2006 AMC 8 Problem 6

Attempt Problem 6 of the 2006 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

6.

The letter T is formed by placing two 2×42 \times 4 inch rectangles next to each other, as shown. What is the perimeter of the T, in inches?
1212
1616
2020
2222
2424
Answer: C
Concepts:perimeterrectangle
Small hint:
Start with the perimeters of the two rectangles separately.
Big hint:
The shared interior segments are counted in the separate perimeters but not in the outside perimeter.
Solution:
If we found the total perimeter of the two rectangles separately, we would have gotten 2(2(2+4))=2⋅2⋅6=24. 2(2(2 + 4)) = 2 \cdot 2 \cdot 6 = 24.
In the letter T, we can see that their intersection removes a piece of length 22 from each of the rectangles. Therefore, the perimeter of the T is 24−2⋅2=20. 24 - 2 \cdot 2 = 20.
Thus, C is the correct answer.
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