2009 AMC 8 Problem 25

Attempt Problem 25 of the 2009 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 2009 AMC 8 solutions, or check the answer key.

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

25.

A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is 12\dfrac{1}{2} foot from the top face. The second cut is 13\dfrac{1}{3} foot below the first cut, and the third cut is 117\dfrac{1}{17} foot below the second cut. From the top to the bottom the pieces are labeled A,A, B,B, C,C, and D.D. The pieces are then glued together end to end in the order C,C, B,B, A,A, DD to make a long solid as shown. What is the total surface area of this solid in square feet?
66
77
41951\dfrac{419}{51}
15817\dfrac{158}{17}
1111
Answer: E
Concepts:surface area3D geometry
Small hint:
Think of surface area by looking from the six directions.
Big hint:
The four pieces stack back to a unit cube when viewed from either side.
Solution:
Look at the solid from the six coordinate directions.
Viewed from either end, the exposed vertical faces have total height equal to the thickness of piece A,A, namely 12\dfrac{1}{2} foot. Thus each end view has area 12\dfrac{1}{2} square foot.
From each side, the four pieces stack to the side view of the original unit cube, so each side view has area 11 square foot.
From the top and bottom, each view shows four 11-by-11 faces, so each has area 44 square feet.
The total surface area is 12+12+1+1+4+4=11\dfrac{1}{2}+\dfrac{1}{2}+1+1+4+4=11 square feet.
Thus, E is the correct answer.
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