Difference between revisions of "2010 AMC 12A Problems/Problem 3"
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<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10</math> | <math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10</math> | ||
− | == Solution == | + | == Solution 1 == |
If we shift <math>A</math> to coincide with <math>E</math>, and add new horizontal lines to divide <math>EFGH</math> into five equal parts: | If we shift <math>A</math> to coincide with <math>E</math>, and add new horizontal lines to divide <math>EFGH</math> into five equal parts: | ||
Line 49: | Line 49: | ||
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}=10</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | From the problem statement, we know that | ||
+ | <cmath>\frac{[ABCD]}{2} = \frac{[EFGH]}{5} \Rightarrow [ABCD]=\frac{2[EFGH]}{5}</cmath> | ||
+ | |||
+ | If we let <math>a = EF</math> and <math>b = AD</math>, we see | ||
+ | <cmath>[ABCD] = 2ab = \frac{2a^2}{5} \Rightarrow b = \frac{a}{5}</cmath>. Hence, <math>\frac{AB}{AD} = \frac{2a}{b} = 2a(\frac{5}{a}) = \boxed{\bold{10}}</math> | ||
==Video Solution 1 (Logic and Word Analysis)== | ==Video Solution 1 (Logic and Word Analysis)== |
Revision as of 16:12, 1 April 2024
Contents
Problem
Rectangle , pictured below, shares of its area with square . Square shares of its area with rectangle . What is ?
Solution 1
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 .
Solution 2
From the problem statement, we know that
If we let and , we see . Hence,
Video Solution 1 (Logic and Word Analysis)
~Education, the Study of Everything
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.