the rectangle

[Marty McFly]

Hi, people. I presented my 9 year old son with this question. he couldn't answer it. he went t ohis math teacher who couldn't either answer it. NO ONE in his school could answer it, not even the eighth grade teacher. Why do you think?

the question:

Find the area of any rectangle then find the perimiter. can the area ever equal to the perimiter? if it does, which rectangles would equal? why those particular ones?

my son (a fourth grader) was coming home with homework like "find the area of this, find the perimiter of that" the teacher should have known the answer, no? he's TEACHING the subject!

whadaya think?

[Steve D'Ippolito]

Hi, people. I presented my 9 year old son with this question. he couldn't answer it. he went t ohis math teacher who couldn't either answer it. NO ONE in his school could answer it, not even the eighth grade teacher. Why do you think?

the question:

Find the area of any rectangle then find the perimiter. can the area ever equal to the perimiter? if it does, which rectangles would equal? why those particular ones?

my son (a fourth grader) was coming home with homework like "find the area of this, find the perimiter of that" the teacher should have known the answer, no? he's TEACHING the subject!

whadaya think?

A 4x4 square is a rectangle (albeit a special case of one). Its area and perimeter are both 16. Basically 4x=x^2 in this case, solve for x by dividing both sides by x. This is the well, DUH!! answer to this question.

In the more general case, 2x + 2y = x*y will give you your answer (x and y are the dimensions of the rectangle). If you manipulate this (divide both sides by x, move the 2y/x to the right hand side, then factor y out on the right and divide both sides by the mess y is multiplied by) you get y = 2/(1-2/x). You should be able to plug any x into this greater than 2 and get an answer (x=2 results in a divide by zero, x<2 results in a negative value for y). For example x=8 gives you y = 8/3; x*y = 64/3 for the area, 2(x+y) gives you 64/3 perimeter.

[Marty McFly]

A 4x4 square is a rectangle (albeit a special case of one). Its area and perimeter are both 16. Basically 4x=x^2 in this case, solve for x by dividing both sides by x. This is the well, DUH!! answer to this question.

In the more general case, 2x + 2y = x*y will give you your answer (x and y are the dimensions of the rectangle). If you manipulate this (divide both sides by x, move the 2y/x to the right hand side, then factor y out on the right and divide both sides by the mess y is multiplied by) you get y = 2/(1-2/x). You should be able to plug any x into this greater than 2 and get an answer (x=2 results in a divide by zero, x<2 results in a negative value for y). For example x=8 gives you y = 8/3; x*y = 64/3 for the area, 2(x+y) gives you 64/3 perimeter.

hey you lost me with the factoring. I know the square one is easy, because the equation would be 4x=xsquared 4x would be the perimiter because there are 4 sides to any square. the x squared wold be the area. the only possible number is 4. even my son understood that. BUT

when I did 2x+2y=xy I couldn't get anywhere with that... I mean, I devided by x and got 2y/x=y but then what happened to the other 2? so I'm lost. help! Oh, and I forgot how to factor... is it like doing 2x+2y and changing it to

2(x+y) ?

also, x cannot equal 8!

[softwareNerd]

If he's 9, does he do complicated algebra?

I assume that the answer they're looking for is a 4x4 square, like Steve said.

"Visually" (x * x = 4 * x) therefore x = 4?

[Steve D'Ippolito]

hey you lost me with the factoring. I know the square one is easy, because the equation would be 4x=xsquared 4x would be the perimiter because there are 4 sides to any square. the x squared wold be the area. the only possible number is 4. even my son understood that. BUT

when I did 2x+2y=xy I couldn't get anywhere with that... I mean, I devided by x and got 2y/x=y but then what happened to the other 2? so I'm lost. help! Oh, and I forgot how to factor... is it like doing 2x+2y and changing it to

2(x+y) ?

also, x cannot equal 8!

Why can't X=8? I just made it equal 8 in my previous response and got a 64/3 x 8 rectangle

Anyhow: Start with

2x + 2y = x*y (left side is the perimeter, right side is the area)

divide both sides by x

2 + 2y/x = y

subtract 2y/x

2 = y - 2y/x

factor out y

2 = y* (1 - 2/x)

divide by that mess on the right

2/(1-2/x) = y.

Simple algebra; I don't know why your son's math teacher could not do it.

[punk]

What an appalling question:

Can the area ever equal the perimeter?

The correct answer is "No, never" for the simple reason that the perimeter is measured in (say) meters, and the area is measured in (say) meters squared.

A (one dimensional) meter will never equal a (two dimensional) meter squared.

Of course they want you to over look this and set:

x*y = 2*(x+y)

But (of course there must be an implicit length measurement "a" in this so it is really:

x*y = 2*a*(x+y)

[Marty McFly]

What an appalling question:

Can the area ever equal the perimeter?

The correct answer is "No, never" for the simple reason that the perimeter is measured in (say) meters, and the area is measured in (say) meters squared.

OK, ok, let me rephrase: can the NUMBER of square (inches) in the area ever equal the NUMBER of flat (inches) of the perimiter (of the same rectangle)?

as you already saw, 4x4 works.

Steve D'Ipolito also, x cannot equal 8 by 63/3 what's 63/3? 21, no? so the rectangle you are talking about is 8x21? the area doesn't equal the perimiter of that rectangle. so therefore x cannot equal 8.

[Adrian Hester]

Why can't X=8? I just made it equal 8 in my previous response and got a 64/3 x 8 rectangle

Nope; looks like you made an error in your arithmetic somewhere. With x=8, y=8/3, so the rectangle would have an area of 64/3 (in units of L^2) and a perimeter of 64/3 (in units of L).

2/(1-2/x) = y.

A more convenient form of the equation for x>2 is y=2x/(x-2).

[Steve D'Ippolito]

Nope; looks like you made an error in your arithmetic somewhere. With x=8, y=8/3, so the rectangle would have an area of 64/3 (in units of L^2) and a perimeter of 64/3 (in units of L).

No, I did something far sillier than an arithmetic error: I misquoted myself. In my first post I had it right.

Mary McFly, I don't see where you got that I said ANYTHING about 63/3rds.

In any case all confusion which I caused aside, 8 does indeed work and gives the "other" length of 8/3.

[Capitalism Forever]

whadaya think?

I think it must be a public school. Where do I collect my prize?

[Marty McFly]

I think it must be a public school. Where do I collect my prize?

nope, my son attends private school (a very cheap one, but still)

In any case all confusion which I caused aside, 8 does indeed work and gives the "other" length of 8/3.

fine explain how 8 works? there is no rectangle with a length or width of 8 that has the same number perimiter as area

[softwareNerd]

Does your son do the type of algebra that would enable him to take the problem posed and get to something like

( x * x = 4 * x) ?

Edited November 23, 2007 by softwareNerd

[Adrian Hester]

fine explain how 8 works? there is no rectangle with a length or width of 8 that has the same number perimiter as area

Of course there is! With x=8 and y=8/3, you have area=x*y=64/3 and perimeter=2x+2y=16+(16/3)=(48+16)/3=64/3. I think what you meant to say is that there's no *square* with side of length 8 whose area has the same value as its perimeter.

[Marty McFly]

Does your son do the type of algebra that would enable him to take the problem posed and get to something like

( x * x = 4 * x) ?

no, but I showed it to him, it's not hard to understand. when he says that there are only four sides in a square, and therefore the perimiter will be times 4, I flipped out a pencil and showed him how it works with algebra.

[Steve D'Ippolito]

fine explain how 8 works? there is no rectangle with a length or width of 8 that has the same number perimiter as area

Geez, Marty, I used 8 in my ORIGINAL REPLY as an example, then you denied that it works and keep asking over and over.

8 * 8/3 = 64 /3 (Area) (note that is 64 not 63)

Perimiter 8 + 8 + 8/3 + 8/3 = 24/3 + 24/3 + 8/3 + 8/3 = 48/3 + 16/3 = 64/3.

Do you wish to continue asserting that 8 does not work?

[Marty McFly]

Geez, Marty, I used 8 in my ORIGINAL REPLY as an example, then you denied that it works and keep asking over and over.

8 * 8/3 = 64 /3 (Area) (note that is 64 not 63)

Perimiter 8 + 8 + 8/3 + 8/3 = 24/3 + 24/3 + 8/3 + 8/3 = 48/3 + 16/3 = 64/3.

Do you wish to continue asserting that 8 does not work?

these are not real numbers, they wouldn't work in measuremnts. how long is a line that's 2.666666666666666666666666666666666666666666666666666666666666666? this is not an exact measurement. I meant precission. like 4x4

[John McVey]

People people people...

First, when X and Y are not constrained to be integers there are an infinite number of rectangles with different proportions that fit the question. There is a solution to Y for every single X, except X=2. Yes, that includes X=8. Of course, as X gets ever larger, Y gets ever closer to 2.

Second, in between integer and non-integer, I suspect there's also an infinite number of solutions with numbers that are expressible as regular fractions, though I think that after a short while many would have larger components than a 9yo child can handle calmly.

Third, even when X and Y are constrained to be integers there are two unique solutions, not just the one at 4x4. The other is 3x6 = 18 = 3 + 6 + 3 + 6.

Lastly, I doubt very much if the kind of algebra you guys are invoking is what's pitched at 9yo's. Rather, this was the level I came across when I was tutoring maths to a 14yo. So, I instead agree with SoftwareNerd that a solution was intended to be found visually - especially with today's dumbed down education. Heck, I suspect that the teacher wanted to be given a bunch of only 4x4 answers and would get a tad flustered by students who found 3x6, but I could be being too cynical.

Can we close this one now before it degenerates into acrimony!??

JJM

[softwareNerd]

So, I instead agree with SoftwareNerd that a solution was intended to be found visually - especially with today's dumbed down education.

My son's 9 too, and they're posed some problems like this even though they haven't been taught the necessary algebra. I think the expectation is that they will form something equivalent to the idea represented by (4 * ?) = ( ? * ?), in their minds, and then they will start to plug in some possibilities: 1... no, 2... no, 3... no, 4... hey! that works.

I don't know if this is "dumbing down". I think not. More accurately, I think it would be if things remained at this level. If done right, I think this type of approach can actually ground the abstractions of algebra. Once the kid figures that 4 works, if he is actually shown (4 * ?) = (? * ?), it's kinda "obvious" that ? = 4. At that level of knowledge, the thought that there could be other answers does not even occur to the child.

(I plan to try this one on my own 9 year old, to see how he tackles it. Will report back later.)

Edited November 25, 2007 by softwareNerd

[Adrian Hester]

these are not real numbers, they wouldn't work in measuremnts. how long is a line that's 2.666666666666666666666666666666666666666666666666666666666666666? this is not an exact measurement. I meant precission. like 4x4

Doesn't matter; you can always exactly trisect any given line segment in geometry. Take a line segment eight units long, trisect it with compass and straight-edge, project that length perpendicular to the two ends of the original line segment, and connect the other ends of the two perpendicular segments. The resultant rectangle, which is very easily constructed, will have the same perimeter in units of length as its area in units of length squared. Alternately, take a rectangle 8 feet by 24 feet--its area in square yards will have the same value as its perimeter in yards. In any case, the same objection holds for a square of side 4--either you accept an infinite precision in measurement (which is usually indicated by writing 4 and not 4.0), in which case 8/3 is just as valid a length as 4, or else you cannot say the side of your square is exactly 4.00000... units and its perimeter 16.00000... units and area 16.00000... units squared.

Edited November 25, 2007 by Adrian Hester

[Marty McFly]

People people people...

Third, even when X and Y are constrained to be integers there are two unique solutions, not just the one at 4x4. The other is 3x6 = 18 = 3 + 6 + 3 + 6.

Lastly, I doubt very much if the kind of algebra you guys are invoking is what's pitched at 9yo's. Rather, this was the level I came across when I was tutoring maths to a 14yo. So, I instead agree with SoftwareNerd that a solution was intended to be found visually - especially with today's dumbed down education. Heck, I suspect that the teacher wanted to be given a bunch of only 4x4 answers and would get a tad flustered by students who found 3x6, but I could be being too cynical.

Can we close this one now before it degenerates into acrimony!??

JJM

actually, this was the answer I was hoping my son would give. (!)

4x4 and 3x6 these are the ONLY rational numbers that work on a rectangle where the area will equal the perimiter.

the most important question that follows that, is what my son hadn't yet figured out, is WHY are these the only possible answers? why can no other rectangle equal the perimiter # with the area #?

My son's 9 too, and they're posed some problems like this even though they haven't been taught the necessary algebra. I think the expectation is that they will form something equivalent to the idea represented by (4 * ?) = ( ? * ?), in their minds, and then they will start to plug in some possibilities: 1... no, 2... no, 3... no, 4... hey! that works.

EXACTLY!! thank you! finally someone who can understand the mind of a 9 year old. try it with your son, it's alot of fun (if he likes goemitry) tell me how it goes?

Edited November 25, 2007 by Marty McFly

[Marty McFly]

Alternately, take a rectangle 8 feet by 24 feet--its area in square yards will have the same value as its perimeter in yards.

8x24=192

8+8+24+24=64

so area here does not equal perimiter. also, 4.000000000000000 is IS a precise number. it's an actual number you count one.. to.. three.. four.. inches/feet/meters/paper-clips etc.

2.66666666666666666666666666666666666666666666666666666666666666666666-up to infinity is not a number you can measure with.

[Capitalism Forever]

(emphasis added)

Alternately, take a rectangle 8 feet by 24 feet--its area in square yards will have the same value as its perimeter in yards.

8x24=192

8+8+24+24=64

so area here does not equal perimiter.

Please have the decency to read the exact post before replying to it. 192 sqare feet = 21 1/3 square yards; 64 feet = 21 1/3 yards.

it's an actual number you count one.. to.. three.. four.. inches/feet/meters/paper-clips etc.

2.66666666666666666666666666666666666666666666666666666666666666666666-up to infinity is not a number you can measure with.

??!!

[Laure]

Marty,

Go to this site, GRAPH SKETCHER, and key in 2*x/(x-2) to see a graph of the function. It shows that it's a hyperbola, with the x (or y) approaching 2 as y (or x) approaches infinity. This is why the only positive integer pairs that work are (4,4) and (3,6) (or (6,3)).

I think you ought to review some definitions and concepts, so that you don't confuse your son. The fact that you say that 8/3 is not a real number, and not a rational number, shows that you either have very little mathematical training, or you have forgotten what you did have. Look up the definitions of real number, rational number, integer, whole number, irrational number, imaginary number. You need to use these terms the same way they are traditionally used in mathematics, or your son will be very confused. Also, 2.66666 repeating is just as precise as 4.0000 repeating. Any real number is just a precise as any other. Think about a right triangle with sides of length 1 and 2. The hypotenuse has a length of exactly the square root of 5, even though that's an irrational number.

Edited November 25, 2007 by Laure

[Adrian Hester]

8x24=192

8+8+24+24=64

so area here does not equal perimiter.

Your error in reading has already been corrected--in yards and square yards they are equal in value. But think about the question of yards and feet a bit more. How long is four feet in yards? 1.333333.... Four feet, by your argument, is precisely four feet and precisely measurable. Yet in yards it's not what you consider a measurable number--it's a number you say can't be used in measurement. Yet yards too are precisely measurable by your argument--three feet exactly. (And in fact, since a yard is defined as three feet, then it is precisely three.) If it's precisely measurable in one unit (feet), then if you change to a second unit (yards) precisely measurable in terms of the first unit, then surely it should be precisely measurable in that unit too, right? Yet suddenly by changing from feet to yards you get a (supposedly) unmeasurable length.

also, 4.000000000000000 is IS a precise number. it's an actual number you count one.. to.. three.. four.. inches/feet/meters/paper-clips etc.

No, that would be 4--without a decimal expansion it's the symbol for the integer in counting, not a measurement. If you write out the decimal expansion, then you're admitting a finite degree of precision, and no matter how many zeroes you write out, it's not infinitely precise.

2.66666666666666666666666666666666666666666666666666666666666666666666-up to infinity is not a number you can measure with.

Again, false. You're being thoroughly misled essentially by the fact that we have ten fingers. As I already mentioned, it's quite easy to trisect a line segment, which means that you can divide a ruler into thirds, ninths, twenty-sevenths, and so on right down as far as you want with as much precision as you want (there's nothing special at all about dividing a ruler into tenths, hundredths, and so on instead or thirds), so in fact 2.6666.... is a number you can measure with just as readily as 4.000... The thing is, 3 is not a divisor of 10, so its decimal expansion is infinitely repeating. (And it's because it's infinitely repeating, roughly, which is what "rational number" means, that you can base a ruler on units of three. The case is different if they're infinitely many non-repeating digits.) But it still corresponds to a definite length precisely measurable by trisecting a unit if that unit is precisely measurable (though not one that will match up with any of the lengths ticked off on a ruler by subdividing it forever and ever by tenths, which is what the infinitely repeating threes means geometrically); otherwise you're in the position of saying that only some lengths can be precisely and meaningfully trisected, all others being meaningless lengths (essentially any lengths not formed from the original unit by combinations of repeated subdivisions by halves and fifths)--and this by simple virtue of the fact that you've chosen one essentially arbitrary length as your basic unit of length. If you choose a different unit, then many of these unmeasurable lengths will in general suddenly become measurable.

This isn't meant to give you a hard time. Rather, it's worth going through because all of this has been an issue in the philosophy of mathematics since the Ancient Greeks, though in a slightly different form. How do mathematical entities connect to reality? How does abstract geometry relate to actual measurements? A common Greek reply was essentially Platonic, that they are pure forms or what-not imperfectly reflected in reality. (Though at least one Greek, Protagoras, took essentially your approach, which is to say that only what is concretely real is true--thus, he argued that because in physical reality all circles and lines have finite thickness, then no lines or circles can really intersect in only one point.) With the adoption of decimal notation, we moderns have additional questions--how do numbers in general relate to the integers, and more generally how do algebra and geometry connect up with each other? The way it's handled in modern mathematics is to use bare numbers--integers and pure fractions--to indicate what is countable and hypothesized to be precisely measurable, decimal expansions to indicate measurements to a certain degree of precision; and the pure, abstract mathematics is tied to real measurements by indicating the degree of precision in any subsequent measurement in terms of the precision of the first measurement.

Edited November 26, 2007 by Adrian Hester

[Steve D'Ippolito]

By the way Marty I've seen you use "real" number and "rational" number incorrectly.

Real is any number on the number line--even pi, e, the square root of 2....

Rational means any number that can be expressed as a fraction of two integers.

You were probably meaning to say either whole number or integer.