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

(Created page with "== Problem == The letter T is formed by placing two <math> 2 \times 4 </math> inch rectangles next to each other, as shown. What is the perimeter of the T, in inches? <asy> siz...")
 
 
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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
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/KS7M9uzD9SI
 +
 +
==See Also==
 +
{{AMC8 box|year=2006|num-b=5|num-a=7}}
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{{MAA Notice}}

Latest revision as of 13:21, 29 October 2024

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

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

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