Difference between revisions of "2012 AMC 10B Problems/Problem 21"
| Line 1: | Line 1: | ||
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 
