Difference between revisions of "1954 AHSME Problems/Problem 27"
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== Solution == | == Solution == | ||
Because the circle has the same radius as the sphere, the cylinder and sphere have the same radius. Then from the volume of cylinder and volume of a sphere formulas, we have <math>\frac{1}{3} \pi r^2 h= \frac{2}{3} \pi r^3 \implies h=2r\implies \frac{h}{r}=2</math> <math>\boxed{(\textbf{D})}</math> | Because the circle has the same radius as the sphere, the cylinder and sphere have the same radius. Then from the volume of cylinder and volume of a sphere formulas, we have <math>\frac{1}{3} \pi r^2 h= \frac{2}{3} \pi r^3 \implies h=2r\implies \frac{h}{r}=2</math> <math>\boxed{(\textbf{D})}</math> | ||
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+ | ==See Also== | ||
+ | |||
+ | {{AHSME 50p box|year=1954|num-b=26|num-a=28}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 01:30, 28 February 2020
Problem 27
A right circular cone has for its base a circle having the same radius as a given sphere. The volume of the cone is one-half that of the sphere. The ratio of the altitude of the cone to the radius of its base is:
Solution
Because the circle has the same radius as the sphere, the cylinder and sphere have the same radius. Then from the volume of cylinder and volume of a sphere formulas, we have
See Also
1954 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
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