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

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==Solution 3==
 
==Solution 3==
  
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Note that the ratio of the two triangle's weights is equal to the ratio of their areas, as the height is negligible. The ratio of their areas is equal to the square of the ratio of their sides. So if <math>x</math> denotes the  weight of the second triangle, we have <cmath>\frac{x}{12}=\frac{5^2}{3^2}=\frac{25}{9}</cmath>. Solving gives us <math>x \approx 33.33</math> so the answer is <math>\boxed{\textbf{(D)}\ 33.3}</math>.
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Note that the ratio of the two triangle's weights is equal to the ratio of their areas, as the height is negligible. The ratio of their areas is equal to the square of the ratio of their sides. So if <math>x</math> denotes the  weight of the second triangle, we have <cmath>\frac{x}{12}=\frac{5^2}{3^2}=\frac{25}{9}</cmath> Solving gives us <math>x \approx 33.33</math> so the answer is <math>\boxed{\textbf{(D)}\ 33.3}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=B|num-b=9|num-a=11}}
 
{{AMC10 box|year=2016|ab=B|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:35, 11 February 2018

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 $AOPS12142015$

Solution 3

Note that the ratio of the two triangle's weights is equal to the ratio of their areas, as the height is negligible. The ratio of their areas is equal to the square of the ratio of their sides. So if $x$ denotes the weight of the second triangle, we have \[\frac{x}{12}=\frac{5^2}{3^2}=\frac{25}{9}\] Solving gives us $x \approx 33.33$ so the answer is $\boxed{\textbf{(D)}\ 33.3}$.

See Also

2016 AMC 10B (Problems • Answer Key • Resources)
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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