2006 AMC 8 Problems/Problem 6

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

Video Solution by OmegaLearn

https://youtu.be/abSgjn4Qs34?t=1531

~ pi_is_3.14

Video Solution by WhyMath

https://youtu.be/KS7M9uzD9SI

2006 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2006 AMC 8 Problems/Problem 6

Contents

Problem

Solution

Video Solution by OmegaLearn

Video Solution by WhyMath

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