Difference between revisions of "2012 AMC 8 Problems/Problem 23"
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==Problem== | ==Problem== | ||
| − | An equilateral triangle and a regular hexagon have equal perimeters. If the area | + | An equilateral triangle and a regular hexagon have equal perimeters. If the triangle's area is 4, what is the area of the hexagon? |
<math> \textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}4\sqrt3\qquad\textbf{(E)}\hspace{.05in}6\sqrt3 </math> | <math> \textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}4\sqrt3\qquad\textbf{(E)}\hspace{.05in}6\sqrt3 </math> | ||
| − | ==Solution== | + | ==Solution 1== |
Let the perimeter of the equilateral triangle be <math> 3s </math>. The side length of the equilateral triangle would then be <math> s </math> and the sidelength of the hexagon would be <math> \frac{s}{2} </math>. | Let the perimeter of the equilateral triangle be <math> 3s </math>. The side length of the equilateral triangle would then be <math> s </math> and the sidelength of the hexagon would be <math> \frac{s}{2} </math>. | ||
A hexagon contains six equilateral triangles. One of these triangles would be similar to the large equilateral triangle in the ratio <math> 1 : 4 </math>, 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 <math> 1 </math>. The area of the hexagon is then <math> 1 \times 6 = \boxed{\textbf{(C)}\ 6} </math>. | A hexagon contains six equilateral triangles. One of these triangles would be similar to the large equilateral triangle in the ratio <math> 1 : 4 </math>, 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 <math> 1 </math>. The area of the hexagon is then <math> 1 \times 6 = \boxed{\textbf{(C)}\ 6} </math>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/SctoIY1cbss ~savannahsolver | ||
| + | |||
| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/j3QSD5eDpzU?t=2101 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
==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}} | ||
Latest revision as of 09:21, 16 July 2024
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 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
