How to measure square footage for a room with odd wall sizes?

[DrPizza]

Yes, there are formulas for the areas of irregular polygons. Since you're having trouble just breaking it into rectangles and triangles and finding the sums of the areas, I highly doubt you'd be able to use the formulas. In fact, most of those formulas are merely novelties. The only one's I've seen that are of any use are Heron's formula (given the lengths of the three sides, and no angles), and the formula for a convex quadrilateral (I've forgotten the name of the formula - if it even has a name.) - although that one requires one angle. Also, if you're familiar with linear algebra, and can use a coordinate system to define all the corners in your room, (here you go, I googled for an explanation of this one Although, I'm not certain your situation would allow the use of that particular method, since you haven't given a diagram, and the page I linked to is for a convex polygon. (I teach that formula in my pre-calculus class, but actually, I've never done an example for something other than a triangle.)

Regardless, It's pretty pathetic if you're having trouble breaking the area up into manageable pieces - rectangles, triangles, and such - and summing those areas. I teach remedial level 9th grade mathematics. Even my worst students can manage that task. EVEN when they miss the class where I explain the method, the students who were absent seem to be able to figure it out on their own.

 

[skyking]

With all angles 90, I tend to use the maximum dimensions of the rectangle that encompasses the entire region and subtract the cutouts. That seems to go the fastest.

 

[robisc]

Regardless, It's pretty pathetic if you're having trouble breaking the area up into manageable pieces - rectangles, triangles, and such - and summing those areas. I teach remedial level 9th grade mathematics. Even my worst students can manage that task. EVEN when they miss the class where I explain the method, the students who were absent seem to be able to figure it out on their own.

I will repeat again, I have the area and did so by the method just as you mentioned, breaking the room into rectangles for each area, and I even had to do some subtraction, boy that "taking away" was hard to do (sarcasm) even though 9th grade math was way back when even prior to kollige and my gradeat studies. I was simply asking if there was a formula to do so, as I didn't have the time to google, or draw everyhing out on graph paper and such. Finally thanks for the convex polygon formula you linked to as I did in fact figger it out and was able to use it, such a fomula is what I wanted.

 

[Indolent]

I was simply asking if there was a formula to do so, as I didn't have the time to google, or draw everyhing out on graph paper and such.

Well it seems you obviously have enough time to argue about it on an internet forum. :roll:

Using rectangles and triangles it would take a whole 30 seconds if you have all the dimensions.

 

[BlueFlamme]

Originally posted by: skyking
With all angles 90, I tend to use the maximum dimensions of the rectangle that encompasses the entire region and subtract the cutouts. That seems to go the fastest.

Agreed. Start with the footprint that captures all of it and start reducing.

 

[robisc]

I do now have the time to post, I didn't then. But what really ticks me off about this issue is all the smart remarks I get when if one would only just read then they would see what I originally asked for before they opened their mouths or rather their keyboards. I asked for a formula which I did eventually get, so now everybody just shut up....please : )

Well it seems you obviously have enough time to argue about it on an internet forum.

 

[Howard]

I can only shake my head in disbelief.

 

[Ausm]

Use trig

Or do what "Most" real Estate agents do who don't have enough brains to pound sand.

Make a box over the general area and call that close enough


Ausm

 

[MikeyIs4Dcats]

Originally posted by: robisc

Here it is if those numbers are in order. 518.96 ft² If the room is square (probably not), then the last leg is 3.2', equalling 520.24 ft².

Your diagram is pretty close and the numbers are close to mine.

And for the

OP, this guy is trying harder than you are. You suck.

No, you suck. I already have the dimensions now as mentioned in my last post, I just wondered about an easier way to figure it, guess it can;t be done.

Thanks for all the constructive replies.

it's pretty fvcking easy the way you did it.

 

[Armitage]

Originally posted by: robisc

Regardless, It's pretty pathetic if you're having trouble breaking the area up into manageable pieces - rectangles, triangles, and such - and summing those areas. I teach remedial level 9th grade mathematics. Even my worst students can manage that task. EVEN when they miss the class where I explain the method, the students who were absent seem to be able to figure it out on their own.

I will repeat again, I have the area and did so by the method just as you mentioned, breaking the room into rectangles for each area, and I even had to do some subtraction, boy that "taking away" was hard to do (sarcasm) even though 9th grade math was way back when even prior to kollige and my gradeat studies. I was simply asking if there was a formula to do so, as I didn't have the time to google, or draw everyhing out on graph paper and such. Finally thanks for the convex polygon formula you linked to as I did in fact figger it out and was able to use it, such a fomula is what I wanted.

If edro's diagram really is "pretty close", then this formula isn't applicable as your room is not a convex polygon.

 

[DrPizza]

Originally posted by: robisc

Regardless, It's pretty pathetic if you're having trouble breaking the area up into manageable pieces - rectangles, triangles, and such - and summing those areas. I teach remedial level 9th grade mathematics. Even my worst students can manage that task. EVEN when they miss the class where I explain the method, the students who were absent seem to be able to figure it out on their own.

I will repeat again, I have the area and did so by the method just as you mentioned, breaking the room into rectangles for each area, and I even had to do some subtraction, boy that "taking away" was hard to do (sarcasm) even though 9th grade math was way back when even prior to kollige and my gradeat studies. I was simply asking if there was a formula to do so, as I didn't have the time to google, or draw everyhing out on graph paper and such. Finally thanks for the convex polygon formula you linked to as I did in fact figger it out and was able to use it, such a fomula is what I wanted.

You should have just let it rest... but, sarcastic replies must be responded to...

What I need to know is how to figure a room with many different cuts and differing wall lengths, also there is more than 4 wall, anyone?

Doesn't sound like you already had the answer in that post... First response: "Break it up into rectangles."

Your response:

that would work but there must be a formula right?

(Looking for an *easier??!* method?)

I noticed the word "would." You didn't reply with, "I already did that and got an answer of 523.60 square feet, I was just wondering if there was a formula to do it all at once."

There's no reason whatsoever from your first 2 posts in this thread to believe that you were able to find the area without help from this thread.

Finally thanks for the convex polygon formula you linked to as I did in fact figger it out and was able to use it, such a fomula is what I wanted.

I'm glad that formula helped you. I haven't attempted to apply the formula to your room; and I doubt it would have worked in your situation, as your room apparently *isn't* a convex polygon. Yet, that formula is little more than finding the area of even more rectangles than you could have broken the original problem into.

For what it's worth, I just shrug my shoulders at people posting questions such as yours. I realize that we live in a pretty illiterate (strike that word out... wrong word) innumerate society. Most people can't function mathematically beyond 8th or 9th grade level (the average person in this forum not withstanding - the people here tend to be better educated.) Apparently your "kollige" and "gradeat" school did nothing to help the problem. What a shame.

 

[robisc]

you should have just let it rest... but, sarcastic replies must be responded to..

Which is why I will respond one last time to this ridiculous thread that I started but unfortunately went down the wrong track?

Doesn't sound like you already had the answer in that post... First response: "Break it up into rectangles." Your response:

I didn't have the answer then did I say I did? And I was still asking for what? A formula.

(Looking for an *easier??!* method?)

Yes easier and quicker, what is the issue with that? If you have a deadline and need an answer quickly, one that could possibly be solved with a TI-81 then why have to draw out rectangles?

There's no reason whatsoever from your first 2 posts in this thread to believe that you were able to find the area without help from this thread.

Your point being what? I asked a question regarding a formula and finally came up with an answer on graph paper while waiting to see if a formula exists, help from this thread is what I asked for and was thankful to those that responded with help.

I'm glad that formula helped you. I haven't attempted to apply the formula to your room; and I doubt it would have worked in your situation, as your room apparently *isn't* a convex polygon. Yet, that formula is little more than finding the area of even more rectangles than you could have broken the original problem into.

Thanks again for the formula, it kind of helped, my room isn't exactly square as I already mentioned, I used the formula on a portion of the room and it did in fact work (close answer for that portion), maybe it would be more useful in a different situation but it did help, as I checked my work with it more to see if it would in fact work.

For what it's worth, I just shrug my shoulders at people posting questions such as yours. I realize that we live in a pretty illiterate (strike that word out... wrong word) innumerate society. Most people can't function mathematically beyond 8th or 9th grade level (the average person in this forum not withstanding - the people here tend to be better educated.) Apparently your "kollige" and "gradeat" school did nothing to help the problem. What a shame.

What, asking for a formula is innumerate? I'm sure if you are in fact a professor then even you too don't remember all formulas for all problems, or do? If so great, but my studies aren't and weren't in math, but I do quite well in career with my knowledge of all "standard" math issues and problems, since my job or life doesn?t on a regular basis require me to have much use for the Pythagorean theorem. As for my college again I ask were my studies supposed to provide me with the formula I asked for, and if so should I have been able to retrieve it from the back of my head at a moment's notice? Oh and FWIW I would hate to have had a professor like yourself if the way you respond to a question is by incriminating the person asking.

We'll I guess now I'll get back to counting these toothpicks and using my fingers and just keep practicing my match skills so to never have to ask a math related question again.

 

[PlatinumGold]

Originally posted by: edro
Here it is if those numbers are in order.

518.96 ft²

If the room is square (probably not), then the last leg is 3.2', equalling 520.24 ft².

very nice.