Associated Topics || Dr. Math Home || Search Dr. Math

### 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

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

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