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>.
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{{AMC8 box|year=2006|num-b=5|num-a=7}}

Revision as of 14:59, 21 November 2011

Problem

The letter T is formed by placing two $2 \times 4$ 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]

$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$

Solution

If the two rectangles were seperate, the perimeter would be $2(2(2+4)=24$. It easy to see that their connection erases 2 from each of the rectangles, so the final perimeter is $24-2 \times 2 = \boxed{\textbf{(C)}\ 20}$.

2006 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AJHSME/AMC 8 Problems and Solutions