Difference between revisions of "2016 AMC 12B Problems/Problem 8"
m (→Solution 2) |
m (added minor sidenote) |
||
Line 12: | Line 12: | ||
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> | ||
+ | |||
+ | Sidenote - If we didn't know that <math>\left\frac{a}{b}\right^2</math> for an equilateral triangle, using the area formula, <math>\frac{1}{4}\sqrt{3}a^2</math>, in proportion <math>\frac{1/4(sqrt{3}a^2)}{1/4(sqrt{3}b^2)}</math>, just cancel and simplify to get $\frac{a^2}{b^2}. | ||
==Solution 2== | ==Solution 2== |
Revision as of 13:20, 19 January 2021
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
Sidenote - If we didn't know that $\left\frac{a}{b}\right^2$ (Error compiling LaTeX. Unknown error_msg) for an equilateral triangle, using the area formula, , in proportion , just cancel and simplify to get $\frac{a^2}{b^2}.
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.