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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 162

A rectangle with height 8 and length 24 is wrapped around a cylinder with height 8. The rectangle perfectly covers the curved surface of the cylinder without overlapping itself at all. What is the volume of the cylinder?

A plane intersects a sphere, forming a circle that has area 24pi. If this plane is 5 units from the center of the sphere, then what is the surface area of the sphere?

A sphere is inscribed in a cylinder so that it is tangent to both bases of the cylinder, and tangent to the curved surface of the cylinder all the way around. If the volume of the cylinder is 54pi, then what is the volume of the sphere?

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,469

A rectangle with height 8 and length 24 is wrapped around a cylinder with height 8. The rectangle perfectly covers the curved surface of the cylinder without overlapping itself at all. What is the volume of the cylinder?

A plane intersects a sphere, forming a circle that has area 24pi. If this plane is 5 units from the center of the sphere, then what is the surface area of the sphere?

A sphere is inscribed in a cylinder so that it is tangent to both bases of the cylinder, and tangent to the curved surface of the cylinder all the way around. If the volume of the cylinder is 54pi, then what is the volume of the sphere?

I'll give you a brief outline of a workable method for each of these. After you've given them a try, post back with answers for checking or ask for more help.

(i) The 24 will become the circumference of the cylinder. So you can work out the radius and then the volume.

(ii) Use the 24 pi to work out the radius of the circle. Then use Pythagoras to work out the radius of the sphere.

(iii) The diameter of the cylinder must equal the height. So you can work out the radius from this.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 162

i am stuck on visualiztion for (ii) and i think (i) is 1152/pi

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 162

also, is (iii) 36pi? the radius is 3.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,469

hi cooljackiec,

Excellent answers for (i) and (iii). I agree with both of these.

Diagram for (ii) below. I'm showing the sphere from the side so that you are looking across the plane of the circle. Then it appears as a line.

Little r is the radius of that circle, and big R is the radius of the sphere.

Hope that helps,

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 162

yea my dad said it was easier viewing it in 2D with the plane as a line. then the pythag made radii 7. then it was easy

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**bob bundy****Moderator**- Registered: 2010-06-20
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I didn't get 7. What did you have for little r ?

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 162

my radius of the sphere was 7. what is the little r??

my traingle was a 5-2sqrt6-7. the answer was 196pi

can you help on these two??

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,607

7 is correct for the sphere radius.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,469

Whoops, my mistake. I was working with **circumference** = 24 pi. That leads to r = 12 and a 'nice' Pythagorean triple so I didn't think I had made an error. With **area** = 24 pi I also get R=7.

Should I risk looking at those two problems ???

later edit:

First one: You can use pythag to calculate AF, FH and HA. Heron's formula will give you the area of AFH.

http://en.wikipedia.org/wiki/Herons_formula

Take that as the base of the pyramid. The height will be half of CE. The volume = one third base area times height.

second one. I said earlier in your other thread that the small tetrahedron would be half the size of the large one. I now think that was wrong. Sorry.

Diagram below. Extend CF and BG to meet AD at P. These lines are medians for the triangular faces. So F will divide CP in the ratio 2:1 and similarly G.

So PFG and PCB are similar triangles with FG parallel to CB. And the ratio of sides will be 1:3

So that gives the length scale factor for the small:large tetrahedrons. To get the ratio of volumes you cube the length SF.

Bob

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 162

hmm, it seems i got the answer by subtracting the volume of the prism from the 4 pyramid shaped things. Thanks anyways!

*Last edited by cooljackiec (2013-07-11 05:02:27)*

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