A sphere with radius 3 is inscribed in a conical frustum of slant height 10. (The sphere is tangent to both bases and the side of the frustum.) Find the volume of the frustum.

Could someone help me? I can't find the radius of the cone....

If we look at the frustum from a 2D point of view, we can solve it. (If I could make a diagram it would be easier to explain, sry).

The 2d view of the frustum is a trapezoid. Let's set the top base equal to 2r. Split the trapezoid into a rectangle and two right triangles. Since the hypotenuse is 10(slant height) and the height is 6(the radius is 3), the bottom base is 8+8+2r. So the radius of the base of the cone is r+8.
Let us connect the tangent points from the sides to the center of the circle in the trapezoid. IF we like in the upper right section, we have a "kite" shape with two 90 degree angles. We know three of the side lengths: 3, 3, and r. To find the remaining side we draw a diagonal from the center of the circle to the vertex of the trapezoid that splits the "kite" into two right triangles. They are congruent by HL congruence. that means that the missing side is also r.
Now lets look at the section underneath. We know three of the sides: 3, 3, r+8, and 10-r. The missing side is
the slant height, 10, minus the r that was the fourth side of the "kite" above. Again by, HL congruence, r+8=10-r. That means that the radius of the smaller cone's base (the smaller cone on top of the frustum) is 1, and the radius of the bigger cone is 9 (r+8). Then you just use proportions to find the height, and there you go! Problem Solved!
I hope this wasn't too confusing. I wrote "kite" because it looked like one, but i didn't prove that it was since it's irrelevant to solving the problem. Just an easier way to identify the shape!

I haven't come up with a good geometric solution, but it sems fairly straightforward algebraically.

Draw a cross-section. Place a circle at (0,0) with radius 3. Draw the horizontal lines y=3 and y=-3. These will be the top and bottom of the frustrum.

The radius from (0,0) to (h,k) has slope k/h, so the tangent to the circle at (h,k) has slope -h/k.

The the line tangent to the circle at (h,k) is

y-k = -h/k (x-h)
Now solve that at y=3 and y=-3 to get the values of x. These values are the top radius r and bottom radius R of the frustrum.

The volume of the frustrum is thus 1/3 pi (R62-r^2) * 6

Now, how do we get h and k?

We know that 6^2 + (R-r)^2 = 100
h^2+k^2 = 9
and we have r and R in terms of h and k.

A little algebra, taking care with the details, will how that

(h,k) = (√3,4/√3)
R = 16/√27 + 19/4
r = 16/√27 - 13/4
Note: R-r = 8, as needed

v = 1/3 pi * 6 * (R^2-r^2)
= 2pi * [(13843/432 + 152/√27)-(8659/432 - 104/√27)]
= 2pi (12 + 256/√27)
= 8pi (3 + 64/√27)

Thanks a lot steve it really helps!

To find the radius of the cone, we can use the information given in the problem.

Let's consider the cross-section of the frustum. In the cross-section, we have a large circle (base of the frustum), a smaller circle (top of the frustum), and a triangle connecting the two bases (side of the frustum).

Since the sphere is tangent to both bases and the side of the frustum, it means that the radius of the sphere is equal to the radius of both the large and small circles.

Given that the radius of the sphere is 3, this means that the radius of both the large and small circles is also 3.

Now, let's focus on the triangle connecting the bases. The slant height of the frustum is given as 10. The slant height of a cone is the hypotenuse of a right triangle, and the altitude of the cone (perpendicular distance between the bases) is one of the legs.

In our case, the slant height (10) is the hypotenuse, and the altitude is the other leg. We can use the Pythagorean theorem to find the altitude.

Let's call the altitude h. Using the Pythagorean theorem, we have:

h^2 + (3-3)^2 = 10^2

Simplifying, we get:

h^2 = 100

Taking the square root of both sides, we get:

h = 10

Now that we have the altitude (h) and the radii (r), we can find the volume of the frustum.

The formula for the volume of a frustum is given as:

V = (1/3) * pi * h * (R^2 + r^2 + R*r)

where R is the radius of the large circle and r is the radius of the small circle.

In our case, R = 3 (radius of the large circle) and r = 3 (radius of the small circle).

Plugging in the values, we get:

V = (1/3) * pi * 10 * (3^2 + 3^2 + 3*3)

V = (1/3) * pi * 10 * (9 + 9 + 9)

V = (1/3) * pi * 10 * 27

V = pi * 90

So, the volume of the frustum is 90pi cubic units.