# Thread: Math theory of area calculation

1. ## Math theory of area calculation

So I recently did a lot of area calculations at school and I’ve been wondering…

Since the only way to find out how big an area of a number of square meters really is, is to reverse the formula and calculate it back to meters, why is the area calculation not an average calculating operation? I mean wouldn’t it make so much more sense when speaking of, for example, 5 m˛, if both sides were 5 m? (which would be 1/2*5+5)
As a bonus it would make more sense in relation to other numbers because if you write 2˛ and you mean 4, you don’t write 4˛, either…

Have you ever been wondering the same?

2. Imagine a square with length of side . Using your idea would say that the area is . Now imagine cutting the square in half twice. You now have 4 squares, each with length of side , which means in your system there would now be 4 squares with area . What is the total area of the 4 small squares?

3. Originally Posted by Dirac
Imagine a square with length of side . Using your idea would say that the area is . Now imagine cutting the square in half twice. You now have 4 squares, each with length of side , which means in your system there would now be 4 squares with area . What is the total area of the 4 small squares?
It would be 16m^2, wouldn't it? Nothing got added or taken away from the squares to begin with. It takes up the same amount of space, it's just chopped up. The square with 2m sides is actually 4m^2. If you have 4 of them, and add them together, you get 16m^2. Also, if you visualize all the squares put together, the size of the big square is 4m by 4m, which gets you 16m^2.

Now, if we were talking about the surface area, and treating the perimeter of the squares as the surface area, then we'd have a different answer.

4. Originally Posted by msg_v2
It would be 16m^2, wouldn't it? Nothing got added or taken away from the squares to begin with. It takes up the same amount of space, it's just chopped up. The square with 2m sides is actually 4m^2. If you have 4 of them, and add them together, you get 16m^2. Also, if you visualize all the squares put together, the size of the big square is 4m by 4m, which gets you 16m^2.

Now, if we were talking about the surface area, and treating the perimeter of the squares as the surface area, then we'd have a different answer.
Yes it would be in reality, but not if you use the system proposed by the OP, hence the question.

5. I think Dirac needs to teach a lesson on the mathematical proof technique of reductio ad absurdum!

6. Originally Posted by ferrus
I think Dirac needs to teach a lesson on the mathematical proof technique of reductio ad absurdum!
Actually, I'd be delighted if you could give me a lesson.

7. Originally Posted by msg_v2
Actually, I'd be delighted if you could give me a lesson.

8. Originally Posted by ferrus
Well, looks like somebody needs to pay the troll toll.

9. I think the OP's confusion may stem from a kind of 'bracketing' mistake when hearing areas. It is wrong to think of as . Instead think of it as , as in, there are five squared-meters, not five-meters squared.

Does that makes sense?

10. Originally Posted by Dirac
I think the OP's confusion may stem from a kind of 'bracketing' mistake when hearing areas. It is wrong to think of as . Instead think of it as , as in, there are five squared-meters, not five-meters squared.

Does that makes sense?
I've been confused a lot by the absence of brackets in some calculations that were shown to me, so yeah, this does makes a lot more sense with them, and is basically what I've been trying to say.

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