Difference between revisions of "2007 AMC 10B Problems/Problem 23"
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<cmath>a=\frac{2\sqrt{2}}{\sqrt{2}-1} = \frac{4+2\sqrt{2}}{2-1} = \boxed{\mathrm{(E) \ } 4+2\sqrt{2}}</cmath> | <cmath>a=\frac{2\sqrt{2}}{\sqrt{2}-1} = \frac{4+2\sqrt{2}}{2-1} = \boxed{\mathrm{(E) \ } 4+2\sqrt{2}}</cmath> | ||
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{{AMC10 box|year=2007|ab=B|num-b=22|num-a=24}} | {{AMC10 box|year=2007|ab=B|num-b=22|num-a=24}} | ||
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Revision as of 13:21, 16 February 2019
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)
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. 
