The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

A 'Pyramiddle' Tent Problem

Date: 07/12/99 at 17:13:46
From: Hank Hufnagel
Subject: Algebra


I am a scoutmaster, and, in the course of designing an old-fashioned 
kind of tent, I have bumped into an algebra problem that is beyond 
me. You can find details at:

Hank's Tent Problem   

"...Making liberal use of the Pythagorean Theorem, the four right 
triangles give the equations:

   x^2 = d^2 + h^2
   y^2 = d^2 + 4^2
   z^2 = (r-d)^2 + (h-d)^2
   y^2 = x^2 + z^2

"The idea was then to solve for d in terms of h and r. This would let 
me design a pyramiddle tent from nearly any rectangular tarp...

"Can you solve the problem? Just figure out an equation that yields d 
when values for h and r are inserted."

It would be great if you or someone you know could crack this for me.

Hank Hufnagel

Date: 07/13/99 at 15:31:38
From: Doctor Rick
Subject: Re: Algebra

Hi, Hank.

That's a nice problem! A practical real-life problem with some 
interesting geometry and plenty of Pythagorean Theorem exercise, plus 
solution of multiple simultaneous equations and solution of a 
quadratic. I hope math teachers will get to use it!

Here is how to solve the problem. Your 4 equations are:

(1)  x^2 = d^2 + h^2
(2)  y^2 = d^2 + r^2
(3)  z^2 = (r-d)^2 + (h-d)^2
(4)  y^2 = x^2 + z^2

With r and h given, we have 4 unknowns: d, x, y, and z. First 
eliminate y by equating (2) and (4):

(5) d^2 + r^2 = x^2 + z^2

Next, use (1) to substitute for x^2 in (5):

    d^2 + r^2 = d^2 + h^2 + z^2

Subtract d^2 from both sides:

    r^2 = h^2 + z^2

Now we can solve for z (which is an important quantity, the depth of 
the tent) in terms of the given r and h:

(6) z^2 = r^2 - h^2

Use (6) to substitute for z^2 in (3):

    r^2 - h^2 = (r-d)^2 + (h-d)^2

    r^2 - h^2 = r^2 - 2rd + d^2 + h^2 - 2hd + d^2

    d^2 - 2d(r+h) + h^2 = 0

Now we have a quadratic equation. We can use the quadratic formula to 
solve it:

    d = (r + h +- sqrt((r+h)^2 - 4h^2))/2

Since d can't be greater than r or h, only the negative sign makes 
sense. We can rewrite the square root a little, and we get:

    d = (r+h - sqrt((r+3h)(r-h)))/2

There is your formula; it gives the results you have in your table.

- Doctor Rick, The Math Forum   
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Practical Geometry
High School Triangles and Other Polygons

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.