Question

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 $\dfrac{1}{2}$ foot from the top face. The second cut is $\dfrac{1}{3}$ foot below the first cut, and the third cut is $\dfrac{1}{17}$ foot below the second cut. From the top to the bottom the pieces are labeled $A,$ $B,$ $C,$ and $D.$ The pieces are then glued together end to end in the order $C,B,A,D$ to make a long solid as shown. What is the total surface area of this solid in square feet?

$6$

$7$

$\dfrac{419}{51}$

$\dfrac{158}{17}$

$11$

Solution:

Look at the solid from the six coordinate directions.

From the two ends, the visible areas are each the face of piece $A$ , which has area $\dfrac{1}{2}$ square foot.

From each side, the four pieces stack to the side view of the original unit cube, so each side view has area $1$ square foot.

From the top and bottom, each view shows four $1$ -by- $1$ faces, so each has area $4$ square feet.

The total surface area is $\dfrac{1}{2}+\dfrac{1}{2}+1+1+4+4=11$ square feet.

Thus, E is the correct answer.

2009 AMC 8 Problem 25

Solution:

Problem 25 in Other Years