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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.
hi jjoy
Welcome to the forum.
The algebra is messy but it should work. Sorry, I cannot find a simpler method.
Two tangents to a circle from a point will be equal.
Then the small (removed) cone will be in the same proportion as the large cone. Hence:
I don't have a simple solution to these, but someone else on the forum will probably put the equations through a computer for us and come up with values for x and y.
Then it should be easy (?) to compute the volume.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi jjoy and Bob
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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Thanks Stefy.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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No problem.
The volume is now easy to compute.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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I got 182pi, is that right?
Girls can be just as good as boys at math. We just need to get the same encouragement.
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hi championmathgirl
it is not what I have just calculated. As y is not a whole number, I would not expect the volume to be either.
What did you get for the height and radius of the two cones (starting large cone and the one that's cut away)?
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi Bob;
I think that championmathgirl's answer is correct.
I understand how you got your equations, but haven't worked out yet how stefy arrived at his solutions for x and y. However, they check out in Geogebra.
In the volume formula below I used the following (borrowing some info from your drawing):
The formula for the volume of a conical frustum is:
So...
Last edited by phrontister (2015-07-18 10:09:14)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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hi phrontister
Thank you. I decided to work it out on paper this time, and I get 182pi as well. Whoops!
I shall see if I can get an algebraic solution to the equations as a penance.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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substitute into the third expression (I'll make an equation later):
simplifying and equating to the 'y' expression:
Rejecting the second solution as y cannot be negative in this problem, we have y = 5/4
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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I used a simpler algebra way to find x. Because you could use Pythagorean Theorem and write the base as 16+2x. Then since we also have the base as 20-2x. Setting them equal to each other we get x=1. Then use the ratios y/(y+10)=x(10-x) to find y.
Girls can be just as good as boys at math. We just need to get the same encouragement.
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Hi;
In the image I added DM to create right-angled triangle DMA (for which the lengths of DA and DM are known) in order to obtain the length of AM (by Pythagoras).
Last edited by phrontister (2017-02-27 00:32:04)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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championmathgirl. You are indeed a champion. That's a much better way. I award you this:
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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I used a simpler algebra way to find x. Because you could use Pythagorean Theorem and write the base as 16+2x. Then since we also have the base as 20-2x. Setting them equal to each other we get x=1. Then use the ratios y/(y+10)=x(10-x) to find y.
I don't know which base you guys are talking about. Also, is x the radius of the inscribed circle?
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I understand how you got your equations, but haven't worked out yet how stefy arrived at his solutions for x and y
Here is one way:
FindInstance[
y/(10 + y) == x/(10 - x) == Sqrt[-x^2 + y^2]/(
6 + Sqrt[-x^2 + y^2]), {x, y}, Reals, 5]
And then a tiny ansatz is required.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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I don't know which base you guys are talking about. Also, is x the radius of the inscribed circle?
Don't forget this is a frustum. That is, you start with a big cone, chop off a small cone at the top, and the frustum is what remains. In post 2 I have made a diagram that shows a vertical cross section through the middle of the solids. The red outline is the frustum. It has a circular top radius x and a circular base, radius to be found.
The sphere fits inside the frustum and touches the top and bottom circles and also the sides. The sphere appears in the diagram as a circle because the cross section cuts through the middle of the sphere. It has a radius of 3.
The sloping side of the frustum is 10.
If a pair of tangents are drawn from a point to a circle they will have equal length. To prove this draw a line from the point to the centre of the circle, making two triangles. It is fairly straight forward to show that these triangles are congruent.
So the top radius, x, is repeated down the sloping side, and hence the lower distance is 10-x. By the same tangent length rule this distance is repeated across the bottom. So the base radius is 10-x.
championmathgirl's method is to drop a perpendicular from the top radius down to the bottom making a right angled triangle. It has sides of 10, 10-x-x, and 6. You can use Pythagoras to work out x.
I like to work out the volume of a frustum by calculating the volumes of the large and small cones, and then subtracting.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Oh, so the cross-section is vertical? And that tiny triangle is the top part of the cone?____________________________________________________________________________________________________________________________________
How did you get the base as 20-2x?
____________________________________________________________________________________________________________________________________How do you know that JD=x?
Last edited by RandomPieKevin (2015-07-21 05:44:33)
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Yes and yes.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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How did you get the base as 20-2x?
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Wait, I got 20pi... It's wrong! I got 1/4\pi as the volume of the top and 81/4\pi as the volume of the whole thing.
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How do you know that JD=x?
From a point you can make two tangents to a circle. The lengths of these two tangents (from point to circle) are equal.
How did you get the base as 20-2x?
Using the same rule AJ = AH = 10 - x so the base is AB = 20 - 2x
In triangle DAM, DM = FH = 6 and DA = 10, therefore AM = 8.
AH = 10 - x and also x + 8 => 10 - x = x + 8 => x = 1
Therefore AH = 9.
So the large cone is 9x as big as the small cone.
If the height of the small cone = h, then 9 times small height = large height => 9h = 6 + h => h = 3/4
So small cone
and large cone
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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How do we know that AH=AJ and EF=DM/AM?
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Can you see AGD is a right angled triangle?Reason AG and DG are angle bisectors of angles A & D whose sum is 180.?
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Cut the frustum and inscribed sphere with a vertical plane passing through the center of the sphere and perpendicular to the two bases.
Let the slant height
and the radius of the circleLet
since and , Since . From draw a perpendicular to base at point , then is a right triangle, where and . Since the radius of the smaller base we choose the volume of the frustumOffline
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Last edited by Tindela (2018-01-19 06:17:47)
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