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I entered that code. I do not see any picture.

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,730

Just the code? Did you put in a point E?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

Hi,

I drew up Agnishom's pic, entered your code and got the same for g as you did.

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,730

It is better if you enter IntegralBetween[sqrt(49 - x²), 7, x(E), 0] then he will shade the area.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

It only shaded the left half of the shape and output the area value just for that section.

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,730

Yes but you can see by symmetry that the answer is twice that. I left out the 2 in 2*IntegralBetween[sqrt(49 - x²), 7, x(E), 0] so you can see what geogebra is integrating.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

Yes, I saw the symmetry but wondered if there was a way to shade the whole shape and give the whole area, and so, not knowing what I was doing, I tried reordering components of the expression but always got the same left-half result.

*Last edited by phrontister (2014-03-31 02:46:12)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,730

To get both sides shaded you will have to abandon the sector idea and go to circles as she did.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

Is the maths for that fairly simple, or will it need 'integrals' and/or 'functions', neither of which I've ever learnt about? Or something else not taught in high school, perhaps?

I feel I'm getting out of my depth here.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,730

Geogebra will do it all for you.

Want to try?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

Yes. I've already given it a go but wasn't getting anywhere with it (or maybe I was close, but my nose got in the way).

I think I need a hint.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,730

Draw the square and use the Circle with radius tool.

Pick point A and click it and enter 7 then do the same with point B.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

I already tried that a while back, but no ideas leapt out at me.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,730

It will in just a second. Now use the intersection tool and click on both circles. E and F will be created.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

Done...I hadn't created F before.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,730

F will be on the bottom. If it is not then just rename the top intersection point E.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

The orientation is correct. Still no ideas emerging...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,730

Okay enter this:

IntegralBetween[sqrt(49 - x²), 7, x(E), 0] and IntegralBetween[sqrt(49 - (x - 7)²), 7, 7, x(E)]

color them any way you like. What do you see?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

The shape's left and right shaded halves and their respective area values (same for each).

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,730

Does yours look like this?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

Yes, exactly.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,730

The nice thing is that the equations for the circles are calculated for you. Under conics. With those equations M can now get exact answers. That is what she did.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

You'll have to enlighten me, I think. I've never learnt curve equations and don't understand what they do.

I should really turn in now, though, as it's waaay after midnight. To be continued...(thanks so far )

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,730

Okay, goodnight and we will continue tomorrow.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

Using trig to find that ∠DAE = 30°,

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