Carpeting an Irregular Hexagon
Date: 07/31/2003 at 23:54:07 From: Jamin Hoffman Subject: Area and angles of an irregular hexagon My office is an irregular hexagon. Side A is 10'2"; side B is 2'5"; side C is 6'4", side D is 9'10", side E is 3'0", and side F is 5'10". Is there a way to determine (1) the area of the office, given only the lengths of these sides, and (2) the angles of each of the six corners? Thanks - this is purely a practical question (I want to carpet it, and build some shelves that will fit into some of the corners). I cut some sticks to scale with the side lengths, and put them together to measure the angles - but am I correct in assuming that these sides represent a unique hexagon? Also, since there are no parallel walls, it's hard to make a determination as to the area using the squares plus triangles method (though, in thinking about it, it IS do-able). More thinking (and browsing through similar answers on your site) suggests that this is NOT a unique hexagon, and there may be many answers to this question.
Date: 08/01/2003 at 12:02:36 From: Doctor Rick Subject: Re: Area and angles of an irregular hexagon Hi, Jamin. If you played enough with the sticks, that should convince you that the lengths alone do not define a unique hexagon. If you can't directly measure the angles, perhaps you could measure some diagonals of the room. For instance, if you measure from the first corner to the third, fourth, and fifth corners, this would give us enough information to define a unique hexagon. We could then use the law of cosines to find the angles, and Heron's formula for the area of a triangle to find the area of the room. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Date: 08/01/2003 at 19:26:21 From: Jamin Hoffman Subject: Area and angles of an irregular hexagon Thank you for the information and suggestion! Since I am NOT a mathematician (but a lowly orchestra conductor), can you refresh my memory on the law of cosines and Heron's formula? Thanks again - Jamin Hoffman
Date: 08/01/2003 at 20:47:43 From: Doctor Rick Subject: Re: Area and angles of an irregular hexagon Hi, Jamin. You could use those two phrases to do searches of the Dr. Math Archives. http://mathforum.org/library/drmath/mathgrepform.html In brief, here's what to do. First, to find an angle of a triangle, knowing the lengths of the three sides. I'll call the sides a, b, and c, and the angle opposite side a is angle A. The Law of Cosines says a^2 = b^2 + c^2 - 2*b*c*cos(A) Solving for A, we have A = arccos((b^2 + c^2 - a^2)/(2*b*c)) For example, suppose the first diagonal of your room is 11' 3", so we have a triangle with sides a = 11' 3" = 11' + 3/12' = 11.25' b = 10' 2" = 10' + 2/12' = 10.1667' c = 2' 5" = 2' + 5/12' = 2.4167' Note that I am renaming the sides so that angle A is one of the angles of the hexagon, namely the angle between sides b and c (your A and B). Now, we can calculate a^2 = 11.25*11.25 = 126.56 b^2 = 10.1667*10.1667 = 103.36 c^2 = 2.4167*2.4167 = 5.8404 2*b*c = 2*10.1667*2.4167 = 49.140 (b^2 + c^2 - a^2)/(2*b*c) = (103.36 + 5.8404 - 126.56)/49.140 = -0.353268 arccos(-0.353268) = 110.687 degrees The last step may be unfamiliar to you. On the Windows calculator, with 0.353268 showing in the display, you click the Inv box, then the cos button, to get this function (the arccosine, or inverse cosine - the angle whose cosine is 0.353268). If you measure diagonals from one corner as I suggested, only two of the angles of the room can be found directly in this manner; the other angles will be the sums of angles of the triangles that meet in one corner. Or you could measure each diagonal that cuts off one corner of the room, and use it to find the angle that it cut off. To find the area of a triangle, use Heron's formula: Area = sqrt(s(s-a)(s-b)(s-c)) where s = (a+b+c)/2 which is called the semiperimeter of the triangle (half the perimeter). For the first triangle, we have s = (11.25 + 10.1667 + 2.4167)/2 = 11.9167 s-a = 11.9167 - 11.25 = 0.6667 s-b = 11.9167 - 10.1667 = 1.75 s-c = 11.9167 - 2.4167 = 9.5 s(s-a)(s-b)(s-c) = 11.9167*0.6667*1.75*9.5 = 132.083 and the square root of this is 11.4927. This is the area of the triangle in square feet. If you divide the room into triangles as I suggested, and measure the sides of these triangles, you can calculate the area of each triangle and add them up to get the area of the room. I hope this helps! Is your office in an odd corner of a concert hall? It's certainly an odd shape. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Date: 08/02/2003 at 14:05:58 From: Jamin Hoffman Subject: Thank you (Area and angles of an irregular hexagon) Thank you again for your help - and, yes, you deduced correctly. My office is in an odd corner, tucked between two rehearsal areas, an instrument storage area, and the music library. Thanks again - I think I have enough information to get what I need now!
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.