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Is it length by width or is there more to it???

2007-01-24 14:45:40 · 10 answers · asked by jessica m 1 in Home & Garden Do It Yourself (DIY)

10 answers

Yes it is usually length x width, unless you have a weird area in your room, for instance, its L shaped, then you would break down the room into smaller rectangles/squares, where you would find the area of each smaller area then add the areas together in order to get the total room area.

2007-01-24 20:58:58 · answer #1 · answered by fidel410 5 · 0 0

Sure. You can take the dimensions in inches and multiply them together to get the area in square inches, then divide that number by 144 square inches per square foot to find how many square feet you have. Don't forget, however, that in installing a floor there is always waste and you will need more wood that just the square footage of the floor. Whoever sells you the floor will help with that. They'll also help with the math.

2016-05-24 06:19:42 · answer #2 · answered by Anonymous · 0 0

yea, it's just length by width a.k.a. base x height. The square footage of a room is the room's area, and lxw is the formula to find the area.

2007-01-24 14:57:16 · answer #3 · answered by 19 year old girl 2 · 0 0

Length by width is correct.

2007-01-24 14:56:27 · answer #4 · answered by Anonymous · 0 0

Length x width. A 12x12 room would be 144 sq. ft.

2007-01-24 14:51:38 · answer #5 · answered by mom of 2 5 · 0 0

multiply the length times the width in feet

2007-01-24 14:53:28 · answer #6 · answered by Anonymous · 0 0

Yes it is width times length. finished basement area may not be included in selling a home.

2007-01-24 14:57:27 · answer #7 · answered by blue_eagle74 4 · 0 0

That's all there is kido, just mulitply the width by the length.

2007-01-24 14:51:39 · answer #8 · answered by MT C 6 · 0 0

x times y

x = L
Y = W

= square foot.. :) gotta trust ur instinct sometimes lol funny question though..

2007-01-24 14:55:08 · answer #9 · answered by Vu 3 · 0 0

That is it!

2007-01-24 14:49:24 · answer #10 · answered by Jim C 6 · 0 0

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