2007 AMC 10B Problems/Problem 23

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Problem

A pyramid with a square base is cut by a plane that is parallel to its base and $2$ units from the base. The surface area of the smaller pyramid that is cut from the top is half the surface area of the original pyramid. What is the altitude of the original pyramid?

$\textbf{(A) } 2 \qquad\textbf{(B) } 2+\sqrt{2} \qquad\textbf{(C) } 1+2\sqrt{2} \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 4+2\sqrt{2}$

Solution

Since the two pyramids are similar, the ratio of the altitudes is the square root of the ratio of the surface areas.

If $a$ is the altitude of the larger pyramid, then $a-2$ is the altitude of the smaller pyramid.

\[\frac{a}{a-2}=\frac{\sqrt{2}}{1} \longrightarrow a= a\sqrt{2} - 2\sqrt{2} \longrightarrow a\sqrt{2}-a=2\sqrt{2}\] \[a=\frac{2\sqrt{2}}{\sqrt{2}-1} = \frac{4+2\sqrt{2}}{2-1} = \boxed{\mathrm{(E) \ } 4+2\sqrt{2}}\]

See Also..

2007 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AMC 10 Problems and Solutions

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