Difference between revisions of "2016 AMC 10B Problems/Problem 10"
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| + | ==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 <math></math>\frac{x}{12}=\frac{5^2}{3^2}=\frac{25}{9}<math>. Solving gives us </math>x \approx 33.33<math> so the answer is </math>\boxed{\textbf{(D)}\ 33.3}$. | ||
==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 13:34, 11 February 2018
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 1
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
Also recall that the area of an equilateral triangle is 
so we can give a ratio as follows:
Cross multiplying and simplifying, we get 
Which is 
- Solution by
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 
denotes the weight of the second triangle, we have $$ (Error compiling LaTeX. Unknown error_msg)\frac{x}{12}=\frac{5^2}{3^2}=\frac{25}{9}
x \approx 33.33
\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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
