Difference between revisions of "2006 AMC 8 Problems/Problem 6"

m
Line 13: Line 13:
 
==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}}

Revision as of 00:13, 5 July 2013

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 $24-2 \times 2 = \boxed{\textbf{(C)}\ 20}$.

See Also

2006 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png