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

Revision as of 21:21, 15 April 2022

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 the same. 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 $\boxed{\textbf{(D)}\ 33.3}$ .

Note: In general, the weight would be directly proportional to the volume. However, we are told they have equal height, so we essentially treat the problem as 2D.

Video Solution

https://youtu.be/iAE4sL27on4

~savannahsolver

Solution 4 (Intuition)

We ignore the height in this solution because it's constant with both triangles; thus, it has no effect on the difference in their respective weights. Now here's the intuition: notice how the $3$ by $3$ triangle is simply scaled by a factor of $\frac{5}{3}$ on each length to form the $5$ by $5$ triangle, which means that intuitively on one length the weight is scaled by that factor, and we note that we have 2-dimensional system since we ignored the height; therefore, we proceed by doing $(\frac{5}{3})^2 \cdot (12)$ to get our desired result as $\boxed{\textbf{(D)}\ 33.3}$ .

~triggod

@@ Line 40: / Line 40: @@
 ==Solution 4 (Intuition)==
-We ignore the height in this solution because its constant with both triangles; thus, it has no effect on the difference of their respective weights. Now here's the intuition: notice how the 3 by 3 triangle is simply scaled by a factor of 5/3 on each length to form the 5 by 5 triangle, which means that intuitively on one length the weight is scaled by that factor, and we note that we have 2 dimensional system since we ignored the height; therefore, we proceed by doing (5/3)^2(12) to get our desired result as  <math>\boxed{\textbf{(D)}\ 33.3}</math>.
+We ignore the height in this solution because it's constant with both triangles; thus, it has no effect on the difference in their respective weights. Now here's the intuition: notice how the <math>3</math> by <math>3</math> triangle is simply scaled by a factor of <math>\frac{5}{3}</math> on each length to form the <math>5</math> by <math>5</math> triangle, which means that intuitively on one length the weight is scaled by that factor, and we note that we have 2-dimensional system since we ignored the height; therefore, we proceed by doing <math>(\frac{5}{3})^2 \cdot (12)</math> to get our desired result as  <math>\boxed{\textbf{(D)}\ 33.3}</math>.
 ~triggod
 ==See Also==
 {{AMC10 box|year=2016|ab=B|num-b=9|num-a=11}}
 {{MAA Notice}}

2016 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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