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

(Solution 2)
(Solution 2)
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<math>\frac{\frac{9\sqrt3}{4}}{12}</math> =
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<math>\frac{\frac{9\sqrt3}{4}}{12}</math> = <math>\frac{\frac{25\sqrt3}{4}}{x}</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 12:50, 20 May 2016

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

$(\frac{3}{5})^2=\frac{12}{x}$.

We can then solve the equation to get $x=\frac{100}{3}$ which is closest to $\boxed{\textbf{(D)}\ 33.3}$

Solution 2

Also note that the area of an equilateral triangle is $\frac{a^2\sqrt3}{4}$ So we can give a ratio as follows


$\frac{\frac{9\sqrt3}{4}}{12}$ = $\frac{\frac{25\sqrt3}{4}}{x}$

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
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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