Prime 15 Posted September 10, 2008 Report Share Posted September 10, 2008 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 Quote Link to post Share on other sites

0 HoustonHokie 0 Posted September 10, 2008 Report Share Posted September 10, 2008 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(X^{2}+X^{2}), 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 units^{2}. 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 units^{2})*(0.577X units) = 0.167X units^{3} Quote Link to post Share on other sites

0 Guest Posted September 10, 2008 Report Share Posted September 10, 2008 (edited) 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 Quote Link to post Share on other sites

0 Guest Posted September 10, 2008 Report Share Posted September 10, 2008 (edited) The image looks like a cube sawed of along its longest diagonal. If that is the case shouldnt the volume of the pyramid be a^{3}/2 where a is the length of the side edge. Edited September 10, 2008 by vimil Quote Link to post Share on other sites

0 Guest Posted September 10, 2008 Report Share Posted September 10, 2008 Is that really a pyramid? - I thought a pyramid had a base of four sides... Quote Link to post Share on other sites

0 Guest Posted September 10, 2008 Report Share Posted September 10, 2008 Haha- "Where is John Galt?" I like it. Quote Link to post Share on other sites

0 Prime 15 Posted September 10, 2008 Author Report Share Posted September 10, 2008 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(X^{2}+X^{2}), 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 units^{2}. 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 units^{2})*(0.577X units) = 0.167X units^{3} 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. Quote Link to post Share on other sites

0 Guest Posted September 11, 2008 Report Share Posted September 11, 2008 (edited) 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 Quote Link to post Share on other sites

0 Prime 15 Posted September 11, 2008 Author Report Share Posted September 11, 2008 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. Quote Link to post Share on other sites

0 Guest Posted September 11, 2008 Report Share Posted September 11, 2008 (edited) 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 Quote Link to post Share on other sites

0 Guest Posted September 11, 2008 Report Share Posted September 11, 2008 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? Quote Link to post Share on other sites

## Question

## Prime 15

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

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