maths question

[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

[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

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.

[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.

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!

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...

Still it's good fun playing.

[Wongo]

Hellooo.

[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!!

[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

[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!!

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

[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.