Date: Tue, 22 Sep 1998 10:31:32 -0600 Sender: CURTIS CONSTRUCTION Subject: hexagon
Could someone calculate this for me?
Thank you, Hi Lee, |

No doubt we are looking at a regular hexagon (which is composed of 6 equilateral triangles sharing one vertex at the centre of the hexagon).The diagonal of a regular hexagon, coloured red in the diagram, is twice the side length, so one possible answer to the question is that the sides are 19'4" long. | ||

If, however, the width refers to the distance between parallel sides, labeled W in the diagram, consider the triangle shown with sides of length W, L and 2L. This is a right angled triangle so the theorem of Pythagoras insures that
Solving this equation for L gives In your case W=38.666... feet so L is approximately 38.666/1.732 = 22.3242 feet which is approximately 22 feet 4 inches.
Cheers, |

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