## Answer: Volume of a Pyramid

The problem asked for the volume of a pyramid given the lengths of its six sides. “Pyramid”, by the way, was supposed to mean this thing:

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Let’s start with the simpler problem of finding the area of a triangle given the lengths of its sides. Triangles look like this:
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The area is given by:

$Area = \frac{1}{2}base*height$.

Let the base of the triangle in the above figure be $c$. By definition of the sine function,

$\sin(\alpha) = \frac{h}{b}$

with $h$ the height of the triangle.

Plugging in these values of the base and height, we get

$Area = \frac{1}{2}cb\sin{\alpha}$.

The law of cosines provides the missing information on $\alpha$.

$b^2 + c^2 - 2bc\cos{\alpha} = a^2$.

Solving for $\sin(\alpha)$ and plugging into the area formula, we get

$Area = \frac{1}{4}\sqrt{2(a^2b^2 + b^2c^2 + c^2a^2) - (a^4 + b^4 + c^4)}$.

The internet informs me that this result is known as “Heron’s Formula”. I don’t know who Heron was, but I’m guessing a Babylonian with a long thin neck, webbed feet, and a predilection towards fish.

The dimensions are correct – they’re $length^2$. Also, $a, b,$and $c$ enter into the formula symmetrically.

You can only form a triangle if the shortest side is greater than difference of the other two. That is, any two sides added together need to be longer than the third side. Imagine that isn’t true for some $a, b, c$. We can’t form a triangle with little sticks of those lengths, but we can blindly plug the numbers into the formula. Imagine the sides are listed in increasing length. Then let

$c = a + b + \Delta$.

Substituting into Heron’s Formula gives

$Area = \frac{\Delta}{4} \sqrt{-\left[\Delta^2 + 4(\Delta* a + \Delta* b + a^2 + b^2) \right]}$.

If $\Delta >0$, we get an imaginary area.

The partial derivative of the area with respect to $a$ is

$\frac{a(b^2+c^2-a^2)}{8*area}$.

This is zero when

$a^2 = b^2 + c^2$,

so that a right triangle will not change its area if you make the hypotenuse slightly shorter or longer. If you lengthen one of the shorter legs of any triangle, you always increase the area. If you lengthen the longest leg, you increase the area of an acute triangle and decrease the area of an obtuse triangle.

Imagine the perimeter of the triangle is fixed, so that

$a + b + c = P$.

We want to maximize the area. The derivative of the area with respect to $a$ is now

$\frac{d}{da}(A) = \frac{\partial A}{\partial a} + \frac{\partial A}{\partial b} \frac{db}{da} + \frac{\partial A}{\partial c} \frac{dc}{da}$.

The restriction of constant perimeter means

$\frac{db}{da} + \frac{dc}{da} = -1$.

If the area is extremized, the derivative with respect to the length of any side should be zero, no matter which of the other two sides I chose to shorten to compensate. So first I let

$\frac{db}{da} = 0, \frac{dc}{da} = -1$,

then

$\frac{db}{da} = -1, \frac{dc}{da} = 0$.

The algebra then yields

$a(b^2 + c^2 - a^2) = c(a^2 + b^2 - c^2) = b(a^2 + c^2 - b^2),$

which is true only for

$a = b = c = P/3$.

So an equilateral triangle encloses the greatest area for a given perimeter, as you would expect.

The volume of the pyramid is given by

$V = \frac{1}{3} base*height$.

The factor of $\frac{1}{3}$ can be verified with a bit of calculus. Imagine calculating the volume by integrating over slices parallel to the base. Here’s a picture from Wikipedia. It’s only a little bit relevant, but then again I include these visual aids not so much to help you understand the problem as to mock the blind.

At a given height $x$ beneath the tip of the pyramid, the width and height of the triangle with both be proportional to $x$, meaning the area is proportional to $x^2$. Then the volume of the pyramid is

$\int_0^h dx \, A*\left(\frac{x}{h}\right)^2 = \frac{1}{3} A*h$.

We know the area of the base $A$ because we got it from a Sumerian avian a few paragraphs ago. Now we just need the height. It’s not pretty, but here’s a way to do it.

Look down on the pyramid from above, like this:

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Call those three edges that go from the base to the tip $d, e,$and $f$. What we’re looking at in the picture is not their true lengths, since they really pop up out of the screen. Instead, we see their projected lengths, $D, E,$and $F$. ($D$ is shorter than $d$. The capital is the length projected into the plane of the base, while the lower case is the true length.)

Each of these sides and projected sides, along with the altitude of the pyramid, form a right triangle. (Take the altitude to be the segment from the tip to the base, and perpendicular to the base. Since it’s perpendicular to the base, it must be perpendicular to the projected sides in the base, and therefore forms right triangles with them.) The Pythagorean theorem now gives three relations between the height $h$ and the lengths of the sides and their projections.

$d^2 = D^2 + h^2$.

$e^2 = E^2 + h^2$.

$f^2 = F^2 + h^2$.

The outside edges of that figure are the lengths of the sides of the base, $a, b,$ and $c$. If you imagine $a$ forms a triangle with $D$ and $E$, $b$ with $E$ and $F$, and $c$ with $F$ and $D$, we can start using the law of cosines again. Letting $\alpha$ be the angle between $D$ and $E$, etc., we get

$a^2 = D^2 + E^2 - 2DE\cos(\alpha)$.

$b^2 = E^2 + F^2 - 2EF\cos(\beta)$.

$c^2 = F^2 + D^2 - 2DF\cos(\gamma)$.

Finally,

$\alpha + \beta + \gamma = 2\pi$.

That gives seven equations for seven unknowns $(D, E, F, \alpha, \beta, \gamma, h)$. Solving for $h$ and plugging into the volume formula, along with the area of the base gives (eventually),

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Actually, I just got that from Wikipedia. The algebra was too nasty. I lose.

### 2 Responses to “Answer: Volume of a Pyramid”

1. WW Says:

Which article is that from?

2. Mark Eichenlaub Says:

http://en.wikipedia.org/wiki/Tetrahedron