Difference between revisions of "2012 AMC 8 Problems/Problem 23"
m (→Solution 2) |
|||
Line 33: | Line 33: | ||
The area of a regular hexagon with side length <math>s</math> is <math>\dfrac{3s^2\sqrt{3}}{2}</math>. | The area of a regular hexagon with side length <math>s</math> is <math>\dfrac{3s^2\sqrt{3}}{2}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/SctoIY1cbss ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2012|num-b=22|num-a=24}} | {{AMC8 box|year=2012|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 09:37, 18 May 2022
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
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.