Difference between revisions of "2012 AMC 8 Problems/Problem 23"
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Setting this equal to <math>4</math> gives us <math>\dfrac{4y^2\sqrt{3}}{4}=4\implies 4y^2\sqrt{3}=16\implies y^2\sqrt{3}=4</math>. | Setting this equal to <math>4</math> gives us <math>\dfrac{4y^2\sqrt{3}}{4}=4\implies 4y^2\sqrt{3}=16\implies y^2\sqrt{3}=4</math>. | ||
− | + | Substitute <math>y^2\sqrt{3}</math> into the area of a regular hexagon to yield <math>\dfrac{3(4)}{2}=6</math>. | |
Therefore, our answer is <math>\boxed{\textbf{(C)}\ 6}</math>. | Therefore, our answer is <math>\boxed{\textbf{(C)}\ 6}</math>. |
Revision as of 23:13, 23 July 2020
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
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.