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SATII数学考试快递——Inscribed Solids

An inscribed solid is a solid placed inside another solid, with the edges of the two solids touching. The figures below are, from left to right, a cylinder inscribed in a sphere, a sphere inscribed in a cube, and a rectangular solid inscribed in a sphere.

Math IC questions that involve inscribed solids don’t require any techniques other than those you’ve already learned. These questions do require an ability to visualize inscribed solids and an awareness of how certain line segments relate to both of the solids in a given figure.

Most often, an inscribed-solid question will present a figure of an inscribed solid and give you information about one of the solids. For example, you may be given the radius of a cylinder, and then be asked to find the volume of the other solid, say a rectangular solid. Using the figure as your guide, you need to use the radius of the cylinder to find the dimensions of the other solid so that you can answer the question. Here’s an example:


	In the figure below, a cube is inscribed in a cylinder. If the length of the diagonal of the cube is 4and the height of the cylinder is 5, what is the volume of the cylinder?

The formula for the volume of a cylinder is πr2(h). The question states that h = 5, but there is no value given for r. So in order to solve for the volume of the cylinder, we need to first find the value of r.

The key step in this problem is to recognize that the diagonal of a face of the cube is also the diameter, or twice the radius, of the cylinder. To see this, draw a diagonal, d, in either the top or bottom face of the cube.

In order to find this diagonal, which is the hypotenuse in a 45-45-90 triangle, we need the length of an edge of the cube, or s. We can find s from the diagonal of the cube (not to be confused with the diagonal of a face of the cube), since the formula for the diagonal of a cube is swhere s is the length of an edge of the cube. The question states that the diagonal of the cube is 4so it follows that s = 4. This means that the diagonal along a single face of the cube is 4(using the special properties of a 45-45-90 triangle). Therefore, the radius of the cylinder is 4/ 2 = 2Plug that into the formula for the volume of the cylinder, and you get π (2)2 5 = 40π.