[FutureMedicalDoctor]

(Original post by MathematicsKiller)
I'm afraid instantaneous transportation devices still do not exist. Sorry.

If you manage to create a transporter, pm me and I'll be happy to study with you.

I forgot to mention online. Like carried out through services such as skype, tinychat, scriblink etc.?

[MathematicsKiller]

(Original post by FutureMedicalDoctor)
I forgot to mention online. Like carried out through services such as skype, tinychat, scriblink etc.?

Unfortunately, my internet sucks too much for that.

[SimonM]

(Original post by FutureMedicalDoctor)
I forgot to mention online. Like carried out through services such as skype, tinychat, scriblink etc.?

No, but it's not a bad idea.

[tommm]

I can't believe that 633 people are currently reading the maths forum...

[Piecewise]

(Original post by tommm)
I can't believe that 633 people are currently reading the maths forum...

It's 711 now! O_o

[SimonM]

(Original post by tommm)
I can't believe that 633 people are currently reading the maths forum...

It is the week of STEP, AEA, and the last few FPs and later A-level modules... An interesting time to be on TSR. (If you don't have a ticket to John's)

[Daniel Freedman]

(Original post by SimonM)
It is the week of STEP, AEA, and the last few FPs and later A-level modules... An interesting time to be on TSR. (If you don't have a ticket to John's)

In my opinion, Trinity was better than John's (but obviously I'm biased).

[tommm]

I saw Trinity's fireworks (on a punt, right in the middle, right in front of the fire and water jets) and they were pretty ****ing awesome. Went to Downing's ball last night, was generally awesome but THEY RAN OUT OF BACON AT 4AM

[Astronomical]

How do you do this?

Its off the oxford PAT 2010, and is driving me mad!

[tommm]

(Original post by Astronomical)
How do you do this?

Its off the oxford PAT 2010, and is driving me mad!

I'd probably use integration, not sure if there's a nicer way of doing it.

This isn't the best place to ask this, you should really start a new thread.

[Oh I Really Don't Care]

(Original post by Astronomical)
How do you do this?

Its off the oxford PAT 2010, and is driving me mad!

I have only glanced at this but the first thing that popped in my mind was you can probably make an ellipse out of that ...

[ben-smith]

(Original post by Astronomical)
How do you do this?

Its off the oxford PAT 2010, and is driving me mad!

you know the area of a segment right?
Well, consider the right angled triangle made from the center, the point where the dividing line touches the circumference and the point where the dividing line touches the x axis. you know two of the lengths and it is a right angled triangle so some simple trigonometry will tell you the half the angle over which the area of the segment you want is subtended. Then it's just a matter of plugging away with the formula.
Integration? Elipses? you guys have been doing too much real maths!

[anshul95]

(Original post by ben-smith)
you know the area of a segment right?
Well, consider the right angled triangle made from the center, the point where the dividing line touches the circumference and the point where the dividing line touches the x axis. you know two of the lengths and it is a right angled triangle so some simple trigonometry will tell you the half the angle over which the area of the segment you want is subtended. Then it's just a matter of plugging away with the formula.
Integration? Elipses? you guys have been doing too much real maths!

yes I did it this way too. Thought I was missing something when all these other guys were talking about ellipses - but then I saw your post. Thanks for confirmation. I would like to see how the ellipse way turns out though.

[dnumberwang]

(Original post by ben-smith)
you know the area of a segment right?
Well, consider the right angled triangle made from the center, the point where the dividing line touches the circumference and the point where the dividing line touches the x axis. you know two of the lengths and it is a right angled triangle so some simple trigonometry will tell you the half the angle over which the area of the segment you want is subtended. Then it's just a matter of plugging away with the formula.
Integration? Elipses? you guys have been doing too much real maths!

My first instinct was to integrate too

[ben-smith]

(Original post by dnumberwang)
My first instinct was to integrate too

sledgehammers are fun to use

[Astronomical]

(Original post by ben-smith)
you know the area of a segment right?
Well, consider the right angled triangle made from the center, the point where the dividing line touches the circumference and the point where the dividing line touches the x axis. you know two of the lengths and it is a right angled triangle so some simple trigonometry will tell you the half the angle over which the area of the segment you want is subtended. Then it's just a matter of plugging away with the formula.
Integration? Elipses? you guys have been doing too much real maths!

Thanks, I think I get how to do it now. I will have another go at it later.

(Original post by DeanK22)
I have only glanced at this but the first thing that popped in my mind was you can probably make an ellipse out of that ...

I'd love to see this!

[Oh I Really Don't Care]

(Original post by Astronomical)
I'd love to see this!

Assuming you are familiar with the area of an ellipse with major, minor radius a,b is

pi*a*b

then

pi/2 * R/2 * sqrt(R^2 + R^2/4)

from the diagrama nd pythagoras

[Astronomical]

(Original post by ben-smith)
you know the area of a segment right?
Well, consider the right angled triangle made from the center, the point where the dividing line touches the circumference and the point where the dividing line touches the x axis. you know two of the lengths and it is a right angled triangle so some simple trigonometry will tell you the half the angle over which the area of the segment you want is subtended. Then it's just a matter of plugging away with the formula.
Integration? Elipses? you guys have been doing too much real maths!

(Original post by anshul95)
yes I did it this way too. Thought I was missing something when all these other guys were talking about ellipses - but then I saw your post. Thanks for confirmation. I would like to see how the ellipse way turns out though.

(Original post by Astronomical)
Thanks, I think I get how to do it now. I will have another go at it later.

So, having tried the triangle method, I got:



Is this right?

[Oh I Really Don't Care]

(Original post by Astronomical)
So, having tried the triangle method, I got:



Is this right?

I am unsure as to what you have done because I haven't looked at it but;

area = area of sector - 2*(area of identical right angle triangles).

If that's what you did then yes, you have done it correctly.

[Astronomical]

(Original post by DeanK22)
I am unsure as to what you have done because I haven't looked at it but;

area = area of sector - 2*(area of identical right angle triangles).

If that's what you did then yes, you have done it correctly.

This might help explain what I tried to do:



The formula for A is just that for finding the area of a segment subtended by an angle, and is essentially just the area of the sector, minus that of the triangle, and then factorised. Our textbook derives it as such: