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 <math>b=\sqrt{3a}</math>. | 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 <math>b=\sqrt{3a}</math>. | ||
| − | Drawing the points out, it is possible to have a diagram where | + | Drawing the points out, it is possible to have a diagram where <math>b=\sqrt{3a}</math> |
So, <math>b=\sqrt{3a}</math>, so <math>b:a= \sqrt{3}=(A)</math> | So, <math>b=\sqrt{3a}</math>, so <math>b:a= \sqrt{3}=(A)</math> | ||
Revision as of 19:38, 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 
.
Drawing the points out, it is possible to have a diagram where
So, , so 
