Difference between revisions of "2010 AMC 12A Problems/Problem 3"
m (apologies for that) |
m (Semi-automated contest formatting - script by azjps) |
||
Line 25: | Line 25: | ||
== Solution == | == Solution == | ||
+ | == Problem == | ||
=== Solution 1 === | === Solution 1 === | ||
Line 34: | Line 35: | ||
&x = 2s = 10h\\ | &x = 2s = 10h\\ | ||
&\frac{AB}{AD} = \frac{x}{h} = \boxed{10\ \textbf{(E)}}\end{align*}</cmath> | &\frac{AB}{AD} = \frac{x}{h} = \boxed{10\ \textbf{(E)}}\end{align*}</cmath> | ||
+ | |||
+ | == See also == | ||
+ | {{AMC12 box|year=2010|num-b=NaN|num-a=NaN|ab=A}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
=== Solution 2 === | === Solution 2 === |
Revision as of 21:53, 25 February 2010
Problem
Rectangle , pictured below, shares of its area with square . Square shares of its area with rectangle . What is ?
Solution
Problem
Solution 1
Let , let , and let .
See also
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem NaN |
Followed by Problem NaN |
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 |
Solution 2
The answer does not change 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 |