Difference between revisions of "2016 AMC 12B Problems/Problem 8"
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Another approach to this problem, very similar to the previous one but perhaps explained more thoroughly, is to use proportions. First, since the thickness and density are the same, we can set up a proportion based on the principle that <math>d=\frac{m}{V}</math>, thus <math>dV=m</math>. | Another approach to this problem, very similar to the previous one but perhaps explained more thoroughly, is to use proportions. First, since the thickness and density are the same, we can set up a proportion based on the principle that <math>d=\frac{m}{V}</math>, thus <math>dV=m</math>. | ||
| − | However, since density and thickness are the same and <math>A is proportional to s^2</math> (recognizing that the area of an equilateral triangle is <math>\frac{(s)^2\sqrt{3}}{4}</math>), we can say that <math>m is proportional to s^2</math>. | + | However, since density and thickness are the same and <math>A</math> is proportional to <math>s^2</math> (recognizing that the area of an equilateral triangle is <math>\frac{(s)^2\sqrt{3}}{4}</math>), we can say that <math>m</math> is proportional to <math>s^2</math>. |
Then, by increasing s by a factor of <math>\frac{5}{3}</math>, <math>s^2</math> is increased by a factor of <math>\frac{25}{9}</math>, thus <math>m=12*\frac{25}{9}</math> or <math>\boxed{\textbf{(D)}\ 33.3}</math>. | Then, by increasing s by a factor of <math>\frac{5}{3}</math>, <math>s^2</math> is increased by a factor of <math>\frac{25}{9}</math>, thus <math>m=12*\frac{25}{9}</math> or <math>\boxed{\textbf{(D)}\ 33.3}</math>. | ||
Revision as of 23:22, 1 February 2018
Contents
Problem
A thin piece of wood of uniform density in the shape of an equilateral triangle with side length 
inches weighs 
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 
inches. Which of the following is closest to the weight, in ounces, of the second piece?
Solution
By: dragonfly
We can solve this problem by using similar triangles, since two equilateral triangles are always similar. We can then use
.
We can then solve the equation to get 
which is closest to 
Solution 2
Another approach to this problem, very similar to the previous one but perhaps explained more thoroughly, is to use proportions. First, since the thickness and density are the same, we can set up a proportion based on the principle that 
, thus 
.
However, since density and thickness are the same and 
is proportional to 
(recognizing that the area of an equilateral triangle is 
), we can say that 
is proportional to .
Then, by increasing s by a factor of 
, is increased by a factor of 
, thus 
or .
See Also
| 2016 AMC 12B (Problems • Answer Key • Resources) | |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
