2011 AMC 8 Problems/Problem 19: Difference between revisions
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==Solution== | ==Solution== | ||
The figure can be divided into <math>7</math> sections. The number of rectangles with just one section is <math>3.</math> The number of rectangles with two sections is <math>5.</math> There are none with only three sections. The number of rectangles with four sections is <math>3.</math> <math>3+5+3=\boxed{\textbf{( | The figure can be divided into <math>7</math> sections. The number of rectangles with just one section is <math>3.</math> The number of rectangles with two sections is <math>5.</math> There are none with only three sections. The number of rectangles with four sections is <math>3.</math> <math>3+5+3=\boxed{\textbf{(D)}\ 11}</math> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=18|num-a=20}} | {{AMC8 box|year=2011|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:54, 26 October 2017
Problem
How many rectangles are in this figure?
Solution
The figure can be divided into
sections. The number of rectangles with just one section is
The number of rectangles with two sections is
There are none with only three sections. The number of rectangles with four sections is
See Also
| 2011 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
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