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A cylinder rests on top of a table, with a cone inscribed within, vertex up. Both heights and radii are 8 cm. A hemispherical bowl of radius 8 cm rests nearby on the same table, its circular rim parallel to the table.

Let r be the common radius of the cone, cylinder, and hemisphere. Show that the ring and the disk have the same area, for all positions of the slicing plane.

User Anoyz
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Final answer:

The area of both the ring formed on the cylinder by the slicing plane and the disk is calculated as πr² for all positions of the slicing plane parallel to the bases, where r is the common radius of the cylinder, cone, and hemisphere.

Step-by-step explanation:

The question asks about the area of a 'ring' and a 'disk' that result from slicing a cylinder, cone, and hemisphere with a given plane, all of which have a common radius r.

To approach this problem, we recognize that the disk's area will always be a circle's area with radius r, which can be calculated using the formula A = πr². As for the ring, it is the area of the cylinder's end that is visible from the slicing plane minus the area of the cone that is inscribed within it.

Assuming the slicing plane is perpendicular to the bases of the cylinder and the cone, the area of the ring will be equal to the area of the disk because the area of the top of the cone that gets subtracted from the cylinder's round end will always leave an area that is equal to πr².

This holds true for any position of the slicing plane, as long as it's parallel to the cylinder and cone's bases.

User Guillim
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