a square is formed by 6 rectangles- given the total perimeter of the 6 rectangles is 330cm.
- determain the area of the square
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a square is formed by 6 rectangles- given the total perimeter of the 6 rectangles is 330cm.
- determain the area of the square
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!Quote:
Originally Posted by 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!:DQuote:
Originally Posted by Burnsy
1089 square cm
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
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)
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
So Spirit are you going to come clean and tell us which one you really meant!!!!!!
Chipman
Quote:
Originally Posted by Claw Hama
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.Quote:
Originally Posted by SPIRIT
yeah cmon, what is it?
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.
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
But Burnsy you are the only one who mentioned a box Spirit only said a square.
Is that your funky green machine?
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
I know that is what I mean, you take a first impression and start running without actually reading it and processing it sometimes.Quote:
Originally Posted by Claw Hama
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.
You on drugs Ray??
Interior sides:oo:, must be like interior design, both make no sense to me:rolleyes:Quote:
Originally Posted by Claw Hama
Quote:
Originally Posted by RayG
How about a diagram Ray?
Sounds interesting
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!!
Ha .. is it the caffiene overdose that obvious ! :rolleyes:Quote:
Originally Posted by Claw Hama
Rays diagram
It's ok Ray that was good value have another cupa mate. Hey maybe you're right who knows at this point.
a aQuote:
Originally Posted by Chipman
+--------+--------+
|............|...........| 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...
I think Arty has done a runner. Possible reasons:
Just spent five minutes looking for that last emoticon, never looked at them all before, how's this one :wank: :oo::oo::oo:
- 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:
Ray are you in real estate, what floor area is your house?
I think your on the money Burnsy. Some of the emo's are great fun.
Quote:
Originally Posted by RayG
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
No, and the house is smaller than the shed, (which is too small anyway). if that helps.Quote:
Originally Posted by Claw Hama
What are you doing up at this hour in newcastle anyway!
I have no idea but this has been a funny post. I might just say goodnight to all, have fun.
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.Quote:
Originally Posted by Chipman
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
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 theQuote:
Originally Posted by switt775
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.
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
Quote:
Originally Posted by switt775
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
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
Hellooo.:U
So now we've found the 2 easy ones (555.6...... and 1089 well done boys). I wonder if the third one exists?