Difference between revisions of "2010 AMC 10A Problems/Problem 2"
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Let the length of the small square be <math>x</math>, intuitively, the length of the big square is <math>4x</math>. It can be seen that the width of the rectangle is <math>3x</math>. Thus, the length of the rectangle is <math>4x/3x = 4/3</math> times large as the width. The answer is <math>\boxed{B}</math>. | Let the length of the small square be <math>x</math>, intuitively, the length of the big square is <math>4x</math>. It can be seen that the width of the rectangle is <math>3x</math>. Thus, the length of the rectangle is <math>4x/3x = 4/3</math> times large as the width. The answer is <math>\boxed{B}</math>. | ||
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+ | == See also == | ||
+ | {{AMC10 box|year=2010|ab=A|num-b=1|num-a=3}} |
Revision as of 19:34, 1 January 2012
Problem 2
Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width?
Solution
Let the length of the small square be , intuitively, the length of the big square is . It can be seen that the width of the rectangle is . Thus, the length of the rectangle is times large as the width. The answer is .
See also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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 10 Problems and Solutions |