ABSOLUTELY STUCK!

Re: ABSOLUTELY STUCK!
You don't need any ideas for the problem, you already have that. What you need is an answer and that I will not give. (sounds suspiciously homeworky)

nuh uh sturgis.

I think he's making a tomb for his beloved pet dung beetle, Ramses LXXVII, who has crossed the river styx.

I built a pyramid of beer bottles once, but I didn't measure it.

But, in the spirit of being helpful, the river Styx is part of the Greek mythos and the pyramid is clearly Egyptian, or possibly Mesoamerican.

It's an important symbol in James Joyce. I think it's Finnegan but his stuff is one long blur. It's for something feminine. Some tin-pot Goddess or other. The Bass bottle has it on the label. Or it used too.

well didn't the egyptians cross some kind of river?

maybe the volga?

Egyptians often buried their dead on the west bank of the Nile due to a belief that the underworld was located to the west where the sun set, if that's what you mean.

Or it might've been the Seine, I forget.

I grew up on the Manasquan River, and sometimes a dead fish could be seen bumping up against a piling.

I think that's kind of a greek/egyptian thing. We would poke it with styx.



don't mess around with dead jellyfish though. nooooooo sir, don't do that.

You'd poke a dead fish with a defunct 70s rock band? Have you no shame?

Defunct?

they didn't smell that bad.

you should look over your lecnotes again
- what is the height of a pyramid onto a base
- pythagore theorem

good luck

Since the altitudes of the three faces are the same, think about where the top vertex must be relative to the base triangle. In other words, if you were to look down on the pyramid, viewing a triangle with the top vertex as a point inside the triangle, where inside the triangle does that point reside?

The way to work out the height is to determine the mid-point of the base with said lateral sides. Why the mid-point?

Imagine this pyramid was thin sheet metal and you folded it out so it was flat. Your shape would be a central triangle - with sides lengths you know, and three isoceles triangles off each side of the central triangle - with side lengths each 24".

Now what happens when these three outside triangles are folded over so that there points meet to form the apex of the pyramid? For each triangle, pivoting or rotating along its base line what path does the point of the the connected traingle traverse? Image the Sun was overhead and you tracked the shadow of the point of each traingle as you folded it in - one at a time.

Being isoceles (i.e. for the vertical triangles forming the sides of the pyramid each side is 24") - so the apex of the pyramid must lie along the interone intersection of three lines, where each line must bisect at 90 degrees - exactly dividing the three edges of known size of the central triangle.

Let me say that again. The apex of the pyramid, must be at at point above the central triangle where a line bisecting each of your 3 known edges at 90 degrees meets.

Hence this point can be found and its distance from each point of the central base calculated. You can also then calculate this points distance from the mid point of each side! So now you will have two right angled triangles with a common height - a simple equation to solve to find the common height!

Alternately you could work out the distance of the base's mid points (call it M) from its the corners call them A, B and C and solve

24^2 - AM^2 = height ^2
24^2 - BM^2 = height ^2
24^2 - CM^2 = height ^2

So this hints that to be consistent the distance AM = BM = CM so its the midpoint of your triangle with three known sides. Surely you can find that? Second hint if you draw a lines from A, B or C through M - must not they respectively exactly bisect sides BC, AC, and AB?

So M must be half the length of a line coming from point A to the midpoint of side BC.

If your still stuck - check your notes on Oblique triangles e.g. http://www.clarku.edu/~djoyce/trig/oblique.html

Let AC = 25, BC = 50 and AB = 39. Now as 50 = 25 * 2, bisect BC to get a triangle with two sides equal to 25, and its hight will be your M!

39^2 = 50^2 + 25^2 - 2*25*50*Cos C.

This will give you an isoceles triangle with two sides equal to 25"and a known angle, so you can work out the base. The height = M^ = 25^2 - (1/2 * base)^2, and the base / 25 = sin (1/2 C)

g__day:

"The base edges of a triangular pyramid are 25 inches, 39 inches, and 50
inches long, and all three lateral faces are 24 inches in height. Find the height of this pyramid."

The triangular lateral faces have heights of 24, not sides of length 24.

The "midpoint" you refer to is the circumcenter. As you have pointed out, it is the point where the perpendicular bisectors of the sides meet. If the sides of the lateral faces are all 24, then the apex is directly above the circumcenter.

However, since the faces have heights of 24, the apex is directly above the incenter. The incenter is the point where the angle bisectors meet. Key to this problem is the fact that the incenter is the point where equal length perpendiculars to the base triangle's sides meet.

I'm not quite sure of the meaning the "faces" are of lateral height 24" vs the edges of each 3 triangles. Do they mean they given you the distance from the mid point of each base edge to the apex, not the distance from each base corner to the apex?

Nice catch if I made that overly complex.

This won't change the placement of the mid-point on the base, if you have read this correctly it merely simplifies finding the height.

Looking at the dimensions, I just realized that the sides can't be 24. You can't make a triangle (lateral face) from sides of length 50, 24, and 24.

"This won't change the placement of the mid-point on the base."

Sure it will. The base triangle is obtuse. The circumcenter (your midpoint) is outside the base triangle.

Not unless its curved! Good catch, well that makes the problem even easier!