How to calculate the height of a cut cone (if it is called that in english), if the surface area is 320pi cm2, R = 10 and r = 5. See picture for explanation.

First is this correct:

Code:

P -surface area
P=B+B1+M
B= 10^2*Pi
B1=5^2*pi
M=(R+r)*s*Pi
M=15*s*Pi
so
P=100*pi + 25*Pi + 15*Pi*s
320 = 100 + 25 + 15*s
195 = 15*s
s = 13
H^2 = s^2 - (R - r)^2
H^2 = 169 - 25
H^2 = 144
H = 12

It seems correct but i have a solution which says it is not. Can the solution be wrong?

I'm going to make some assumptions which may be wrong and go from there. Plus, I really like 'cut cone' better than conic section.

Math people have no sense of clarity or beauty sometimes. Me, I have no sense at all.

Code:

Assume
The cut is parallel to the bottom of the cone. ie. a circle.
Note:
You did not give us all the data we need -
We must have either the Length S or an angle-theta
(the angle made by line S and line R). Because without one
or the other H could be infinitely long - if theta is 90 deg
for instance - an infinitely long rod.
Assume we know s
H = sqr( s^2 - (R-r)^2)
Assume we know theta
Assume I remember trig which I never use
H = tan(theta) * (R-r)

If you do not know S, then you have to know an angle. Period. As far as I can see.

If the angle were 90 degrees, you would have a soild shaped like a rod. Agreed? You could then cut the rod anywhere, from 1" out to 4 km out from the starting end. The size of the circle's area is the same no matter how long it is. Therefore you have to stipulate an angle less than 90 degrees.

(I'm making up the results here because I don't have a calculator handy). But. If the angle is 89.99 degrees, H = 50 with r = 1, and
H=3 with r=1 at 80 degrees. So angle matters.

The only other way to solve it is with S having a value.

In other words without any angle and without length S:

|\ \
| \
| \
|___\ sector side = cone side
| 5 \ \
| \ cut cone side
|H \
|__10___\ / /
- A cone side surface is a Sector
- A cut cone side surface is a sector with removed inner sector.
- The sector side is the extrapolated cone side; the crossection of the
cone can be divided into 2 perpendicular triangles, with
same angles => equilateral, and ratio is 1:2 (the figure)
=> The sector side is 2 times cut cone side
=> the sector Area/cone side surface Area is 3/4 since,
<= (the ratio was 2 => the inner sector/outher sector=
1^2*pi*radius^2/2^2*pi*radius^2=1/4)
- the area of a sector is theta*radius^2/2 (theta=sector angle)
- theta = arc/radius (by definition)
=> Sector area= arc*radius/2
- the arc of the sector is the circumference of the bottom circle which is
2*10*pi=20pi
- the radius is 2 times the cone side, 2*sqr(5^2+H^2) (pythagoras)
=> The Cone side is now:
3/4 * 20pi * sqr(25+H^2) = (320-5^2-10^2)pi = 195pi (bottom and top surfaces removed)
=> sqr(25+H^2)=13
=> H=12

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I don't really understand Kedaman's post. Anyway I think you have enough information for a solution since you have the surface area. I also would guess that there are 2 solutions of H for any surface area. Perhaps a surface area of 320*pi cm² is the one value which has only one solution.

First Let us consider a Right circular cone of Height H
and Slant height L and base Radius R

Now the portion formed by the chopping of the top by a
plane parallel to the base is called the frustum of the cone.
Let the Frustum have slant height l, top surface radius r
base radius R and height h.

Now we see from the figure that

R/L = r/(L-l) and R/H = r/(H-h)

Therefore
R(L-l)=rL ==> L=Rl/(R-r)

and similarly H=Rh/(R-r)

Now..

The total surface area of a cone is

= Flat surface area + curved surface area
= pi * (base-radius)^2 + pi * base-radius * slant-height

Now the Flat surface area of the Frustum is
= Pi * r^2 + pi * R^2 (since there are two flat suraces)

The Curved surface area of the frustum

= Curved surface area of the Entire Cone - Curved surface area of the Tip cone

= pi.R.L - pi.r.(L-l)

= pi.R.L - pi.r.L + pi.r.l

= pi.L(R-r) + pi.r.l

substituting L = R.l/(R-r)

= pi.[R.l/(R-r)].(R-r) + pi.r.l

= pi.R.l + pi.r.l

Therefore the total surface area of the frustum

= Flat surface area + curved surface area
= pi.r^2 + pi.R^2 + pi.R.l + pi.r.l

T.S.Area = pi(r^2 + R^2 + l(R+r))

------

We also see that

(H-h)^2 = (L-l)^2 - r^2 (pythagoras theorem)

Substitute H = Rh/(R-r) and L = Rl(R-r) and simplify

we get h^2 = l^2 - (R-r)^2

------

Now the given problem...

T.S.Area = pi * (5^2 + 10^2 + l(10+5)) = pi * 320

==> 25 + 100 + 15(l) = 320 or

l = 13

therefore h = SQR(13^2 - (10-5)^2)

h = SQR(144) = 12

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