Difference between revisions of "2012 AMC 10B Problems/Problem 14"
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==Solution== | ==Solution== | ||
| − | + | <center><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></center> | ||
| − | 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: | + | 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: |
| − | <math>2\times\frac{\sqrt{3}}{4}(2\sqrt{3}-2)^2 = 8\sqrt{3} - 12</math> | + | <math>2\times\frac{\sqrt{3}}{4}(2\sqrt{3}-2)^2 = \boxed{\mathbf{(D)} 8\sqrt{3} - 12}.</math> |
| − | + | ==Solution 2== | |
| + | 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> | ||
| + | |||
| + | ==Video Solution by Math4All999== | ||
| + | https://youtu.be/zAu6aCkxO_g?feature=shared | ||
==See Also== | ==See Also== | ||
Latest revision as of 08:14, 15 December 2023
Problem
Two equilateral triangles are contained in square whose side length is 
. 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?
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]](http://latex.artofproblemsolving.com/1/b/4/1b491bea6ff8007acc151d99ff17f5f0ab905857.png)
Observe that the rhombus is made up of two congruent equilateral triangles with side length equal to GF. Since AE has length 
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 
. The formula for the area of an equilateral triangle of length 
is 
. It follows that the area of the rhombus is:
Solution 2
We know that both of these triangles are congruent with base length 
. Using 30-60-90 triangle rules, we find the height of both of these triangles to be 
. Since the height of the square is , we can see that 
(overlap)
. Solving this we get that the overlap is 
. 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 
. Because the triangles are 30-60-90, we can find the base to be 
. We use the area of a triangle formula to get the area of one of these four triangles to be 
Lastly, since there were four of these triangles in the rhombus, we do 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
