2016 AMC 10A Problem 10

Attempt Problem 10 of the 2016 AMC 10A 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 2016 AMC 10A solutions, or check the answer key.

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

10.

A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is 11 foot wide on all four sides. What is the length in feet of the inner rectangle?
11
22
44
66
88
Answer: B
Concepts:areaarithmetic sequence
Small hint:
Let the inner rectangle have length ll
Big hint:
The three region areas are l,2l+6,2l+14l,2l+6,2l+14
Solution:
Let ll be the length of the inner rectangle. Then the area of the inner rectangle is l.l.
The area of the middle region is going to be (l+2)⋅3−l=2l+6. (l + 2) \cdot 3 - l = 2l + 6. The area of the outer region is (l+4)⋅5−(l+2)⋅3=2l+14. (l + 4) \cdot 5 - (l + 2) \cdot 3 = 2l + 14.
We know that these 33 values form an arithmetic sequence. That means that l+2l+14=2(2l+6)3l+14=4l+12l=2\begin{aligned} l + 2l + 14 &= 2(2l + 6) \\ 3l + 14 &= 4l + 12 \\ l &= 2\end{aligned}
Thus, the correct answer is B .
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