Difference between revisions of "2012 AMC 8 Problems/Problem 23"

Revision as of 17:21, 20 January 2025

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?

$\textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}4\sqrt3\qquad\textbf{(E)}\hspace{.05in}6\sqrt3$

Solution 1

Let the perimeter of the equilateral triangle be $3s$ . The side length of the equilateral triangle would then be $s$ and the sidelength of the hexagon would be $\frac{s}{2}$ .

A hexagon contains six equilateral triangles. One of these triangles would be similar to the large equilateral triangle in the ratio $1 : 4$ , 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 $1$ . The area of the hexagon is then $1 \times 6 = \boxed{\textbf{(C)}\ 6}$ .

Solution 2 (Hard)

Let the side length of the triangle be $2x$ , so the side length of the hexagon is $x$ . By the Pythagorean theorem, the height must be $x\sqrt{3}$ . Therefore, the area is $\frac{x^2\sqrt{3}}{2}$ . We can set up the equation $4 = \frac{x^2\sqrt{3}}{2}$ . Solving for $x$ , we get $x = \frac{2}{\sqrt[4]{3}}$ . Therefore, the side length of the hexagon must be $\frac{2}{\sqrt[4]{3}}$ . Applying the Pythagorean theorem again, we get the height of one triangle of the hexagon is $\frac{\sqrt{3}}{sqrt[4]{3}}$ , so the area for one triangle is $\frac{\frac{\sqrt{3}}{\sqrt[4]{3}}\times\frac{2}{\sqrt[4]{3}}}{2}$ , which is $1$ . Therefore, the area of the hexagon is $\boxed{\textbf{(C)}\ 6}$ .

Video Solution

https://youtu.be/SctoIY1cbss ~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/j3QSD5eDpzU?t=2101

~ pi_is_3.14

@@ Line 8: / Line 8: @@
 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 (Hard)==
+Let the side length of the triangle be <math>2x</math>, so the side length of the hexagon is <math>x</math>. By the Pythagorean theorem, the height must be <math>x\sqrt{3}</math>. Therefore, the area is <math>\frac{x^2\sqrt{3}}{2}</math>. We can set up the equation <math>4 = \frac{x^2\sqrt{3}}{2}</math>. Solving for <math>x</math>, we get <math>x = \frac{2}{\sqrt[4]{3}}</math>. Therefore, the side length of the hexagon must be <math>\frac{2}{\sqrt[4]{3}}</math>. Applying the Pythagorean theorem again, we get the height of one triangle of the hexagon is <math>\frac{\sqrt{3}}{sqrt[4]{3}}</math>, so the area for one triangle is <math>\frac{\frac{\sqrt{3}}{\sqrt[4]{3}}\times\frac{2}{\sqrt[4]{3}}}{2}</math>, which is <math>1</math>. Therefore, the area of the hexagon is <math>\boxed{\textbf{(C)}\ 6}</math>.
 ==Video Solution==

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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