2007 AMC 10B Problems/Problem 23

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.

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

2007 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2007 AMC 10B Problems/Problem 23

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