Difference between revisions of "2012 AMC 10B Problems/Problem 21"
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When you see a and 2a, you could think of 30-60-90 triangles. Since all of the other's lengths are a, you could think that b is root3a. | When you see a and 2a, you could think of 30-60-90 triangles. Since all of the other's lengths are a, you could think that b is root3a. | ||
Drawing the points out, it is possible to have a diagram where b=root3a. | Drawing the points out, it is possible to have a diagram where b=root3a. | ||
| − | So, b=root3a, so B:A=root3a $\textbf{(A)} | + | So, b=root3a, so B:A=root3a $\textbf{(A)}\ sqrt {3}\qquad\textbf |
Revision as of 18:28, 12 March 2012
When you see a and 2a, you could think of 30-60-90 triangles. Since all of the other's lengths are a, you could think that b is root3a. Drawing the points out, it is possible to have a diagram where b=root3a. So, b=root3a, so B:A=root3a $\textbf{(A)}\ sqrt {3}\qquad\textbf
