bigoldalphamale

ok, i am trying to measure the square footage of some office space my company is leasing. one room in particular which has an irregular shape. here are the specs:

wall 1: 53"
wall 2: 64"
wall 3: 90"
wall 4: 122"
wall 5: 95"
wall 6: 37"
wall 7: 93"
wall 8: 131"
ceiling: 96"

if anyone has a formula for calculating the square footage of this area, please lay it on me. i have tried to recall my formulas for finding the area of irregular polygonal shaped containers, but i keep getting ridiculous anwers. any help would be appreciated.

Sticky

Can you post a drawing of the layout?

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Dragonlich

I don't know of any way to calculate this stuff for an irregular shaped polygon, but it might be possible... However, if it is impossible, how's about trying to divide it into squares and triangles? Draw a scale map, use one wall as base line, move lines from there, with 90 degree angles...

Or it may be easier to surround the shape with a square or box, and subtracting the area outside.

Oh, you do need the angles of the walls for this.

(hmm... in theory, you could flood the entire room with water, measure how many gallons you put in, and compensate for the height of the room... )

hrdwareguy

Dragonlich beat me to it.

Make a square/rectangle and find the area of that. Then find the area of the remaining triangles. Since the walls aren't all the same length you may have to make further square and triangle divisions.

When done, add up all the individual areas to get the total area.

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platypus

You need a pencil, paper, measuring tape, calculator, and a penny.

Since the 8 sides of the room are all different lengths, no set of opposite sides will be parallel.
The above encompassing square solution would not be accurate.
However, it may be accurate enough for your purposes.

But, if you want to be exact...

Place penny on the floor in the approximate center of the room.
Drawing the layout as you go, measure the distance from the
penny to each corner of the room. You already know the lengths
of the side walls so this will give you the dimensions of 8 triangles.

Use Heron's formula to calculate the area of each triangle.

Given the sides of a triangle A,B,C first calculate the semi-perimeter S.

S = (A + B + C)/2

The area of the triangle is the square root of (S(S - A)(S - B)(S - C))

Now add together the areas of the 8 triangles. This gives you the area for the irregular octagon room.

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Dragonlich

Platypus' method wins, although I prefer my "fill with water" idea.

etla

If you can supply the layout and which walls are at 90 degrees I could fire up AutoCAD and possibly give you an answer.

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bigoldalphamale

thanks guys! your ideas worked and i got it figured out!

platypus

Actually I prefer the "Fill with Jello" idea.

Turn the room into one giant jello mold. 'Cuz, hey, everybody's got room for jello.

__________________
Out the 10Base-T, through the router, down the T1, over the leased line, off the bridge, past the firewall...nothing but Net.