Difference between revisions of "2006 AMC 8 Problems/Problem 6"
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== Solution == | == Solution == | ||
If the two rectangles were seperate, the perimeter would be <math> 2(2(2+4)=24 </math>. It easy to see that their connection erases 2 from each of the rectangles, so the final perimeter is <math> 24-2 \times 2 = \boxed{\textbf{(C)}\ 20} </math>. | If the two rectangles were seperate, the perimeter would be <math> 2(2(2+4)=24 </math>. It easy to see that their connection erases 2 from each of the rectangles, so the final perimeter is <math> 24-2 \times 2 = \boxed{\textbf{(C)}\ 20} </math>. | ||
| + | |||
| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/abSgjn4Qs34?t=1531 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2006|num-b=5|num-a=7}} | {{AMC8 box|year=2006|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:45, 2 January 2023
Problem
The letter T is formed by placing two 
inch rectangles next to each other, as shown. What is the perimeter of the T, in inches?
![[asy] size(150); draw((0,6)--(4,6)--(4,4)--(3,4)--(3,0)--(1,0)--(1,4)--(0,4)--cycle, linewidth(1));[/asy]](http://latex.artofproblemsolving.com/4/e/5/4e5ee969b5b05027d7b34dc893efcf56dfd272be.png)
Solution
If the two rectangles were seperate, the perimeter would be 
. It easy to see that their connection erases 2 from each of the rectangles, so the final perimeter is 
.
Video Solution by OmegaLearn
https://youtu.be/abSgjn4Qs34?t=1531
~ pi_is_3.14
See Also
| 2006 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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. 
