Question

A truncated cone has horizontal bases with radii 18 and 2. A sphere is tangent to the top, bottom, and lateral surface of the truncated cone. What is the radius of the sphere

Answer 1

Answer:  6 units

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Explanation:

A truncated cone is where we start with a regular 3D cone and chop off the top. The portion up top is a smaller cone, which we'll ignore. The bottom part is the truncated cone portion.

In the real world, a lampshade is one example of a truncated cone.

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Let x be the radius of the sphere.

We'll be focusing on a vertical cross section of the truncated cone. Refer to figure 1 in the diagram below.

We have the following points

• A = center of the sphere
• B = center of the circular base of the truncated cone
• C = point 18 units to the right of point B
• E = point directly above point A, and its the center of the circular top of the truncated cone
• D = point 2 units to the right of point E
• F = the location where the circle touches the slanted curved side of the truncated cone
• G = point directly below point D, and located on segment BC

We'll connect a few of those points forming the dashed lines in figure 1.

To start off, draw a segment from D to G. This forms rectangle BGDE with sides of EB = 2x and BG = 2. The side BC is 18 units, so that must mean GC = BC - BG = 18 - 2 = 16.

Since EB = 2x, this means DG is also 2x.

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Now focus on triangles ABC and AFC. They are congruent right triangles. We can prove this using the HL (hypotenuse leg) theorem. Recall that the radius of a circle is perpendicular to the tangent line, which is why angle AFC is 90 degrees. Angle ABC is a similar story.

Because they are congruent right triangles, this indicates side BC is the same length as side FC. Therefore, FC = 18

Through similar logic, triangles ADE and ADF are congruent as well which leads to ED = FD = 2.

Combine sides FD and FC to get the length of DC

DC = FD+FC = 2+18 = 20

This is the hypotenuse of the right triangle GCD

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After all that, we have the right triangle GCD with the legs of 2x and 16. The hypotenuse is 20. Refer to figure 2 shown below.

As you can probably guess, we'll use the pythagorean theorem to find the value of x.

a^2 + b^2 = c^2

(DG)^2 + (GC)^2 = (CD)^2

(2x)^2 + (16)^2 = (20)^2

4x^2 + 256 = 400

4x^2 = 400-256

4x^2 = 144

x^2 = 144/4

x^2 = 36

x = sqrt(36)

x = 6 is the radius of the sphere.