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.
I really need help with this question, it's very difficult for me! I have a diagram and I set up some equations, but I can't figure out what to do next. Please, please help! Thanks.
You might take a look here:
https://www.mathisfunforum.com/viewtopic.php?id=21271
Thanks! 😊
To find the volume of the frustum, we need to break it down into smaller, more manageable parts: the cone and the smaller cone that is cut off by the sphere.
First, let's find the volume of the larger cone. The formula to find the volume of a cone is V = (1/3)Ï€r^2h, where r is the radius of the base and h is the height of the cone. In this case, the radius of the base is equal to the radius of the sphere, which is 3. To find the height of the cone, we can use the Pythagorean theorem: h^2 = (10 - 2r)^2 + r^2.
Plugging in the values, we have h^2 = (10 - 2(3))^2 + 3^2 = 4^2 + 3^2 = 16 + 9 = 25. Taking the square root of both sides, we get h = 5. So, the height of the cone is 5.
Now we can calculate the volume of the larger cone using the formula V = (1/3)Ï€r^2h. Plugging in the values, we get V = (1/3)Ï€(3^2)(5) = (1/3)Ï€(9)(5) = (1/3)Ï€(45) = 15Ï€.
Next, let's find the volume of the smaller cone. This cone is formed by the part of the larger cone that is cut off by the sphere. The volume of this cone can be found by subtracting the volume of the sphere from the volume of the larger cone.
The formula for the volume of a sphere is V = (4/3)Ï€r^3, where r is the radius. In this case, the radius is 3, so the volume of the sphere is V = (4/3)Ï€(3^3) = (4/3)Ï€(27) = 36Ï€.
Therefore, the volume of the smaller cone is 15Ï€ - 36Ï€ = -21Ï€.
Finally, to find the volume of the frustum, we subtract the volume of the smaller cone from the volume of the larger cone. V = (15Ï€) - (-21Ï€) = 15Ï€ + 21Ï€ = 36Ï€.
So, the volume of the frustum is 36Ï€.