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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.

post-9379-1221076809_thumbgif

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

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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

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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 by Chuck Rampart

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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 by vimil

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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.

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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 by woon

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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! B))

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.

post-9379-1221108304_thumbgif

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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 by vimil

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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?

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