I have an odd shaped room that I need to buy floor covering for. The materials I want to use are expensive so I don’t want to buy more than I really need. Is there an easy way to figure the approximate area of the floor?

The dimensions of the room are 9’6" X 6’3" X 3’4" X 9’0" X 6’3" X 1’6" X 5’6".

Any polygon can be broken up into triangles. I don’t know what you would consider “easy”, however, but the area of any triangle is 0.5 X base X height. Get out a tape measure and a calculator.

To give a simpler example, if you have a room with four sides, all measuring 10 feet–without knowing the angles, all that you know is that it’s between 0 sq ft (a “degenerate” square–two 180º angles, two 0º angles), and 100 sq ft (a perfect square, four 90º angles).

Yes, that would help greatly. With that, the co-ordinates of the corners can be mapped onto a Cartesian system. From there, triangular sections would be taken and their areas calculated. Adding those together gives you the total area.

Most floor coverings are sold in rectangular rolls. If your corners aren’t 90[sup]o[/sup] as you’ve indicated, you’ll have to do some cutting to get pieces to fit. Also, it’s very unlikely that you’ll get an exact fit; i.e., if your heptagonal room happens to have an area of 100 ft[sup]2[/sup], buying a 10 ft x 10 ft piece is not going to guarantee that you can cover your floor.

It’s rather like making a dress from a Simplicity[sub]TM[/sub] pattern. They tell you how to lay out the cutout patterns on the fabric of your choice is such a way that the patterns (if any) on the fabric will join together well. That is, they tell you how to cut the pieces so that you don’t end up with horizontal stripes on the front and vertical stripes on the back. Additionally, all the littler pieces (sleeves, bows, pockets) are laid out as well. These are also mapped out to get the most efficient use of your fabric. But there is always a sizable amount of excess.

Now a floor isn’t as complex as a dress pattern, but I imagine that 5-15% excess might be required if your room is very irregular.

Quite an irregular heptagonal-shaped room you have there.

Without the angles of the vertices (scitalk for corners), there really is no way to figure out the area of your room.

And if you could give the angles in degrees, I think I can figure out the area of your room (rough) in about five minutes. Scientific calculator on my PC and all:)

This may or may not help, since it only works for “lattice polygons”, i.e., polygons such that you can set up a square grid so that each corner of the polygon falls on a corner of the grid; I figured I’d give it a link anyway, 'cause it’s cool.

The planimeter is a drafting tool used by architects and realtors to measure the area of any flat shape, even if has curved sides. However, it relies on an accurate drawing of the area.

You might get the most accurate estimate by covering the area in some inexpensive material like newspaper and measuring that.

Don’t know whether my suggestion might help but it’s an idea:
If you are good in computer programming you could do a computersimulation:
(BTW this method works for ANY flat surface even curved and
irregular);
Place the surface on an imaginary first quadrant; imagine
a unit square around to fit the polygon.
Let’s say in your case a square of 10mx10m. If the unit square fits, this means that your surface is less than 10x10m=100m2. Now, you could make a program that does the following:
Generate 2 random numbers, representing a coordinate X and Y
on the first quadrant with a maximum value of 10. The point
generated is in the polygon or not. Count the successes.
If you divide the successes through the trials, you have an estimate of the surface (e.g. if you have 77% success in our
case of a unit square of 100m2 this means the surface is approx. 77m2. More trials give more accurate information.

You shoyuld be able to use Pick’s theorem as shown. The lattice can really be as close together as you please.

Of course it is easier just to divide it into triangles. You don’t need the angles if you are willing to measure the “third” side to each triangle and use Heron’s theorem (see below) but otherwise you do need a third variable for each triangle (any angle or side). You could also use trigonometry.

(Heron’s Law, IIRC:
A triangle of sides S1, S2, S3 has a perimeter of S1+S2+S3
if s=perimeter/2, area A=sqrt[(S)(S-S1)(S-S2)(S-S3)]

(Trigonometry, a triangle of sides a, b and c has angles A, B and C that are opposite the given side.

A= 1/2 bc sin A = 1/2 ab sin C = 1/2 ac sin B

(This is essentially the same as 1/2bh but slightly easier in practice).

A circle encloses the maximum area for a given perimeter. So the room can’t be larger than 136 square feet. If it were perfectly square the area would be 107 square feet.

So you’re probably looking at more than one square of materials. (One square is 100 square feet.) Understand the room’s odd dimensions will give you some wastage. A lot depends on what material you plan on using. If you’re just laying down plywood, get 4 4x8 sheets and start cutting. Carpet I think is sold in squares so you may be stuck with buying 2. Tile would be easy too; you can generally return unused tile, and a single box doesn’t represent a huge investment.