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A sphere is inscribed in a cone with height 3 and base radius 4. What is the ratio of the volume of the sphere to the volume of the cone?

The original is wrong for some reason.

 

 May 4, 2024
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Let the sphere have radius r and consider a right triangle through the center of the sphere and the base of the cone, with one leg being the radius of the base, 4, and the hypotenuse being the slant height of the cone.

 

By the Pythagorean Theorem, the other leg (which is the height of the cone minus the radius of the sphere) has length 32−r2​.

 

 

 

Since this right triangle is similar to a right triangle with legs 4 and 3 (i.e. a 4-3-5 triangle), we have [\frac{3-r}{3} = \frac{r}{4},]so r=43​.

 

The volume of the cone is 31​π(42)(3)=16π, and the volume of the sphere is 34​π(43​)3=83​π, so the ratio of the volume of the sphere to the volume of the cone is 3/128.

 May 4, 2024

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