Difference between revisions of "2013 AMC 10B Problems/Problem 15"
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<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{\sqrt{6}}{2}\qquad\textbf{(C)}\ \sqrt{3} \qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{3\sqrt{2}}{2} </math> | <math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{\sqrt{6}}{2}\qquad\textbf{(C)}\ \sqrt{3} \qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{3\sqrt{2}}{2} </math> | ||
| − | ==Solution== | + | ==Solution 1== |
Using the formulas for area of a regular triangle <math>(\frac{{s}^{2}\sqrt{3}}{4})</math> and regular hexagon <math>(\frac{3{s}^{2}\sqrt{3}}{2})</math> and plugging <math>\frac{a}{3}</math> and <math>\frac{b}{6}</math> into each equation, you find that <math>\frac{{a}^{2}\sqrt{3}}{36}=\frac{{b}^{2}\sqrt{3}}{24}</math>. Simplifying this, you get <math>\frac{a}{b}=\boxed{\textbf{(B)} \frac{\sqrt{6}}{2}}</math> | Using the formulas for area of a regular triangle <math>(\frac{{s}^{2}\sqrt{3}}{4})</math> and regular hexagon <math>(\frac{3{s}^{2}\sqrt{3}}{2})</math> and plugging <math>\frac{a}{3}</math> and <math>\frac{b}{6}</math> into each equation, you find that <math>\frac{{a}^{2}\sqrt{3}}{36}=\frac{{b}^{2}\sqrt{3}}{24}</math>. Simplifying this, you get <math>\frac{a}{b}=\boxed{\textbf{(B)} \frac{\sqrt{6}}{2}}</math> | ||
| + | |||
| + | ==Solution 2== | ||
| + | The regular hexagon can be broken into 6 small equilateral triangles, each of which is similar to the big equilateral triangle. Each of the small equilateral triangle's area is <math>\frac{1}{6}</math>th of the area of the big equilateral triangle. Therefore each side of the big triangle is <math>\sqrt{6}</math> times the side of the small triangle. The desired ratio is $\frac{3\sqrt{6}}{6}=\frac{\sqrt{6}}{2}\Rightarrow\Box{B}. | ||
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== See also == | == See also == | ||
{{AMC10 box|year=2013|ab=B|num-b=14|num-a=16}} | {{AMC10 box|year=2013|ab=B|num-b=14|num-a=16}} | ||
Revision as of 16:23, 25 December 2013
Contents
Problem
A wire is cut into two pieces, one of length 
and the other of length 
. The piece of length is bent to form an equilateral triangle, and the piece of length is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is 
?
Solution 1
Using the formulas for area of a regular triangle 
and regular hexagon 
and plugging 
and 
into each equation, you find that 
. Simplifying this, you get 
Solution 2
The regular hexagon can be broken into 6 small equilateral triangles, each of which is similar to the big equilateral triangle. Each of the small equilateral triangle's area is 
th of the area of the big equilateral triangle. Therefore each side of the big triangle is 
times the side of the small triangle. The desired ratio is $\frac{3\sqrt{6}}{6}=\frac{\sqrt{6}}{2}\Rightarrow\Box{B}.
See also
| 2013 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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. 
