2012 AMC 10B Problems/Problem 21: Difference between revisions
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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=<math>\sqrt{3a}</math> | Drawing the points out, it is possible to have a diagram where b=<math>\sqrt{3a}</math> | ||
So, | So, <math>b=\sqrt{3a}</math>, so <math>b:a= \sqrt{3}=(A)</math> | ||
Revision as of 18:37, 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, 
, so 
