Difference between revisions of "2012 AMC 8 Problems/Problem 23"
m (→Solution 1) |
Pi is 3.14 (talk | contribs) |
||
(3 intermediate revisions by 3 users not shown) | |||
Line 7: | Line 7: | ||
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>. |
==Solution 2== | ==Solution 2== | ||
Line 15: | Line 15: | ||
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>. | ||
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 | ||
+ | |||
+ | ==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}} | {{MAA Notice}} |
Latest revision as of 21:41, 2 January 2023
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.