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### Volume of a Trapezoidal Solid

```
Date: 11/15/2000 at 23:08:25
From: Greg
Subject: Area/volume of a trapezoid

I have a volume that is 75 ft long. The front of the figure is 57 ft
wide, 35 ft high, while the rear is 72 ft wide, 12 ft high. I know how
to figure the volume L*W*H and the area L*W, but how do I account for
the slope of the ceiling and the opposite widths?
```

```
Date: 11/16/2000 at 08:56:15
From: Doctor Peterson
Subject: Re: Area/volume of a trapezoid

Hi, Greg.

It sounds like your shape can be thought of as a 57x35 foot rectangle
at the front, joined by planes to a 72x12 foot rectangle at the back,
75 feet away. Here is a formula for the volume of that shape, which I
will draw as a frustum-like figure with rectangular top and bottom
rather than front and back (since that is the form in which I have
most often dealt with it):

a2
+---------------+
b2/               / \---------
+---------------+   \      |
/                |    \     |
/.................|.... +    | h
/                  |    /     |
/                   |   /      |
/                    |  /b1 ------
/                     | /
/                      |/
+-----------------------+
a1

V = [a1*b1 + a2*b2 + (a1*b2 + a2*b1)/2] * h/3

Here the bottom is a1 x b1, and the top is a2 x b2, with the a's
parallel and the b's parallel (this is important).

An alternative version of the formula, using the average length and
width, is:

V = [a1*b1 + a2*b2 + 4((a1+a2)/2 * (b1+b2)/2)] * h/6
\___/   \___/     \___________________/
area of  area of          area of
bottom    top             middle
rectangle rectangle       rectangle

The "middle" rectangle has sides that are the average of the sides of
the top and bottom rectangles:

a2
+---------------+
b2/               / \---------
+---------------+   +      |
/                |  / \     |
/.................|./.. +    | h
/                  |/   /     |
+-------------------+   /      |
/      (a1+a2)/2     |  /b1 ------
/                     | /
/                      |/
+-----------------------+
a1

a1 = 57
b1 = 35
a2 = 72
b2 = 12
h  = 75

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 08/27/2001 at 12:39:03
From: Jason
Subject: Gallons/inch of height for a trapezoidal solid

I have the same shape. Getting the volume is the easy part, but I need
to know if I were to fill this shape with a liquid, how many gallons
would there be per inch of height?  I understand how to do this with a
rectangular volume, but I can't get the vol/height ratio for this type
of geometric solid.

Thanks a lot!
```

```
Date: 08/27/2001 at 16:31:41
From: Doctor Peterson
Subject: Re: gallons/inch of height for a trapezoidal solid

Hi, Jason.

If by "gallons per inch of height" you just mean the volume of a full
container divided by its height, just use the formula (with dimensions
in inches) to get cubic inches, divide by 231 to get gallons, and
divide that by the height.

But that doesn't mean much. For any container other than a cylinder,
the ratio will vary with depth, and I think you want to vary the
depth. I'm going to assume that you mean you are partially filling the
container, and want to know the volume at different levels. That is,
you want the volume as a function of height.

To review the situation, here's the picture:

a2
+---------------+
b2/               / \---------
+---------------+   \      |
/                |    \     |
/.................|.... +    | h
/                  |    /     |
/                   |   /      |
/                    |  /b1 ------
/                     | /
/                      |/
+-----------------------+
a1

V = [a1*b1 + a2*b2 + (a1*b2 + a2*b1)/2] * h/3

Now, the width in the "a" direction changes linearly from a1 to a2,
making it

a = a1 + (a2-a1)k = (1-k)a1 + ka2

at kh units up from the bottom (where k varies from 0 to 1); likewise,

b = b1 + (b2-b1)k = (1-k)b1 + kb2

The volume of the solid with height kh, base a1 by b1 and top a by b
is

V(k) = [a1*b1 + a*b + a1*b/2 + a*b1/2] * kh/3

= [a1*b1 + ((1-k)a1 + ka2)((1-k)b1 + kb2) +
a1((1-k)b1 + kb2))/2 + ((1-k)a1 + ka2)b1/2] * kh/3

= [a1*b1 + (1-k)^2 a1*b1 + (1-k)ka1*b2 + (1-k)ka2*b1 +
k^2 a2*b2 + (1-k)a1*b1/2 + ka1*b2/2 + (1-k)a1*b1/2 +
ka2*b1/2] * kh/3

= [(1 + (1-k) + (1-k)^2)a1*b1 +
((1-k)k + k/2)(a1*b2 + a2*b1) +
k^2a2*b2] * kh/3

= [(3-3k+k^2)a1*b1 + k^2a2*b2 +
(3k-2k^2)(a1*b2 + a2*b1)/2] * kh/3

If we use the actual liquid level L instead of the ratio k, we can
replace

k = L/h

and get

V(L) = [(3h^2-3Lh+L^2)a1*b1 + L^2a2*b2 +
(3Lh-2L^2)(a1*b2 + a2*b1)/2] * L/(3h^2)

This gives volume as a function of depth.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry

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