2011 AMC 8 Problems/Problem 19: Difference between revisions
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<math> \textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12 </math> | <math> \textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12 </math> | ||
==Solution== | ==Solution 1 (Splitting It Up)== | ||
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> | 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> | |||
==Solution 2 (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 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 14:52, 2 February 2025
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 1 (Splitting It Up)
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
Solution 2 (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 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
