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Sat 23 Sep, 2006 05:07 pm
Hey!!
I have three Geometry questions and would very much appreciate any help/suggestions!
1. A pyramid of volume 56 cc is cut by a plane that is parallel to the base of the pyramiud. The cutting plane bisects the altitude of the pyramid. What is the volume of the frustum?
2. A regular hexagon with edges of length 12 cm is the base of a pyramid. The combined area of the lateral faces of the pyramid is 480 square cm.
a) What is the volume of the pyramid.
b) To the nearest tenth of a degree, calculate the size of the dihedral angle formed by the hexagonal base and a lateral face of the pyramid.
3. What is the area of a regular hexagon with edges of length 12 cm?
Thank you,
Tom
I also want to let everyone here know that I'm not lazy and I'm not just trying to get you to do my homework! These three questions are from a set of 30, and I really have NO idea how to set up any of them, especially the frustum question.
Google search immediately gave
V=1/3 Base area x Height.
1. Frustrum height reduces to 1/2 therefore volume reduces to 1/8 by similar figure ratios.
2. Use pythagoras to work out missing "heights" of equilateral and isosceles triangles, then use trig to find angle. Note that for the base you are dealing with the standard ratios for a 30-60-90 triangle i.e. sides 1:2:sqrt(3)/ The slanting isosceles triangles height comes from the area of one of the faces. Hence the height of the pyramid.
3. is part of ans 2 i.e. 6 triangles each area 36 sqrt (3)
Can you explain what you mean by, "Frustrum height reduces to 1/2 therefore volume reduces to 1/8 by similar figure ratios?" How did you come up with 1/8?
I don't understand?
A property of similar figures is that
If Ratio of corresponding lengths = a:b
Then Ratio of areas = a squared: b squared
& Ratio of volumes = a cubed: b cubed
so Heights 2:1 => Volumes 8:1
(You can prove this to yourself by playing with the dimensions of a cuboid)
So just to clarify....the answer would be that the volume of the frustum is 7
HELP!!!!!!! I still really don't understand problem #2. Can someone tell me the answer or go through it more in depth actually using the numbers. I really don't get it!
Thanks!
aspiring
You seem to be unfamiliar with solid geometry. Its really a matter of drawing different planes and using elementary pythagoras and trigonometry. Giving you the answer won't help you.
1.Find the area of the base which is six equilateral trangles with height 6sqrt (3) by pythag.
2. Find the perpendicular of a slanting face isosceles triangle from one sixth of the given area by making p the subject of the triangle area.
3. Draw the diagram for the vertical triangle through the centre of the pyramid. By pythagoras use p as the hypotenuse to get height of pyramid (half triangle base = 6 sqrt(3) from 1 above). Use trig to get elevation of slant side
4. Use volume formula
Can we do it the other way round?
You start on it, and then tell us, where you get stuck.
Is this school work?
Then I am sure, you must know the basic geometric formulae.
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