Difference between revisions of "2016 AMC 10B Problems/Problem 10"
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<math>\textbf{(A)}\ 14.0\qquad\textbf{(B)}\ 16.0\qquad\textbf{(C)}\ 20.0\qquad\textbf{(D)}\ 33.3\qquad\textbf{(E)}\ 55.6</math> | <math>\textbf{(A)}\ 14.0\qquad\textbf{(B)}\ 16.0\qquad\textbf{(C)}\ 20.0\qquad\textbf{(D)}\ 33.3\qquad\textbf{(E)}\ 55.6</math> | ||
| − | ==Solution== | + | ==Solution 1== |
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 | ||
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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> | ||
| + | |||
| + | ==Solution 2== | ||
| + | Also note that the area of an equilateral triangle is <math>\frac{a^2\sqrt(3)}{4}</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 13:45, 20 May 2016
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 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 note that the area of an equilateral triangle is 
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. 
