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 
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?
Solution
Since the two pyramids are similar, the ratio of the altitudes is the square root of the ratio of the surface areas.
If 
is the altitude of the larger pyramid, then 
is the altitude of the smaller pyramid.
![\[\frac{a}{a-2}=\frac{\sqrt{2}}{1} \longrightarrow a= a\sqrt{2} - 2\sqrt{2} \longrightarrow a\sqrt{2}-a=2\sqrt{2}\]](http://latex.artofproblemsolving.com/e/f/6/ef62ca4b8fd755100e6cd0e0c1930640a1e5b24b.png)
![\[a=\frac{2\sqrt{2}}{\sqrt{2}-1} = \frac{4+2\sqrt{2}}{2-1} = \boxed{\mathrm{(E) \ } 4+2\sqrt{2}}\]](http://latex.artofproblemsolving.com/2/6/4/264d4491567ef04ed2a23bab78614ffa56585a64.png)
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 
because an original height of would give the ratio 
for the surface area and only choice 
is larger than , so the answer is .
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
