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Difference between revisions of "2013 AMC 10B Problems/Problem 15"
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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>
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== See also ==
 +
{{AMC10 box|year=2013|ab=B|num-b=14|num-a=16}}

Revision as of 16:02, 27 March 2013

Problem

A wire is cut into two pieces, one of length $a$ and the other of length $b$. The piece of length $a$ is bent to form an equilateral triangle, and the piece of length $b$ is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is $\frac{a}{b}$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{\sqrt{6}}{2}\qquad\textbf{(C)}\ \sqrt{3} \qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{3\sqrt{2}}{2}$

Solution

Using the formulas for area of a regular triangle $(\frac{{s}^{2}\sqrt{3}}{4})$ and regular hexagon $(\frac{3{s}^{2}\sqrt{3}}{2})$ and plugging $\frac{a}{3}$ and $\frac{b}{6}$ into each equation, you find that $\frac{{a}^{2}\sqrt{3}}{36}=\frac{{b}^{2}\sqrt{3}}{24}$. Simplifying this, you get $\frac{a}{b}=\boxed{\textbf{(B)} \frac{\sqrt{6}}{2}}$

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
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