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?
![[asy] unitsize(8mm); defaultpen(linewidth(.8pt)); draw((0,0)--(4,0)--(4,4)--(0,4)--cycle); draw((0,3)--(0,4)--(1,4)--(1,3)--cycle); draw((1,3)--(1,4)--(2,4)--(2,3)--cycle); draw((2,3)--(2,4)--(3,4)--(3,3)--cycle); draw((3,3)--(3,4)--(4,4)--(4,3)--cycle); [/asy]](http://latex.artofproblemsolving.com/d/0/9/d09caec6074d6abf81a6e3a7755b2eecc103bc41.png)
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 | ||
