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Difference between revisions of "2012 AMC 10B Problems/Problem 14"
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== Problem ==
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Two equilateral triangles are contained in square whose side length is <math>2\sqrt 3</math>. The bases of these triangles are the opposite side of the square, and their intersection is a rhombus. What is the area of the rhombus?
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<math>
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\text{(A) } \frac{3}{2}
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\qquad
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\text{(B) } \sqrt 3
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\qquad
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\text{(C) } 2\sqrt 2 -  1
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\qquad
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\text{(D) } 8\sqrt 3 - 12
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\qquad
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\text{(E)}  \frac{4\sqrt 3}{3}
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</math>
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[[Category: Introductory Geometry Problems]]
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==Solution==
 
==Solution==
−
[[File:2012_AMC-10B-14.jpg‎]]
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<center><asy>
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size(8cm);
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pair A, B, C, D, E, F, G, H, BF, AF;
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A = (0,0);
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B = (1,0);
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C = (1,1);
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D = (0,1);
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E = (1/2,0);
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H = (1/2,1);
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G = (1/2,1/2^(1/2));
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F = (1/2,1-(1/2^(1/2)));
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AF = (3^(1/2)/2,1/2);
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BF = (1-3^(1/2)/2,1/2);
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draw(A--B--C--D--A--AF--D);
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draw(C--BF--B);
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draw(H--E,linetype("8 8"));
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label("$A$",A,SW);
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label("$B$",B,SE);
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label("$C$",C,NE);
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label("$D$",D,NW);
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label("$E$",E,S);
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label("$H$",H,N);
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label("$G$",G+1/20,E);
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label("$F$",F-1/20,W);
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</asy></center>
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 +
Observe that the rhombus is made up of two congruent equilateral triangles with side length equal to GF. Since AE has length <math>\sqrt{3}</math> and triangle AEF is a 30-60-90 triangle, it follows that EF has length 1. By symmetry, HG also has length 1. Thus GF has length <math>2\sqrt{3} - 2</math>. The formula for the area of an equilateral triangle of length <math>s</math> is <math>\frac{\sqrt{3}}{4}s^2</math>. It follows that the area of the rhombus is:
 +
 
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<math>2\times\frac{\sqrt{3}}{4}(2\sqrt{3}-2)^2 = \boxed{\mathbf{(D)} 8\sqrt{3} - 12}.</math>
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==Solution 2==
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We know that both of these triangles are congruent with base length <math>2\sqrt{3}</math>. Using 30-60-90 triangle rules, we find the height of both of these triangles to be <math>3</math>. Since the height of the square is <math>2\sqrt{3}</math>, we can see that <math>3+3-</math>(overlap)<math>=2\sqrt{3}</math>. Solving this we get that the overlap is <math>6-2\sqrt{3}</math>. This overlap is one of the diagonals of the rhombus. Now, it is convenient to understand that the rhombus is made up of 4 congruent 30-60-90 triangles. The height of one of these triangles is half the diagonal that we have found already. Dividing by two, we get the height of one triangle <math>= 3-\sqrt{3}</math>. Because the triangles are 30-60-90, we can find the base to be <math>\sqrt{3}-1</math>. We use the area of a triangle formula to get the area of one of these four triangles to be <math>\dfrac{(3-\sqrt{3})(\sqrt{3}-1)}{2}=\dfrac{4\sqrt{3}-6}{2}=2\sqrt{3}-3</math> Lastly, since there were four of these triangles in the rhombus, we do <math>2\sqrt{3}\times 4 = \boxed{\mathbf{(D)} 8\sqrt{3} - 12}.</math>
  
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Observe that the rhombus is made up of two congruent equilateral triangles with side length equal to GF. Since AE has length <math>\sqrt{3}</math> and triangle AEF is a 30-60-90 triangle, it follows that EF has length 1. By symmetry, HG also has length 1. Thus GF has length <math>2\sqrt{3} - 2</math>. The formula for the area of an equilateral triangle of length s is <math>\frac{\sqrt{3}}{4}s^2</math>. It follows that the area of the rhombus is:
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==Video Solution by Math4All999==
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https://youtu.be/zAu6aCkxO_g?feature=shared
  
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<math>2\times\frac{\sqrt{3}}{4}(2\sqrt{3}-2)^2 = 8\sqrt{3} - 12</math>
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==See Also==
  
−
Thus, answer choice D is correct.
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{{AMC10 box|year=2012|ab=B|num-b=13|num-a=15}}
 +
{{MAA Notice}}

Latest revision as of 08:14, 15 December 2023

Problem

Two equilateral triangles are contained in square whose side length is $2\sqrt 3$. The bases of these triangles are the opposite side of the square, and their intersection is a rhombus. What is the area of the rhombus?

$\text{(A) } \frac{3}{2} \qquad \text{(B) } \sqrt 3 \qquad \text{(C) } 2\sqrt 2 - 1 \qquad \text{(D) } 8\sqrt 3 - 12 \qquad \text{(E)} \frac{4\sqrt 3}{3}$

Solution

[asy] size(8cm); pair A, B, C, D, E, F, G, H, BF, AF; A = (0,0); B = (1,0); C = (1,1); D = (0,1); E = (1/2,0); H = (1/2,1); G = (1/2,1/2^(1/2)); F = (1/2,1-(1/2^(1/2))); AF = (3^(1/2)/2,1/2); BF = (1-3^(1/2)/2,1/2); draw(A--B--C--D--A--AF--D); draw(C--BF--B); draw(H--E,linetype("8 8")); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",E,S); label("$H$",H,N); label("$G$",G+1/20,E); label("$F$",F-1/20,W); [/asy]

Observe that the rhombus is made up of two congruent equilateral triangles with side length equal to GF. Since AE has length $\sqrt{3}$ and triangle AEF is a 30-60-90 triangle, it follows that EF has length 1. By symmetry, HG also has length 1. Thus GF has length $2\sqrt{3} - 2$. The formula for the area of an equilateral triangle of length $s$ is $\frac{\sqrt{3}}{4}s^2$. It follows that the area of the rhombus is:

$2\times\frac{\sqrt{3}}{4}(2\sqrt{3}-2)^2 = \boxed{\mathbf{(D)} 8\sqrt{3} - 12}.$

Solution 2

We know that both of these triangles are congruent with base length $2\sqrt{3}$. Using 30-60-90 triangle rules, we find the height of both of these triangles to be $3$. Since the height of the square is $2\sqrt{3}$, we can see that $3+3-$(overlap)$=2\sqrt{3}$. Solving this we get that the overlap is $6-2\sqrt{3}$. This overlap is one of the diagonals of the rhombus. Now, it is convenient to understand that the rhombus is made up of 4 congruent 30-60-90 triangles. The height of one of these triangles is half the diagonal that we have found already. Dividing by two, we get the height of one triangle $= 3-\sqrt{3}$. Because the triangles are 30-60-90, we can find the base to be $\sqrt{3}-1$. We use the area of a triangle formula to get the area of one of these four triangles to be $\dfrac{(3-\sqrt{3})(\sqrt{3}-1)}{2}=\dfrac{4\sqrt{3}-6}{2}=2\sqrt{3}-3$ Lastly, since there were four of these triangles in the rhombus, we do $2\sqrt{3}\times 4 = \boxed{\mathbf{(D)} 8\sqrt{3} - 12}.$

Video Solution by Math4All999

https://youtu.be/zAu6aCkxO_g?feature=shared

See Also

2012 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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 AMC 10 Problems and Solutions

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