2016 AMC 10B Problems/Problem 10

Revision as of 19:23, 14 February 2017 by Blazeboy (talk | contribs) (Solution 1)

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
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All AMC 10 Problems and Solutions

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