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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.
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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>.
 
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, <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 18:38, 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 $b=\sqrt{3a}$. Drawing the points out, it is possible to have a diagram where b=$\sqrt{3a}$ So, $b=\sqrt{3a}$, so $b:a= \sqrt{3}=(A)$

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