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Square Inside a Square


Date: 01/30/2001 at 16:58:52
From: Jon Kohn
Subject: Geometry

Thank you folks for being here!  

As a math coach, I couldn't figure out:

Imagine a square with eight compass points marked at each corner and 
midpoints of the sides. Create a smaller square inside by drawing 
these 4 lines: N-SW, NE-S, E-NW,and SE-W. How do the areas of the two 
squares compare, and why?  

Thanks again for directing me to a simple prior Dr. Math answer.


Date: 01/30/2001 at 18:52:24
From: Doctor Greenie
Subject: Re: Geometry

Hi, Jon -

The area of the small square is one-fifth the area of the large 
square.

Here's a quick intuitive way to see this (completing the mathematical 
justification for this method might take some work - I didn't pursue 
it very far...).

Draw the figure as you have described it, then mentally "cut and 
paste" the figure by cutting out the small triangular regions in the 
corners and then moving each of them by "hinging" them at points N, E, 
S, and W, and pivoting them so that, for example, the segment from N 
to NW coincides with the segment from N to NE. 

If you do this with all four triangles, you end up with a cross shape 
made up of five squares, each the size of the square whose measure you 
are looking for. Since the total area of these five squares is the 
area of the original square, the area of each small square is 
one-fifth the area of the large square.

Here's a more rigorous derivation of the answer, with a picture of one 
portion of your figure:

  A *
    *   *    D
    *       *
    *      *    *
    *     *         *
    *    *              *    E
    *   *                   *
    *  *                   *    *
    * *                   *         *
    **                   *              *
  B ***************************************** C
                        F

Let the side of the large square be 2x.

Then AB = BF = FC = x, because each is half a side of the large 
square.

Then AC = x * square root of 5 by the Pythagorean theorem.

BD and FE are perpendicular to AC, so all the triangles in the figure 
are similar triangles; and in fact triangles ADB and FEC are congruent 
(because AB = FC).

It follows that CE = ED, and that AD is one-half as long as each of 
these.

Therefore DE (the side of the small square in our problem) is two-
fifths of AC, or (2/5)*x*square root of 5.

Then the area of the small square is (4/5)x^2; while the area of the 
original square is 4x^2.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Triangles and Other Polygons
Middle School Geometry
Middle School Triangles and Other Polygons

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