Difference between revisions of "2011 AMC 8 Problems/Problem 19"
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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+5+3=\boxed{\textbf{(D)}\ 11}</math> | + | 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> |
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| + | <math>3+5+3=\boxed{\textbf{(D)}\ 11}</math> | ||
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| + | ==Count by Reduction== | ||
| + | We can remove the 3 big blocks of rectangles one by one. 7 (left) + 3 (bottom) + 1 = 11 are removed in total. | ||
| + | |||
| + | ~aliciawu | ||
| + | |||
| + | ==Video Solution 1 by SpreadTheMathLove== | ||
| + | https://www.youtube.com/watch?v=mYn6tNxrWBU | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/IEeJsGh3ltk | ||
==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}} | ||
Latest revision as of 10:06, 18 November 2024
Contents
Problem
How many rectangles are in this figure?
![[asy] pair A,B,C,D,E,F,G,H,I,J,K,L; A=(0,0); B=(20,0); C=(20,20); D=(0,20); draw(A--B--C--D--cycle); E=(-10,-5); F=(13,-5); G=(13,5); H=(-10,5); draw(E--F--G--H--cycle); I=(10,-20); J=(18,-20); K=(18,13); L=(10,13); draw(I--J--K--L--cycle);[/asy]](http://latex.artofproblemsolving.com/7/1/1/7111cf9af97b6671111a8bda634b2877ecb06289.png)
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
Count by Reduction
We can remove the 3 big blocks of rectangles one by one. 7 (left) + 3 (bottom) + 1 = 11 are removed in total.
~aliciawu
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=mYn6tNxrWBU
Video Solution by WhyMath
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
