cone with frustrum problem.

hey all,

so, I have encountered a problem in my studies that I am having trouble with. I was hoping the members of this community could help me out.

The problem is based on a right cone. The cone has been sliced to create a frustrum. The height of the entire unsliced cone is 20 inches, with 12 inches above the frustrum and 8 inches within the frustrum. The other piece of information that is known is the slant height of the frustrum, which is 10 inches.

No other information has been given. The problem is asking for the radius of the original cone and the volume of the frustrum.

I would also like to note that the volume of a frustrum equation has not yet been discussed in my class. So, I am under the impression that this problem can be solved without it, though I am struggling to see how.

Thanks in advance for your input.

Re: cone with frustrum problem.

Hello, hellothisismyname!

Quote:

There is a right circular cone, which has been sliced to create a frustum.

The height of the entire unsliced cone is 20 inches, with 12 inches above the frustum and 8 inches below.

The other piece of information that is known is the slant height of the frustrum, which is 10 inches.

Find the radius of the original cone and the volume of the frustum.

Consider the cross-section of the cone.

Code:

` A`

*

/|\

/ | \

/ | \

/ | \

/ |12 \

/ | \

/ | \

D * - - - + - - - * F

/ E| |\

/ |8 8| \10

/ | | \

B * - - - - - * - - - * - * C

G H

We see that: .$\displaystyle \Delta FHC \sim \Delta AEF$

We note that $\displaystyle \Delta FHC$ is a 3-4-5 right triangle.

. . Hence:.$\displaystyle HC = 6$

Since $\displaystyle \Delta AEF$ is a 3-4-5 right triangle:.$\displaystyle EF = 9$

. . Hence:. $\displaystyle GC \:=\:9+6 \:=\:15$ . (radius of original cone)

Cone $\displaystyle ABC$ has radius $\displaystyle GC = 15$ and height $\displaystyle AG = 20$

. . Hence:.$\displaystyle V_{ABC} \:=\:\tfrac{\pi}{3}(15^2)(20) \:=\:1500\pi\text{ in}^3$

Cone $\displaystyle ADF$ has radius $\displaystyle EF = 9$ and height $\displaystyle AE = 12$

. . Hence:.$\displaystyle V_{ADF} \:=\:\tfrac{\pi}{3}(9^2)(12) \:=\:324\pi\text{ in}^3$

Therefore:.$\displaystyle V_{\text{frustum}} \;=\;1500\pi - 324\pi \;=\;1176\pi\text{ in}^3$

Re: cone with frustrum problem.

alternately we have in triangle AEF and AGC

angle EAF = angle GAC also

angle AEF = angle AGC [ right angle ]

Thus we have triangle AEF similar to triangle AGC

Hence we get AE/AG = AF/AC OR 12/20 = AF / (AF + FC )

that is 3/5 = AF / (AF + 10 ) That will give AF = 15

Now we can get required lengths by using Pythagoras theorem and proceed as explained by Soroban

Re: cone with frustrum problem.

Thanks for your responses. I should have drawn the 3-dimensional shape into 2-dimensions and then I would have seen that 8 is also the length of one leg of a right triangle. Since it was the altitude in the drawing, it wasn't clear that it could also represent the right-triangle leg.

Re: cone with frustrum problem.

also, looking at my work I did get the original radius to be 15 in, but the answer key in my book says it is 20in. huh? perhaps the book is just wrong in this instance.