Difference between revisions of "2016 AMC 12B Problems/Problem 8"

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<math>\textbf{(A)}\ 14.0\qquad\textbf{(B)}\ 16.0\qquad\textbf{(C)}\ 20.0\qquad\textbf{(D)}\ 33.3\qquad\textbf{(E)}\ 55.6</math>
 
<math>\textbf{(A)}\ 14.0\qquad\textbf{(B)}\ 16.0\qquad\textbf{(C)}\ 20.0\qquad\textbf{(D)}\ 33.3\qquad\textbf{(E)}\ 55.6</math>
 
==Solution==
 
==Solution==
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We can solve this problem by using similar triangles, since two equilateral triangles are always similar. We can then use
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<math>(\frac{3}{5})^2=\frac{12}{x}</math>.
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We can then solve the equation to get <math>x=\frac{100}{3}</math> which is closest to <math>\boxed{\textbf{(D)}\ 33.3}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=B|num-b=7|num-a=9}}
 
{{AMC12 box|year=2016|ab=B|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:20, 21 February 2016

Problem

A thin piece of wood of uniform density in the shape of an equilateral triangle with side length $3$ inches weighs $12$ ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of $5$ inches. Which of the following is closest to the weight, in ounces, of the second piece?

$\textbf{(A)}\ 14.0\qquad\textbf{(B)}\ 16.0\qquad\textbf{(C)}\ 20.0\qquad\textbf{(D)}\ 33.3\qquad\textbf{(E)}\ 55.6$

Solution

We can solve this problem by using similar triangles, since two equilateral triangles are always similar. We can then use

$(\frac{3}{5})^2=\frac{12}{x}$.

We can then solve the equation to get $x=\frac{100}{3}$ which is closest to $\boxed{\textbf{(D)}\ 33.3}$

See Also

2016 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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 12 Problems and Solutions

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