[Prime]

A trilateral pyramid has all three angles at the apex equal to 90 degrees each:

AHB=AHC=BHC=90;

The side edges are equal a, b, and c:

AH = a; BH = b; CH = c.

Find the volume of the pyramid.

If you recall, volume of a pyramid is equal to one third of its height (dropped from the apex to the base) times the area of the base.

I remember this one from 9th or 10th grade, when math teacher gave it to the class and 10 minutes to solve it

[HoustonHokie]

To find the volume, you need two pieces of information - the area of the base and the height of the pyramid.

To get the area of the base, start by observing that each of the 3 upright triangles are 45-45-90 isosceles right triangles. So if we assign a value of X to the equal length sides a, b, & c, we can say that the length of the third side is sqrt(X²+X²), which is approximately 1.41X. The base is an equilateral triangle, and it is easy to calculate that the area of the base is approximately 0.866X units².

To obtain the height of the pyramid, construct a new triangle with points H, A, and the midpoint between B & C (call it D). The length of HA is known (X), and we can calculate the other two lengths based on right triangles (HD = 0.707X and AD = 1.224X). Calculating the distance of a line from H perpendicular to AD will give us the height of the pyramid, which is 0.577X.

So the volume is (1/3)*(0.866X units²)*(0.577X units) = 0.167X units³

[Guest]

Without looking at HH's answer:

Wouldn't that shape just be 1/2 the volume of a rectangular prism with sides a, b, and c?

So the volume is abc/2.

EDIT: after looking at it some more, I'm pretty sure this is wrong. Oh well.

EDIT: in spoiler

Edited September 10, 2008 by Chuck Rampart

[Guest]

The image looks like a cube sawed of along its longest diagonal. If that is the case shouldnt the volume of the pyramid be a³/2 where a is the length of the side edge.

Edited September 10, 2008 by vimil

[Guest]

Is that really a pyramid? - I thought a pyramid had a base of four sides...

[Guest]

Haha- "Where is John Galt?" I like it.

[Prime]

To find the volume, you need two pieces of information - the area of the base and the height of the pyramid.

To get the area of the base, start by observing that each of the 3 upright triangles are 45-45-90 isosceles right triangles. So if we assign a value of X to the equal length sides a, b, & c, we can say that the length of the third side is sqrt(X²+X²), which is approximately 1.41X. The base is an equilateral triangle, and it is easy to calculate that the area of the base is approximately 0.866X units².

To obtain the height of the pyramid, construct a new triangle with points H, A, and the midpoint between B & C (call it D). The length of HA is known (X), and we can calculate the other two lengths based on right triangles (HD = 0.707X and AD = 1.224X). Calculating the distance of a line from H perpendicular to AD will give us the height of the pyramid, which is 0.577X.

So the volume is (1/3)*(0.866X units²)*(0.577X units) = 0.167X units³

Where did isosceles triangles come from? a, b, and c could be three entirely different numbers.

I did mention, the teacher gave only 10 minutes to solve this problem.

[Guest]

Almost been fooled!

When you rotate 3-dimensionally the pyramid to let the ABH be the base,

It is exactly part of a cuboid with the base axb and height = c.

So, my answer is V = (1/3) * (ab/2) * c = (abc)/6

Edited September 11, 2008 by woon

[Prime]

Almost been fooled!

When you rotate 3-dimensionally the pyramid to let the ABH be the base,

It is exactly part of a cuboid with the base axb and height = c.

So, my answer is V = (1/3) * (ab/2) * c = (abc)/6

PRECISE ANSWER!

I can only add an illustration:

If you flip the pyramid on its side, one of the side edges becomes the height, and the other two -- the legs of a right angle triangle at the base.

[Guest]

I think 4 of these pyramids can be joined together to form a cuboid. So the volume of the pyramid must be abc/4

Edited September 11, 2008 by vimil

[Guest]

I think 4 of these pyramids can be joined together to form a cuboid. So the volume of the pyramid must be abc/4

If you try to arrange four of them to make a rectangular prism, it looks pretty good. If you put the four right-angled corners at four opposite corners of the 3D shape, you can cover the whole surface area. You leave a void in the center, though (whose volume is, apparently, twice that of the original pyramid.

Is there a way to assemble six of these to make one abc rectangular prism?