Square Inside a SquareDate: 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/ |
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