Difference between revisions of "2016 AMC 10B Problems/Problem 10"

(Solution 2)
m (Solution 1)
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We can solve this problem by using similar triangles, since two equilateral triangles are always similar. We can then use
 
We can solve this problem by using similar triangles, since two equilateral triangles are always similar. We can then use
  
<math>(\frac{3}{5})^2=\frac{12}{x}</math>.
+
<math>\left(\frac{3}{5}\right)^2=\frac{12}{x}</math>.
  
 
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>
 
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>

Revision as of 20:23, 14 February 2017

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 1

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

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

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

Solution 2

Also recall that the area of an equilateral triangle is $\frac{a^2\sqrt3}{4}$ so we can give a ratio as follows:


$\frac{\frac{9\sqrt3}{4}}{12}$ $=$ $\frac{\frac{25\sqrt3}{4}}{x}$

Cross multiplying and simplifying, we get $12 \cdot \frac{25}{9}$

Which is $33.\overline{3}$ $\approx$ $\boxed{\textbf{(D)}\ 33.3}$

  • Solution by $AOPS12142015$

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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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