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Rheghead
13-Sep-05, 12:56
This one has got me stumped.

1 Draw a 10cm square.

2 Draw 4 10cm 90 degree radius arcs from each corner of the square, from one corner inside the square to the other.

What is the area of the the middle shape which is a wonky sided circle/square thing?

My niece has asked me this but it has got me. I think it has to be answered by integration rather than trigonometry.

DrSzin this must appeal to you?

Whitewater
13-Sep-05, 15:12
Hi Rhegers, I havn't worked it out but should it not be the difference between the area of the 10cm circle and the area of the 10cm square, (you have four quarters of the circle)

Whitewater
13-Sep-05, 16:01
Hi Rhegers (again) after having sketched out your puzzle what I've assumed is rubbish. sorry!!

Rheghead
13-Sep-05, 17:25
You should have a square with a four point star in it

http://members.aol.com/rheghead/puzzle.gif

jimag
13-Sep-05, 17:35
Try 31.514674 square centimeters for starters.

Rheghead
13-Sep-05, 17:37
Try 31.514674 square centimeters for starters.

How did you work it out?

jimag
13-Sep-05, 17:38
I must have drawn it wrongly because I do not have a star. (Dont feel one either)

jimag
13-Sep-05, 17:39
Rheghead

You tell me how you are going about it.

Rheghead
13-Sep-05, 17:46
I tried deducting (pi r^2)/4from 100 then I took that answer from (pi r^2)/4, but that just gave me an area for the pointy oval shape. :confused

Tristan
13-Sep-05, 19:27
The answer should be 14.15926536 cm2

The area of the square is 100 cm2

If a whole circle was drawn the area would be Pi*100 = 314.1592654cm2
1/4 of the circle is 78.53981634cm2
subtract this from the area of the square 100-78.5398 = 21.46018366cm2
This is the area that is between the outside of the circle but still within the square
There are 4 of these so 4*21.46018366 = 85.84073464cm2 (the area outside all 4 aches
subtract this from 100cm2 and you have the area of the inner bit
100-85.84073464 =
14.15926cm2

George Brims
13-Sep-05, 19:44
The area is 31.5147 square cm as jimag said. I drew the figure in AutoCAD and used the area tool. Now somewhere around 1968 I could have worked it out from Geometry but if my employer is going to buy me $5000 worth of software and a nice computer, why should I trouble my ever-declining brain?

hereboy
13-Sep-05, 19:45
or so you would think...

but the four areas which are between each quarter circle and the rest of the square intersect and overlap each other... so the area outside the shape is not 85 but less than that, which would make the area inside the shape greater than 14...

try again.... ;)

PS. my reply was to Tristan.

Rheghead
13-Sep-05, 20:23
As always when we are back to school, the teacher wants to see the working out. :D

It has got me beat!!

DrSzin
13-Sep-05, 20:28
Try 31.514674 square centimeters for starters.
That's what I get too! Did I really get it right first time? Amazing!

A quick calculation gives, for a square of side R:

Area = (Pi/3 + 1 - sqrt(3)) R^2

where Pi = 3.14159..., sqrt means square root, and R^2 is R-squared.

I haven't checked my calculation but it surely can't be a coincidence that my answer agrees with jimag's to 8 significant figures. :cool:

I'll provide working for teacher if he really wants it. I just integrated over the plane area as he suggested. I suspect there's a sneaky geometrical solution, but I'm hopeless at geometry.

Rheghead
13-Sep-05, 20:34
jeepers, how did you come up with the solution?

George Brims
13-Sep-05, 20:36
Where did you get that formula DrSzin? I am not sure what you mean by "a square of side R".

Anyway I just wasted a good half hour drawing extra lines on that thing and putting in angles, and I did finally manage to work it out instead of having the computer do it. It's not as hard as I at first thought. Bit hard to explain here in words without the benefit of pictures though.

DrSzin
14-Sep-05, 00:27
Where did you get that formula DrSzin? I am not sure what you mean by "a square of side R".
I meant a square whose side has length R (R = 10cm in the original version).



Anyway I just wasted a good half hour drawing extra lines on that thing and putting in angles, and I did finally manage to work it out instead of having the computer do it. It's not as hard as I at first thought. Bit hard to explain here in words without the benefit of pictures though.
You're right, it is indeed easy to work out using simple geometry. But at least my solution above was correct, even if I did use a sledgehammer to crack a nut. :roll:

And it's too late at night to explain the simple solution in words...

scotsboy
14-Sep-05, 09:17
I had a totally different shape as I drew the "arcs" on the outside of the square :lol: It wasn't till Rheghead posted the wee graphic that I realised they should be inside the box - a picture is worth a thousand words......or at least a few hundred ;)

jimag
14-Sep-05, 09:53
DrSzin

Sledge hammer or not you are to be commended for your effort, but where did you get the formula. Would you like to show us how it was derived. I, like our friend across the pond, took the easy route and drew it out using CAD.

Rheghead
14-Sep-05, 16:43
Teacher needs both, how you got the math solution and the geometry soutioln.

:confused

pleeeeease!! :lol:

I think i am more stumped after knowing the answer. Life can be like that.

George Brims
14-Sep-05, 17:30
I'm tied up with a meeting today, but I will try to write a coherent note on the geometric solution later.

DrSzin
16-Sep-05, 09:39
I'm tied up with a meeting today, but I will try to write a coherent note on the geometric solution later.
I can save you a job George -- I've already typed up my geometric solution. I still have to check it for typoes, and I'd better add the calculus version to keep teacher happy. It's a pdf file, so I'll stick it on the web and post a link here.

On the other hand, it might be interesting to see if we did it the same way!

----------------------------

I didn't get around to posting my solutions yesterday. Here they are (http://www.geocities.com/drszin/puzzle.pdf).

Please let me know if you spot any errors or typoes.

fred
16-Sep-05, 18:38
I didn't get around to posting my solutions yesterday. Here they are (http://www.geocities.com/drszin/puzzle.pdf).


Not just how I did it.

I started out the same with the triangle OAB and subtracted the area of the triangle from the area of the circle segment. The answer is then the area of a square with sides AB plus 4X the result.

jimag
18-Sep-05, 21:40
I agree with you Fred. After working it out on CAD, this is how I saw the solution evolving but came bogged down proving the included angle was 30 degrees and ran out of time. You've saved me some work.

fred
18-Sep-05, 23:19
I agree with you Fred. After working it out on CAD, this is how I saw the solution evolving but came bogged down proving the included angle was 30 degrees and ran out of time. You've saved me some work.

That's what put me off posting earlier, I knew the angle was 30 degrees but I couldn't easily prove it either.

DrSzin
24-Sep-05, 13:10
If anyone's interested, I can add fred's solution to my notes. I think my original solution is a bit shorter, but I might have missed a few shortcuts in working out his version.

In a moment of boredom, I realised there's a neat solution using Stokes'/Green's Theorem (http://mathworld.wolfram.com/GreensTheorem.html) to write the area as a line integral around the shape ABC. It's neat because the integrals along AC & CB are trivial, as is the one along BA using plane polars. I might add this to my notes if I get really bored...

scorrie
24-Sep-05, 21:56
If anyone's interested, I can add fred's solution to my notes. I think my original solution is a bit shorter, but I might have missed a few shortcuts in working out his version.

In a moment of boredom, I realised there's a neat solution using Stokes'/Green's Theorem (http://mathworld.wolfram.com/GreensTheorem.html) to write the area as a line integral around the shape ABC. It's neat because the integrals along AC & CB are trivial, as is the one along BA using plane polars. I might add this to my notes if I get really bored...

I came across two easy solutions. Number one was to weigh a cut out of the original square, then draw the desired shape and cut it out and weigh that. As the original area is known, one can calculate the area of the "funny" shape in direct weight proportion, as paper is a constant.

Solution two is more complex but SO rewarding. Take the original square and set it alight. Measure the change in temperature and calculate the energy generated by a full square. Next, cut out the "funny" shape as before and burn it, again measuring the change in temperature. The relationship between the relative rises in temperature will give the area of the "funny" shape. Sadly the postie opened the door during my experiment and let a negative temperature drift affect my calculations. Luckily, I was able to offset this with an allowance in Physics known as the "Hall Effect"

Yours in Science, Jimmy Neutron

hereboy
24-Sep-05, 22:54
Number one was to weigh a cut out of the original square, then draw the desired shape and cut it out and weigh that. As the original area is known, one can calculate the area of the "funny" shape in direct weight proportion, as paper is a constant.


This also works if you bake a square of shortbread and then cut out the desired shaped and weigh. Weighing by difference works well - let me explain.

A. Using a bathroom scales weigh yourself plus the square of shortbread.
B. Then weigh yourself plus the cut out shape.
C. Finally, just weigh yourself.

Weight of shape = B-C, weight of original square = A-C.

Ratio of Weights = (B-C)/(A-C)
Therefore area of shape = 100*((B-C)/(A-C)).

Can someone check my algebra and arithmetic to confirm?

PS. It is important to wait until the shortbread has cooled to a constant temperature before making these measurements, otherwise the result will be skewed by the "Walker Effect".

PPS. A variation on this would be to cut out a shape which is one quarter segment of the desired shape (as described in earlier solutions) This new shape becomes B and calculate the weight-area ratio of this shape using the same method as above. The equation then becomes:

4*100*((B-C)/(A-C))

This variation is known as the "Petticoat" method.

;)

scorrie
25-Sep-05, 01:32
I think you are coming the bag here, Hereboy. The Hall Effect is a real phenomenon in Physics, whereas the Walker Effect is actually obesity caused by over-indulgence in crisps. The Petticoat Method is clearly a ploy used by courting males and has nothing to do with calculations. Tall Tails methinks.

Today's word for discussion is:- Gyroscope

Definition:- The potential that one's Social Security Payment holds in respect to maintaining one through a weekend of imbibing in the Camps and punting in the local Bookays.

Yours in Science, Jimmy Neutral.

Rheghead
25-Sep-05, 09:39
Actually Hereboy has a valid method though I'm not sure about the shortbread. I remember having to get rid of a pre-computational gamma spectroscope from work. To calculate the area under any peak on the printout the operator had to cut out the peak with scissors and weigh it on a calibrated balance. We could do the same here, weigh the 10cmX10cm square and weigh it, then cut out the shape and weigh it. Then the area is wt2/wt1 X100cm2.