Definition of Area of a Figure

Your first statement.
The area of a plane figure is defined as (perpendicular length * width)
is wrong.
Go and sit on the naughty step and think about it.
Del

What I mean is a shape like a rectangle,square. A=l*w.

A point has no dimensions.

A line has one dimension, length, but no thickness.

A rectangle or square or similar shape has two dimensions, but no thickness in the third dimension.

A solid has all three dimensions that a human(and any living being) can see.

A temporal solid is one which exists over time and can be seen in all its glorious four dimensions by a time traveller. However, i believe that there is a fifth dimension, which even i cannot see (or imagine).

Thanks kvsridhar, but what about my areas?!

Ah..sorry.

The product of length and width is called AREA.

The product of length, width and thickness is called VOLUME.

They are just names. Like length is a name. You can call it any name you like. Others won't understand though.

As you are not a temporal traveller, you can't know the four-dimensioned one.

kvsridhar,

when I was young, it puzzled me as to how people knew the names of inanimate things, the things couldn't tell them,,,so how did they know they were right, the adults assured me that they absolutely were the correct names.

....A rose, by any other name.....

It's not the names one needs to grapple with, but the time....

jt.

If you found one Euro in my left pocket and another Euro in my Right pocket, what would you have?

I guess, " Somebody's else's Trousers?"

Thanks to all, but what I mean is: what is the interpretation of area?

Is it a product of two values? Does it represent just space (not volume). Does the product have a physical interpretation? I mean what is the proof of the "concept" that A=l*w?

Consider this scenario: a string is stretched out in the shape of a square. If the same string is re-stretched into a rectangle, the area is different. If it is stretched into a triangle again the area is different and so on for all different 2D shapes.

Similarly: if you had say a pound of funny putty. If it were shaped like a sphere it would have a specific volume. If shaped into a square, the volume is same, if shaped like a cylinder, the volume is same. So how does reshaping a two dimensional figure cause it to "lose" or "gain" area; but in the other case, the volume or "space occupied is constant." How is this possible?

You're confusing area with perimeter. The perimeter (length) of the string stays the same, while the area changes. If you form the string into a loop, then pull the string tight using a pair of toothpicks the area will be almost zero, but the perimeter remains the same.

Conversely, the volume of your funny putty remains the same while the exposed surface changes depending on the shape. You could flatten it as thin as you like and get an enormous surface area, but the volume remains the same. Conceivably, you could insert a straw into and it an blow it into a balloon having an enormous volume; the surface of the balloon would also be enormous; but the volume of the funny putty itself would remain the same.

That's just part of the fun of math and geometry.

The concept of 'length' sort-of depends on the idea of a specific unit length, like 1 centimeter. By marking off 1-cm lengths along your string, you can get the total length compared to that initial unit.

Likewise with area. Define a unit area of 1cm x 1cm where the two unit lengths are perpendicular to each other (length x width). You can mark off a series of unit areas all over the area you want to measure and find out how much the total area is, compared to that unit area.

Why are the two lengths perpendicular? Well, theoretically, they don't have to be; you could use non-orthogonal axes -- but that complicates the math. The world we live in seems Euclidian over any reasonable area or volume you want to measure, so using axes at right angles seems to just work right.

Have you done integral equations? You have a line of infinitetismal small width (0) let's say across the "w" dimension and you will find many of them across the "l" dimension. When you sum up all those w's across 0 to "l" you get the entire area which is w x l. If you integrate w dw across l you also will get l x w.

You are looking at this from the wrong end. we needed and invented the words and definitions to facilitate life. We needed to define how much rope, we invented length. We needed to define an amount of flooring for a house, we invented "area".

I wonder what will happen when you get to questioning momentum, kinetic energy, and inertia, etc.

Usbport: you correctly said it: "Define a unit area of 1cm..." what is the logical reasoning behind this definition? Why do we compare how many 1cmxcm exists in an area? I mean it could have been any arbitrary choice a cm x a cm? What I fail to se is why volume (of funny putty) can stay constant, but area changes? Do we need to look at things from a 4th, or 5th dimension! I mean a 3D volume is constant, but a 2D area changes? Sorry to sound like I am negating your answers, but I greatly appreciate your input. This area thing just baffled me!

To Phys: thanks for that. But even in calculus, the smallest area is considered to be a rectangle dx * y or something similar, which is a reference to area as axb. Which comes back to assuming this is the formula for area (product of the two dimensions).

To passintongreen: i think you hit the nail on the head. We invented the fundamental basics to deals with the large wide scale applications.

You seem to continually shoot yourself in the foot. (or are having us on?)

If the volume is fixed, it's fixed; whatever shape you make with it.

If the area is fixed, it's fixed, and can also take many shapes? No dis-similarity.

jt

How do you get Picachu on a bus? - Poke him on!

To usbport: yes, that is an interesting analogy. However, note that in the coins case, the coins themselves hold their individual shapes. So their area is indeed unchanged. However, with a string, the "shape: of the area changes! With coins, we are just rearranging their geometry using the discrete (quantized?) area of a coin as the unit for measurement. What would happen of the coins were made of some plastic/flexible material, so they could stretch and contact? What would be the effect then? Hey a new thought: are we indeed quantizing area? Planck's length at play? Are we on the brink of a new discovery?

Okay, maybe here's the breakthrough: Think of each coin as being composed of even smaller coins with sticky edges, and these mini-coins are as tiny as you want them to be even to the point where the edges of these ultra-tiny coins are invisible. So each large coin has a fixed area of tiny coins held next to each other by sticky edges. So you can change the shapes of the coins as much as you like, but the total area remains fixed.

The point is, you need to chose which property is fixed and which one is malleable.

For example, if you take a sheet of rubber, you can stretch it various ways and get different areas. You could measure just the edges and get different perimeters. But the volume of the rubber sheet is fixed.

Perhaps you'd be interested in a book on topology. You might check Amazon.com, and see if there are any popular books on topology and pick one with good reviews.

Actually everything in math beyond small discrete numbers like one or two (zero is negotiable), that your eye can "see" and differentiate at one time is an abstraction, i.e. a reality logic representation (and generalization) effort in your brain. As things get more complicated, so does the abstraction, and we can fool ourselves about getting "answers" and reasoning on the "roots" of that abstraction (worse as you move up the ladder) but it can be nothing but axiomatic. S.M.

Looks like someone needs to go back to Grade School and relearn the basics.....sit up and listen to the teacher when they discuss the basics....

Some simple examples may illustrate:

A 2"x8" rectangle has a perimeter of 20" and contains 16 1" squares (area = 16).
A 3"x7" rectangle has a perimeter of 20" and contains 21 1" squares (area = 21).
A 4"x6" rectangle has a perimeter of 20" and contains 24 1" squares (area = 24).
A 5"x5" rectangle has a perimeter of 20" and contains 25 1" squares (area = 25).

Giving credit to mathewkyle who seems to be dabbling in advanced mathematics, (Planck's line?), all the responses seem well below his/her intellectual level. What he/she wants to know perhaps is how a few lines can define an area when they themselves have no dimensions. Hmm. i need to speak to some politicians in Bangalore. They have managed to acquire large areas of real estate without toeing a single (legal/ethical/moral) line.

In English language prepositions, above ≠ below.

Was opening a new line of thought. Below ? Above? someone's got a few mix-ups!

The shortest perimeter with the largest area is a circle, this can be seen with a drop of oil in water. The oil flattens out into a plane in the shape of a circle. The surface tension pulls on the perimeter squeezing the area into an optimized shape.

Area is as you say length by width. A line of length 10 has it's length, when we start to consider area we say that has an area of 10. This is because it is length 10 and width 1; 10*1. This is 10 squares in a row, increasing the width adds another row of squares to the width; 10*(1*x).

This is a graphical representation that is a proof of how area is defined. If you want a mathematical proof, you can probably google for one.

Drew K

Yes, of course. The definition comes from:... how many... squares 1unit*1unit fits in...

All I wanted was a physical explanation/representation of the fact!

Area is a measure of the surface enclosed in a closed shape. The shape can be on a plane surface or on a not plane surface the principle is the same. The perimeter is the length of the "string" which build the shape again on a plane or not plane surface. there is NO direct (or biunivoque) relation between perimeter and enclosed area.

The value of the area is obtained by an integration (sum) of elements dx*dy*sin(dx-dy) where dx and dy are the elements of a net on the surface and dx-dy is the angle between the two coordinates. A =∫∫dx*dy*sin(dx-dy) this integral is limited by the shape which defines the area. If the coordinate system is Cartesian i.e. the two directions are perpendicular to each other the angle is 90° and the sinus =1. Thus the integral becomes : A=∫∫dx*dy, as before written limited by the "shape". Let us consider a rectangular shape in a Cartesian reference system: if we integrate along "x" the integral becomes A= B*∫dy = B*H. The same way can be obtained the "area" of any other shape. Same principle can be applied to a non plane surface as for example a cylinder. We consider the "x" in tangential direction and the "y" in axial direction. The dx is an arc whose length is the angular element dφ*R so that the area equation becomes: A= ∫∫ R*dφ*dy. Assuming that the angle is 30° (in the cross section of the cylinder) the B= pi/6*R so that the result will be A= pi*R*H/6.

Now the perimeter is as above P=∫ds but ds= √(dx²+dy²). One sees that between perimeter and area no relation can exist.

For a given perimeter it can be demonstrated that the highest area value is obtained by a circular shape of given perimeter. As example let "L" be the perimeter. For a rectangle L= 2*(B+H) and the area is A=B*H = B*(L/2-B) = L*B/2-B². This function presents a maximum at B=L/4 i.e. when the rectangle is a square. The area will be A= (L/4)²= L²/16. For a circle L= pi*d and A= pi/4*d² = pi/4*(L/pi)²= L²/(4*pi). Or 4*pi<16 thus for same perimeter "L" circle area > square area. the reciprocity is valid as well for a given area the circle has the smallest perimeter.

An example how 2 shapes can have same area and very different perimeters.

Dear nick name: thanks for that detailed explanation.

...dx*dy*sin(dx-dy)

However, this is simply stating the product of two perpendicular vectors since dy*sin(dx-dy)=dH is perpendicular to dx, giving: dx*dy*sin(dx-dy) =dx*dH

I apologise for questioning again, but what I want to know is that why is it that the area is the product of there two perpendicular vectors? I know all the calculus can help us obtain the areas of generally any shape.

However, we still do a form of dx*dH to get the area of an element (on the net of the surface)? I am questioning the logic behind the definition of this product? I mean to ask, can we say area refers to how many perpendicular lengths there are for every height, etc? Some explanation of this form?

Just simply consider a rectangle, and the space it occupies.
If you double the rectangle's width, the occupied space also doubles.
If you double the rectangle's length, the occupied space also doubles.

The only formula that encapsulates both of these facts is A = lw. (Which is simply by definition, anyway.)

There are many interesting "paradoxes" about the areas of complex non-convex plane figures, and various deceptive geometric dissections, but you haven't asked about such things. In the form you have posed, your questions are just about the most subhumanly stupid I have ever heard. Like CaptMoosie, I ask, where in hell were you when 4th grade happened?

If you understood my explanation which is not low level then you were in school and in school in the first class this is explained so that I think you play a game and are NOT serious.

Tempers flaring already? Perhaps I should close this post.

.....Which is simply by definition, "anyway!!!!"

Please do close this. As for flaring tempers, I have been extraordinarily forbearing for much too long.

Thanks, consider it closed. However the arguments have not convinced me.

What are you trying to do, argue with a dictionary? That can be done, but only by qualified persons.

I do not want to prolong this, but I feel it is not right for anyone to make that statement:..what are you trying to do? Is it not inappropriate in such a forum? I have not made any personal attack in anyway. Do I deserve that sentence from Tornado?

What statement? To which post, and what statement, are you replying?

Like I said Mr Tornado, I do not wish to continue with this, but I am referring to you.