with side length is
.
See Also
2012 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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