The Math Forum

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

Cross-Cornering a Shape to Make it Square

Date: 12/19/2002 at 22:52:31
From: Tim Barker
Subject: Cross cornering a shape to make it square

At my work we lay out buildings. Lets say we lay out a building that 
is 30' x 40'. When you cross corner it to see if it is "square" there 
is a 6" difference. What is the equation to find how far to move one 
side to make the shape "square"? 

Date: 12/20/2002 at 10:41:47
From: Doctor Rick
Subject: Re: Cross cornering a shape to make it square

Hi, Tim.

This is a good question! It will take a lot of math to derive the 
answer; feel free to skip down to the end to see the formula I come up 

In geometric terms, if you've measured so all the sides are the 
correct length (opposite sides are the same length), then you have a 
parallelogram, and you want to know how far the angles are from right 
angles, based on the difference in the diagonals.

We can use the formulas for parallelograms in the Dr. Math FAQ: 

There we find formulas for the diagonals p and q in terms of angles A 
and B. (Diagonal p joins the vertices with angle B, and diagonal q 
joins the vertices with angle A; a and b are the two side lengths.)

  p = sqrt[a^2+b^2-2ab cos(A)]
  q = sqrt[a^2+b^2-2ab cos(B)]

We know that angles A and B are supplementary, that is, their sum is 
180 degrees. Thus B = 180-A, and cos(B) = cos(180-A) = -cos(A). Now we 
can express the difference in the diagonals in terms of A:

  q-p = sqrt[a^2+b^2+2ab cos(A)] - sqrt[a^2+b^2-2ab cos(A)]

It's quite a chore to solve this equation for the unknown angle A. But 
I'll assume that the difference is small, like 6 inches out of 50 
feet. Then the angle is close to 90 degrees, and the cosine of the 
angle is approximately equal to the difference between A and 90 
degrees, converted to radians. (180 degrees equal pi radians, that is, 
3.14159... radians.) Calling this angle difference x, the equation 

  q-p = sqrt(a^2+b^2+2abx) - sqrt(a^2+b^2-2abx)

It's still tough, but I can make another approximation. Since x is 
small, I can factor out the term (a^2+b^2):

  q-p = sqrt(a^2+b^2)(sqrt(1+2abx/(a^2+b^2))-sqrt(1-2abx/(a^2+b^2)

and approximate sqrt(1+e) as (1+e/2) where e = +-2abx/(a^2+b^2) is a 
small quantity:

  q-p = sqrt(a^2+b^2)(1+abx/(a^2+b^2)-1+abx/(a^2+b^2))
      = sqrt(a^2+b^2)*2abx/(a^2+b^2)
      = 2abx/sqrt(a^2+b^2)

Now we can solve for x:

  x = (q-p)sqrt(a^b+b^2)/(2ab)

There's one more step. You probably aren't interested in the angle, 
but in how far you need to move one end of a side. Let's say you want 
to move side b. The distance you need to move it is b times the angle 
in radians, which is x. Thus

  distance = bx = (q-p)*sqrt(a^2+b^2)/(2a)

A final simplification uses the fact that sqrt(a^2+b^2) is pretty 
close to the average of the diagonals, (p+q)/2.

  distance = (q-p)((p+q)/2)/(2a)


In words: Multiply the difference in the diagonals by the average of 
the diagonals. Then divide by the twice the length of the side you're 
not moving. This is the distance to move. If the diagonal from the 
stationary end of this side is the longer diagonal, you'll want to 
move the side out (increasing angle A); if this diagonal is the 
shorter diagonal, move the side in.

In your example, if you want to move side b:

  side a = 40 feet
  side b = 30 feet
  difference of diagonals, q-p = 6 inches = 0.5 foot
  average of diagonals, (p+q)/2 = 50 feet
  distance to move = 0.5 * 50 / 80
                   = 0.3125 feet
                   = 3.75 inches

I'll just check this result. If I was 0.3125 feet off, the angle was 
off by (0.3125/30)*180/pi = 0.5968 degrees; the angles in the 
parallelogram were A = 89.4032 and B = 90.5968 degrees. Using the 
formulas for the diagonals, I calculate p = 49.749 feet and 
q = 50.249 feet. The difference in the diagonals is thus 0.500 foot. 
The formula worked.

- Doctor Rick, The Math Forum 
Associated Topics:
College Geometry
College Triangles and Other Polygons
High School 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.