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2007 AMC 10B Problems/Problem 23

Revision as of 09:39, 9 August 2021 by Bakedpotato66 (talk | contribs) (→‎Solution 2 (Playing with the answers))

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.

\[\frac{a}{a-2}=\sqrt{\frac21}=\frac{\sqrt{2}}{1} \longrightarrow a= a\sqrt{2} - 2\sqrt{2} \longrightarrow a\sqrt{2}-a=2\sqrt{2}\] \[a=\frac{2\sqrt{2}}{\sqrt{2}-1} = \frac{4+2\sqrt{2}}{2-1} = \boxed{\textbf{(E) \ } 4+2\sqrt{2}}.\]

Solution 2 (Playing with the answers)

Instead of actually solving this problem, we can play with the answers. The ratios of the altitudes squared is the ratio of the surface areas, so the answer choice has to be larger than $4$ because an original height of $4$ would give the ratio $1:4$ for the surface area and only choice $E$ is larger than $4$, so the answer is $E$.

Solution 3 (Note)

Like in Solution 1, we used the fact that the ratio of surface area is the ratio of side lengths squared. However, with this, we can go two ways.

1. (Solution 1) The ratio of the surface area of the large pyramid to the surface area of the small pyramid is $\frac 21$, so the ratio of their altitudes is $\sqrt{\frac21}=\frac{\sqrt{2}}{1}=\sqrt2$.

2. (Another way) The ratio of the surface area of the small pyramid to the surface area of the large pyramid is $\frac 12$, so the ratio of their altitudes is $\sqrt{\frac12}=\frac{1}{\sqrt2}=\frac{\sqrt2}{2}$.


Then, we can set up variables based on these ratios and solve for the variable, hence getting the altitude of the large triangle, or $\boxed{\textbf{(E)}~4+2\sqrt 2}.$

See Also

2007 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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