Consider a frustum of a cone. We know the radii of upper and lower surfaces (R1, R2), and the height (h). Now we cut it circularly at the distance d from the top. What’s the radius of the new surface?

Hmmmm... that is a tough homework question. What do you think the answer is?

Quote:

---

Originally Posted by odinn7

Hmmmm....that is a tough homework question. What do you think the answer is?

---

I don't know. That's why I'm asking.

We're not allowed to give you the answer. You have to try to figure it out first, post your answer and how you got it, then someone will come along and help you out with it.

Quote:

---

Originally Posted by odinn7

We're not allowed to give you the answer. You have to try to figure it out first, post your answer and how you got it, then someone will come along and help you out with it.

---

I never knew. All right. I tried. I think we should use the equation for calculating the volume of frustum: V= h*pi/3*{R^2 + R*r + r^2}
Let r1 be the radius of the area. The volume of the upper section is:
V1 = h1*pi/3*{r^2 + r1*r + r1^2} ,
and the volume of the lower section is:
V2 = h2*pi/3*{r1^2 + r1*R + R^2}
We know V1 + V2 = V and h1 +h2 = h.
So, h1*pi/3*{r^2 + r1*r + r1^2} + h2*pi/3*{r1^2 + r1*R + R^2} = V
After distributing and simplifying we get:
h*r1^2 + r1*{r*h1 + R*h2} +h1*r^2 + h2*R^2 - 3*V/pi = 0

This is a quadratic equation of the form ax2 + bx + c = 0 where,
a = h
b = {r*h1 + R*h2}
c= h1*r^2 + h2*R^2 - 3*V/pi

x = (-b + sqrt{b^2 - 4ac})/2a , x = (-b - sqrt{b^2 - 4ac})/2a

By substituting a, b, c in these solutions, we should get a parametric answer for
radius r1. I haven't done that yet.
So is this correct?

Quote:

---

Originally Posted by odinn7

We're not allowed to give you the answer. You have to try to figure it out first, post your answer and how you got it, then someone will come along and help you out with it.

---

It sounds like I sometimes like to do things in difficult ways.
Obviously, Frustum is a trapezoid. Let's look at the lower angle. The tangent is h/(R-r).
So tan(a) = h/(R-r)
a = arctan(h/(R-r))
This angle is the same as the ange when we draw a lin parallel to the base.
tan(a) = h1/(r-r1)
r-r1= tan(a)/h1
r1= r - tan(a)/h1

So if we cut a frustum with a plane parallel to the base, the radius of this plane is r - tan(a)/h1.
r is the radius of the upper surface and h1 is the distance from top to the cutting plane.
I hope I am correct.