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### Constructing a Pyramid

```
Date: 05/28/99 at 10:09:24
From: daniel
Subject: Constructing angles for a pyramid

I am trying to build a 6 ft. pyramid out of copper. I have an 8 ft.
base but I am having trouble getting the right angle, 51 deg 51 sec,
for the height I want. I tried cutting the four 8 ft. sections to
exactly 6 ft. but that doesn't work.
```

```
Date: 05/28/99 at 16:58:45
From: Doctor Peterson
Subject: Re: Constructing angles for a pyramid

Hi, Daniel. Sounds like a big project!

I'm not sure what your 51 degree angle is, but here's how I understand
the dimensions you gave: base 8 x 8 feet, height 6 feet.

*
/|\
/  | \\
/   ||  \ \
/     ||   \  \
/      | |    \   \
/        | |6     \   +
/         |  |   /    \  \
/           |/ |         \  \
/       /    |   |           \ \
/   /          |   |            \ \
+               |    |             \ \
\              |    |               \\
\            |     +                \\
\           |      \                 \
\         |        \                 \
\        |         \                 +
8 \      |           \           /
\     |            \     /
\   |              +
\  |         /     8
\|     /
+

We can figure out the dimensions of each of the four side panels
(faces) by a series of right triangles. First we look at the triangle
formed by the altitude of the triangle (6), the perpendicular from the
foot of the altitude to one side of the base (4), and the slanted
altitude of the face (a):

*
/|\
/  | \\
/   ||| \ \
/     |||  \  \
/      | ||   \   \
/        | | |    \   +
/         | 6| | /    \  \
/           |/ | |       \  \
/       /    |   |  |        \ \
/   /          |   |  |a        \ \
+               |    |  |          \ \
\              |    |   |           \\
\            |     +   |            \\
\           |      \  |              \
\         |        \  |              \
\        |        4\ |               +
8 \      |           \|          /
\     |            \|    /
\   |              +
\  |         /     8
\|     /
+

Using the Pythagorean Theorem,

a^2 = 6^2 + 4^2 = 36 + 16 = 52

so

a = sqrt(52) = 7.2111 feet.

Now we can use the right triangle formed by half of this face:

+
/|\
/ | \
/  |  \
/   |   \
/    |    \
/     |     \
/      |      \
s/       |       \s
/        |        \
/         |         \
/          |a         \
/           |           \
/            |            \
/             |             \
/              |              \
/A      4       |       4      A\
+----------------+----------------+

We can find the length of the sides of this triangle (s) by the
Pythagorean Theorem again:

s^2 = 4^2 + a^2 = 16 + 52 = 68

so

s = sqrt(68) = 8.25 feet.

If you want the angle A, we can find that using trigonometry:

A = arctan(a/4) = arctan(1.8) = 60.98 degrees

Since I don't see any 51 degree angles, I may be misinterpreting what
you want to do, but this should at least give you some ideas for how
to do it. If you need more help, let me know where I'm wrong so I can
try again.

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

```
Date: 06/12/99 at 07:14:50
From: daniel f gordon jr
Subject: Re: Constructing angles for a pyramid

Thank you very much. It helped a lot.
```
Associated Topics:
High School Geometry
High School Polyhedra
High School Practical Geometry

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