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= | + | Drawing the points out, it is possible to have a diagram where b=<math>\sqrt{3a}</math> |
So, b=<math>\sqrt{3a}</math>, so B:A= <math>\sqrt{3}</math> | So, b=<math>\sqrt{3a}</math>, so B:A= <math>\sqrt{3}</math> | ||
Revision as of 18:34, 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=
So, b=, so B:A= 
