2012 AMC 8 Problems/Problem 23
Contents
Problem
An equilateral triangle and a regular hexagon have equal perimeters. If the triangle's area 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 .
Substitute into the area of a regular hexagon to yield .
Therefore, our answer is .
Solution 3
Let the side length of the triangle be and the side length of the hexagon be . As explained in Solution 1, , or . The area of the triangle is and the area of the hexagon is . Substituting in for , we get .
Notes
The area of an equilateral triangle with side length is .
The area of a regular hexagon with side length is .
Video Solution
https://youtu.be/SctoIY1cbss ~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/j3QSD5eDpzU?t=2101
~ pi_is_3.14
See Also
2012 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 | ||
All AJHSME/AMC 8 Problems and Solutions |
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