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### 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!
```
Associated Topics:
High School Triangles and Other Polygons

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