Difference between revisions of "2010 AMC 12A Problems/Problem 3"
Flamedragon (talk | contribs) (→Solution 1) |
m (→Solution) |
||
Line 48: | Line 48: | ||
This helps us to see that <math>AD=a/5</math> and <math>AB=2a</math>, where <math>a=EF</math>. | This helps us to see that <math>AD=a/5</math> and <math>AB=2a</math>, where <math>a=EF</math>. | ||
− | Hence <math>\dfrac{AB}{AD}=\dfrac{2a}{a/5}=10</math>. | + | Hence <math>\dfrac{AB}{AD}=\dfrac{2a}{a/5}=\boxed{(E) 10}</math>. |
== See also == | == See also == |
Revision as of 17:04, 3 August 2020
Problem
Rectangle , pictured below, shares of its area with square . Square shares of its area with rectangle . What is ?
Solution
If we shift to coincide with , and add new horizontal lines to divide into five equal parts:
This helps us to see that and , where . Hence .
See also
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 2 |
Followed by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.