maths question

[SPIRIT]

a square is formed by 6 rectangles- given the total perimeter of the 6 rectangles is 330cm.


determain the area of the square

[tea lady]

Are all the rectangles the same? Or doesn't it matter?:?

[Burnsy]

Are all the rectangles the same? Or doesn't it matter?:?
All rectangles can't be the same, a box made out of all rectangles the same becomes a cube made out of squares!

[Burnsy]

Oops, maybe I am looking at this wrong, I assumed Arty was talking about a three dimensional box made of six rectangles. More likely that he is talking about a two dimensional square made up of six rectangles?

[tea lady]

Oops, maybe I am looking at this wrong, I assumed Arty was talking about a three dimensional box made of six rectangles. More likely that he is talking about a two dimensional square made up of six rectangles?
Yeeeeeeeeeeees!:D

[Canetoad]

1089 square cm

[Burnsy]

6 rectangles each 6 times longer than they are wide

perimeter of each rectangle is 6+6+1+1=14

times six rectagles = 84 units in perimeter all up

330 /84 = 3.92857 cm per unit

3.92857x6 = 23.5714cm (length of one side of square)

Area = 23.5714 x 23.5714
= 555.6cm square

[Claw Hama]

A square who cares what it's made of, 330cm total perimiter.
330 div by 4 = 82.5 for one side x 2 = 6806.25 sqcm (Area)

[Chipman]

It depends how you arrange them or if you are being tricky!

If you mean that the perimeter is 330 for all the rectangles once they are put together then 330 is the perimeter of the square and the area would be 6806.25 cm squared as calculated by claw hama the tricky sneaky one (No not you Claw hama, I mean spirits question!!!!!!!)

If all the rectangles are the same size, then there are two ways to arrange them

6 x 1 and the answer would be 555.6 cm squared as calculated by burnsy


3 x 2 and the answer would be 1089 cm squared as calculated by canetoad

[Chipman]

So Spirit are you going to come clean and tell us which one you really meant!!!!!!



Chipman

[Burnsy]

A square who cares what it's made of, 330cm total perimiter.
330 div by 4 = 82.5 for one side x 2 = 6806.25 sqcm (Area)


a square is formed by 6 rectangles- given the total perimeter of the 6 rectangles is 330cm.


determain the area of the square

Total perimeter of the 6 rectangles so each rectangles perimeter is 330 /6 = 55 divide that by the perceived perimeter of a rectangle that is the right proportion to make a square when six of them are put together which is 14 (rectangle is 6 units long and 1 unit wide) = 3.9285. Multiply that by the perceived length of the square 6 (square is 6 units x 6 units) = 23.57. Area equals 23.57 x 23.57 = 555.6cm square.

[Burnsy]

yeah cmon, what is it?

[Claw Hama]

Yep ambiguous, is it the total perimiter of the individual rectangles or of the square?
GIven that you can get multipul answers going with the rectangles maybe it's the perimiter of the square? Owwe it's too late at night for this game.

[Burnsy]

I am meant to be writing reports - due Monday, this is much more fun. The year sevens think I am mad when I get up in front of the class and start running off about how great maths logic is:D

Very easy to run off on a completely wrong tack though - check my first post

[Claw Hama]

But Burnsy you are the only one who mentioned a box Spirit only said a square.
Is that your funky green machine?

[RayG]

The perimeter of the square is 330/2 = 165 cm since half of the sides
are "interior" to the square. so the side length is 165/4 = 41.25cm
and the area is 41.25*41.25 = 1701.563 sq cm


Correction, the center rectangles will only have one external edge, so
if rectangle is a * b then we get

2a = 3b
12a + 12b = 330

solving for b..

b = 7.666cm , a = 11.5 cm

square side is 3b, oe 23 cm

square area is 23*23 = 529 sq cm



QED

[Burnsy]

But Burnsy you are the only one who mentioned a box Spirit only said a square.
Is that your funky green machine?

I know that is what I mean, you take a first impression and start running without actually reading it and processing it sometimes.

Use to be mine, sold it to a 15 year old two months ago to make more space in the shed for my woodworking and to reduce the hobbies/toys - first born is on the way.

[Claw Hama]

You on drugs Ray??

[Burnsy]

You on drugs Ray??

Interior sides:oo:, must be like interior design, both make no sense to me:rolleyes:

[Chipman]

The perimeter of the square is 330/2 = 165 cm since half of the sides
are "interior" to the square. so the side length is 165/4 = 41.25cm
and the area is 41.25*41.25 = 1701.563 sq cm

QED


How about a diagram Ray?

Sounds interesting

[Claw Hama]

It's a shame when you don't have room in your life for all your loves. Both time and space are limited, come on science.
Drugs, must be doing drugs!!

[RayG]

You on drugs Ray??

Ha .. is it the caffiene overdose that obvious ! :rolleyes:

[Claw Hama]

Rays diagram

[Claw Hama]

It's ok Ray that was good value have another cupa mate. Hey maybe you're right who knows at this point.

[RayG]

How about a diagram Ray?

Sounds interesting

a a
+--------+--------+
|............|...........| b
|............|...........|
+--------+--------+
|............|...........| b
|............|...........|
+--------+--------+
|............|...........| b
|............|...........|
+--------+--------+


for it it be square 2a must = 3b,

adding up the edges of all the rectangles 2*6*a + 2*6*b = 330

so 2*6*3b + 2*6*b = 330
b = 7.666.

so side of square must be 23 cm and area is 23*23 = 529 sq cm

More caffiene.. need more caffiene...

[Burnsy]

I think Arty has done a runner. Possible reasons:

Does not have the answer
Typo in the first place meant the question is wrong
Is busy copying the the possible anwers into his assignment (he'll get extra marks for Ray's anwer)
Just likes to give us all the :poop:Just spent five minutes looking for that last emoticon, never looked at them all before, how's this one :wank: :oo::oo::oo:

[Claw Hama]

Ray are you in real estate, what floor area is your house?

[Claw Hama]

I think your on the money Burnsy. Some of the emo's are great fun.

[Chipman]

The perimeter of the square is 330/2 = 165 cm since half of the sides
are "interior" to the square. so the side length is 165/4 = 41.25cm
and the area is 41.25*41.25 = 1701.563 sq cm


Correction, the center rectangles will only have one external edge, so
if rectangle is a * b then we get

2a = 3b
12a + 12b = 330

solving for b..

b = 7.666cm , a = 11.5 cm

square side is 3b, oe 23 cm

square area is 23*23 = 529 sq cm



QED


Ray, its late and I could be wrong but if you combine these two equations, b = 11

so is it irish coffee we are on tonight?????

your approach assumes he wants rectangles arranged 3 x 2 but you can do 6 on top of each other too so I am not sure what spirit wants?

Oh well back to my coffee!!!!!!!

Chipman

[RayG]

Ray are you in real estate, what floor area is your house?

No, and the house is smaller than the shed, (which is too small anyway). if that helps.


What are you doing up at this hour in newcastle anyway!

[Claw Hama]

I have no idea but this has been a funny post. I might just say goodnight to all, have fun.

[Chipman]

Ray, thanks for the diagram and your substitution is a bit off if 2a =3b then

12a + 12b =330 and
6 * 2a + 12b = 330
thus

6*3b + 12b = 330
30b = 330
b =11

I really should get back to my marking or better still go to bed:doh:

[RayG]

Ray, thanks for the diagram and your substitution is a bit off if 2a =3b then

12a + 12b =330 and
6 * 2a + 12b = 330
thus

6*3b + 12b = 330
30b = 330
b =11

I really should get back to my marking or better still go to bed:doh:

Yep, you are correct.

BTW it doesn't matter if the rectangles arranged 1x6 or 2x3 it still works
out the same (although a and b are different values)

So, finally the area is... 1089 sq cm

[switt775]

An interesting and rather frustrating question. Both Burnsy and Canetoad are correct, even though their answers are completely different (don't you just love maths?)

There are two possible solutions, one using the 2 x 3 pattern RayG drew up, which is the one I used first.

If the total perimeter is 330 cm for 6 rectangles, then each rectangle is 330 / 6 = 55cm.

Represent the sides of the rectangle by X and Y.
Total perimeter of each rectangle is 2X + 2Y.

So 2X + 2Y = 55.



When you draw that up like RayG, you find your square has one side of 2X and one side of 3Y. Because it's a square, those two are equal.

So 2X = 3Y

Since they are equal, we can substitute 3Y in the 1st equation, giving us
3Y + 2Y = 55
5Y = 55
Y=11

That's the short side of a our rectangles. Put that back into the 1st equation, and we have the following:

2X + 22 = 55
2X = 33
X = 16.5


The rectangles are each 11 x 16.5, meaning our square is 33 x 33.

Total area is 1089.

So then I looked at Burnsy's answer, and realised his answer while different is correct also. He solved it based on making a square using 1 x 6 rectangles, instead of 2 x 3. But there was nothing in the question that stated the lengths had to be whole numbers.:cool::cool:

[RayG]

An interesting and rather frustrating question. Both Burnsy and Canetoad are correct, even though their answers are completely different (don't you just love maths?)

There are two possible solutions, one using the 2 x 3 pattern RayG drew up, which is the one I used first.

If the total perimeter is 330 cm for 6 rectangles, then each rectangle is 330 / 6 = 55cm.

Represent the sides of the rectangle by X and Y.
Total perimeter of each rectangle is 2X + 2Y.

So 2X + 2Y = 55.



When you draw that up like RayG, you find your square has one side of 2X and one side of 3Y. Because it's a square, those two are equal.

So 2X = 3Y

Since they are equal, we can substitute 3Y in the 1st equation, giving us
3Y + 2Y = 55
5Y = 55
Y=11

That's the short side of a our rectangles. Put that back into the 1st equation, and we have the following:

2X + 22 = 55
2X = 33
X = 16.5


The rectangles are each 11 x 16.5, meaning our square is 33 x 33.

Total area is 1089.

So then I looked at Burnsy's answer, and realised his answer while different is correct also. He solved it based on making a square using 1 x 6 rectangles, instead of 2 x 3. But there was nothing in the question that stated the lengths had to be whole numbers.:cool::cool:

The assumption is that the rectangles are all the same, which simplifies the
solution, but I suspect that the answer will be the same for different sized
rectangles as well. Not that easy to prove without more caffiene....

Thanks for the mental exercise.

[switt775]

Actually, it's easy to prove that different sized rectangles will give different answers.

What is the greatest square or rectanglar area you can enclose with 40M of fencing, assuming this includes a gate which is 2M long?

You might not think it makes much difference what shape you use, because after all you will always have the same amount of fencing, but work out some examples. The gate gives you a minimum length of the short side of 2M, so here we go:

2M x 18M = 36 M2

3M x 17M = 51M2

All the way up to:

10M x 10M = 100M2

Same perimeter of 40M, but the area enclosed varies from 36M2 to 100M2

That's why using rectangles of different relative dimensions gives different answers.

Given that a square is still a rectangle, and since no one said we couldn't use rectangles of various sizes, there are other equally correct answers, one of which would use the following model:

Create 4 squares each of dimensions A * A. Arrange them into a square which has overall dimensions of 2A * 2A.

Add a rectangle along one side, with dimensions of 2A * B. To complete the square, add another rectangle along the other side. It will have dimensions of (2A + B) * C.

So now we need to solve for 3 variables. Too hard without a bit of caffeine in me, maybe someone else can tackle this and let us know the answer.

Just had another quick look at my diagram. B and C are the same (told you this was way to hard before coffee). So the dimensions of the final rectangle are (2A + B) * B. This gives an overall square of (2A+ B) * (2A + B),

or

4*A*A + 4*A*B + B*B

[Chipman]

Actually, it's easy to prove that different sized rectangles will give different answers.

What is the greatest square or rectanglar area you can enclose with 40M of fencing, assuming this includes a gate which is 2M long?

You might not think it makes much difference what shape you use, because after all you will always have the same amount of fencing, but work out some examples. The gate gives you a minimum length of the short side of 2M, so here we go:

2M x 18M = 36 M2

3M x 17M = 51M2

All the way up to:

10M x 10M = 100M2

Same perimeter of 40M, but the area enclosed varies from 36M2 to 100M2

That's why using rectangles of different relative dimensions gives different answers.

Given that a square is still a rectangle, and since no one said we couldn't use rectangles of various sizes, there are other equally correct answers, one of which would use the following model:

Create 4 squares each of dimensions A * A. Arrange them into a square which has overall dimensions of 2A * 2A.

Add a rectangle along one side, with dimensions of 2A * B. To complete the square, add another rectangle along the other side. It will have dimensions of (2A + B) * C.

So now we need to solve for 3 variables. Too hard without a bit of caffeine in me, maybe someone else can tackle this and let us know the answer.


It is hard to play a game without rules!:rolleyes::rolleyes::rolleyes:

If we are allowed to use rectangles of different sizes (including squares) at one time then the number of solutions will be a lot, lot, more than 3! You have to know the rules before you can get the solution

Chipman

[switt775]

It's a bit like Calvin and Hobbes playing baseball, where nobody can agree where 12th base is...:doh:

Still it's good fun playing.:U

[Wongo]

Hellooo.:U

[Wongo]

So now we've found the 2 easy ones (555.6...... and 1089 well done boys). I wonder if the third one exists?

[Wongo]

I need to go. I want the third solution by time I get back OK!! :cool2:


:D

[Burnsy]

Well I just checked in and see we are still going - without Spirit. However it is a beautiful day here in sunny Perth so it's shed time not maths or report writing time:2tsup:

[tea lady]

Stay off the coffee guys. I can lead to all sorts of things. (How many legs was that again?)

[Chipman]

I need to go. I want the third solution by time I get back OK!! :cool2:


:D
Claw hama already found it last night... check out page 1!:U

[Wongo]

A square who cares what it's made of, 330cm total perimiter.
330 div by 4 = 82.5 for one side x 2 = 6806.25 sqcm (Area)


This one?

I don't understand one bit. Can someone please explain it to me.

[Chipman]

This one?

I don't understand one bit. Can someone please explain it to me.

Some of us suspect that Spirit was trying to be tricky... apart from the obvious 3x2 or 6x1 pattern, what if he meant the perimeter of the shape once all the rectangles were put together in a square...the edges of the rectangles "disappear" on the inside of the square once they are put together. Thus 4 sides - 330 cm etc as claw hama said

Chipman

[Wongo]

Don't forget this boys. There are 3 simple steps in maths.

1. Read the question
2. Think
3. Answer the question

C'mon boys don't jump steps. Let's start from step 1 again. :D

Let's read it together

A square is formed by 6 rectangles- given the total perimeter of the 6 rectangles is 330cm.

determain the area of the square

[Wongo]

No it is not tricky. It is a very simple question.

A square is formed by 6 identical rectangles- given the total perimeter of the 6 rectangles is 330cm.

Why am I so sure they are identical?? Because if they are not then there will infinite number of solutions.

[SPIRIT]

74891

[Wongo]

Spirit, you are not kidding are you?

[SPIRIT]

why

[Wongo]

Why didn't you show the diagram from the begining?

1. 1089
2. x*x / 100

[Wongo]

why

Because with the picture your question has only 1 answer. Without the picture your question has countless answers.

[SPIRIT]

how did you work it out?

[Chipman]

Hey Wongo,

I checked out Spirit's other posts and to me this is his style.... he reveals things bit by bit! But give him credit, he sure gets us all interested.

None-the-less I agree, the diagram should have been their right from the start!!!!

A picture is worth a thousand words

Chipman

[Wongo]

how did you work it out?

http://www.woodworkforums.ubeaut.com.au/attachment.php?attachmentid=74895&stc=1&d=1212838522

[Chipman]

http://www.woodworkforums.ubeaut.com.au/attachment.php?attachmentid=74895&stc=1&d=1212838522
Wongo, I really like the way you solved that!!!!

Excuse the pun but that really is thinking outside the box!

Chipman

[AlexS]

Well done, Wongo.
The thing with these problems is, you need to see the whole problem! If Spirit is going to drip feed us, there's no point.

[Wongo]

Sorry I made a mistake. The answer to the second part of the question is

(x/10)^2 or x*x / 100

[Wongo]

Well done, Wongo.
The thing with these problems is, you need to see the whole problem! If Spirit is going to drip feed us, there's no point.


Well Alex I think without the picture the question becomes far more interesting. With it, it is too easy.

[RayG]

Nice work Wongo, brilliant solution.

With your method it's also easy to see that the size of
the rectangles can be any size for a 2x3 and the answer will be the same.

Different answers for 1x6 and other arrangements.

[SPIRIT]

well done for who ever joined in the fun :D see you next time :U

[Wongo]

C'mon spirit we want something harder. We want a challenge.:U

[AlexS]

C'mon spirit we want something harder. We want a challenge.:U - but the whole challenge, please.