Difference between revisions of "2012 AMC 10B Problems/Problem 21"

(Created page with "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. The final diagram looks something like thi...")
 
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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.
The final diagram looks something like this:
+
Drawing the points out, it is possible to have a diagram where b=root3a.
.---b---.
 
a  2a  a
 
.  a  .
 
  a  a
 
    .
 
 
 
 
So, b=root3a, so B:A=root3a=(A)
 
So, b=root3a, so B:A=root3a=(A)

Revision as of 18:26, 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=(A)

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