2006 AMC 12A Problem 6

Attempt Problem 6 of the 2006 AMC 12A 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 12A solutions, or check the answer key.

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

6.

The 8×188 \times 18 rectangle ABCDABCD is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is y?y?
66
77
88
99
1010
Answer: A
Concepts:areasquare (geometry)
Small hint:
The two hexagons reassemble into a square of area 8⋅188\cdot 18
Big hint:
The staircase splits the width into three equal horizontal pieces that sum to 1818
Solution:
The two hexagons form a square of area 8⋅18=144,8 \cdot 18 = 144, so the square has side 12.12.
The staircase cut splits the width into three equal horizontal pieces of length y,y, which together span the full width: y+y+y=18,y + y + y = 18, so y=6.y = 6. (The two vertical steps each rise 12−8=4,12 - 8 = 4, building the extra height of the square.)
Thus, the correct answer is A.
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