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Volume of a Pyramid


Date: 5/20/96 at 17:56:33
From: Anonymous
Subject: Geometry

Hi. I have a problem that I'm trying to solve. 

The problem reads: All six edges of a triangular pyramid are 4 inches 
long. Find the volume of the pyramid.

I can't figure out a way to find the altitude.

I tried the Pythagorean theorem by creating a rectangle. I used four 
as the hypotenuse and half of the bottom edge which would be 2. That 
is what I am not sure about. I came up with the answer of (4/3) times 
the square root of 13. Is that right?
                                        Thanks,
                                              Tim


Date: 7/17/98 at 17:20:53
From: Doctor Ken
Subject: Re: Geometry

[Note: one of our volunteers answered this question, but his answer
contained a mathematical error, and was incorrect.  A while later, a person
reading our archives wrote to us to point out the error, and gave us a
correct solution.  This solution follows.]

The volume of a tetrahedron in which all six edge lengths are 
represented by x is [(x^3)/12]*sqrt(2).  If the edge lengths are 
6, then the volume is 18*sqrt(2).  In the given problem the edge 
lengths are 4.  Therefore the volume should be [16/3]*sqrt(2).  I 
will begin by proving the statement in the first sentence of this 
paragraph.

Call the length of an edge of the regular tetrahedron x.
Volume = (1/3)*[Area of base]*[Height]  (for any pyramid)

First we'll find the area of the base.

Base has altitude = (x/2)*sqrt(3)       (it's a 30-60-90 triangle)
Area of base = (1/2)*[edge]*[altitude]  (formula for area of a triangle)
Area of base = (1/2)*x*(x/2)*sqrt(3)    (substitution)
Area of base = [(x^2)/4]*sqrt(3)        (simplification)

Notice that the Area of the base has a factor of sqrt(3).

The center of base divides the altitude into two segments, whose lengths are
in a 2 to 1 ratio.  Therefore, the distance from the center of base to any
vertex of the base is 2/3 of one of the base's altitudes, or
(2/3)*(x/2)*sqrt(3) = x/sqrt(3).

Thus:
(Height)^2 = x^2 - [x/sqrt(3)]^2        (by Pythagorean Theorem)
(Height)^2 = x^2 - [x^2]/3              (apply the square to the last term)
(Height)^2 = [x^2]*[1 - (1/3)]          (factor out x^2)
(Height)^2 = [2/3]*[x^2]                (simplify)
Height = [sqrt(2)/sqrt(3)]*x            (take square root of both sides)

Notice that the Height has a denominator of sqrt(3).

The two instances of sqrt(3) should disappear in the computation.

Volume = (1/3)*[Area of base]*[Height]
Volume = (1/3)*[(x^2/4)*sqrt(3)]*[(sqrt(2)/sqrt(3))*x]
Volume = [(x^3)/12]*sqrt(2)

Thus, the volume of the pyramid in your problem is [16/3]*sqrt(2).


Robert H. Becker
rbecker@writeme.com
    
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry

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