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    I did it by getting the equation for the segment via the equation and taking the angle of the sector as pi/3, I then took 2(segment) and 2(sector) away from the area of the semi circle to find the shaded area. Hope this helps.
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    (Original post by notnek)
    Firstly, AB=r is incorrect.

    I think you're assuming that unshaded region ABO is a sector when it isn't. A sector is enlosed by one arc and two radii. This region is enclosed by one radius and two arcs.
    Surely AB is equal to r? The circle centre A has the same radius as the circle centre O, and B is a point on the circle centre A?
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    (Original post by AG98)
    But OB is the arc of circle radius r centre A, so AB is a radius = r?
    Sorry - my mistake.
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    (Original post by AG98)
    But OB is the arc of circle radius r centre A, so AB is a radius = r?
    You don't need to know AB for any of the calculations though. The area of the segments and sectors can be found with the radius R and the angle pi/3.
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    (Original post by lizard54142)
    Surely AB is equal to r? The circle centre A has the same radius as the circle centre O, and B is a point on the circle centre A?
    That's what I thought :confused:

    But I've clearly gone wrong somewhere
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    (Original post by Terminatoring)
    If may not be AQA but that doesn't mean it's any less important or valuable.
    The thing about maths is we're all equal, regardless of the exam board the maths remains the same.

    That's why i think you should help me




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    Nah, MEI OCR is 10x harder than Edexcel, other than that I don't know enough to comment. But Edexcel is so boringly easy it's unfair.
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    (Original post by AG98)
    That's what I thought :confused:

    But I've clearly gone wrong somewhere
    You're correct, no worries
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    (Original post by lizard54142)
    Someone's jumping straight to the mark scheme
    Hahaha:happy2:
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    (Original post by joe12345marc)
    Nah, MEI OCR is 10x harder than Edexcel, other than that I don't know enough to comment. But Edexcel is so boringly easy it's unfair.
    It's easy to say that when you aren't studying it.
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    (Original post by Louisb19)
    It's easy to say that when you aren't studying it.
    Do a June 2014 M2 paper in both, mark it and come back to me and try persuade me that Edexcel is harder :P
    Nah seriously, MEI OCR just tricks people, it's horrible. Edexcel is more focused on your understanding, which is way easier than understanding attempts to catch you out.
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    (Original post by lizard54142)
    You're correct, no worries
    But I still don't get why there's a factor of (3sqrt3 - pi) instead of just (pi), as I wrote in my attempted solution
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    (Original post by AG98)
    But I still don't get why there's a factor of (3sqrt3 - pi) instead of just (pi), as I wrote in my attempted solution
    Scroll up and look at the working out that I linked. Pi is not a factor of everything because in order to find the area of the segment you need to find the area of a triangle (which does not contain pi in the calculation).
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    (Original post by AG98)
    But I still don't get why there's a factor of (3sqrt3 - pi) instead of just (pi), as I wrote in my attempted solution
    (Original post by notnek)
    I think you're assuming that unshaded region ABO is a sector when it isn't. A sector is enlosed by one arc and two radii. This region is enclosed by one radius and two arcs.
    What notnek said here! AOB as it is, is not a sector. It is the sector described by the lines OA and OB, and the segment described by the line OB and the arc OB.
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    (Original post by Louisb19)
    Did this question earlier today. Here is my working, if you still don't understand tell me.

    Attachment 398759
    This was very helpful, thanks a lot

    Btw - is that a different equation for the segment? (As opposed to the theta - sintheta one)


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    (Original post by notnek)
    Sorry - my mistake.
    Thanks for the help


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    (Original post by lizard54142)
    Consider the line OC, this gives a sector of the circle, you can work out this area easily. You're then left with the segment of arc OC and line OC. Then draw the line OD, notice you have an equilateral triangle OCD. You can work out this area, and then the area of the segment from this, etc... repeat similarly for AOB!

    EDIT: Beaten to it by two people! Gosh I should I typed quicker
    Thanks for the help


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    (Original post by Terminatoring)
    Thanks for the help


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    No problem. Did you get there in the end?
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    (Original post by Terminatoring)
    This was very helpful, thanks a lot

    Btw - is that a different equation for the segment? (As opposed to the theta - sintheta one)


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    The equation I used was Segment = Sector - Triangle (inside the sector)
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    (Original post by lizard54142)
    What notnek said here! AOB as it is, is not a sector. It is the sector described by the lines OA and OB, and the segment described by the line OB and the arc OB.
    I'm probably being extremely dumb here, so sorry for that

    But I still don't get why you can't just get the area of half the circle and take away the areas outlined in red here:

    I understand that you could work out the area of the triangle OCB and take away the unshaded segments, but that just seems like a more complicated way of doing it, and while it does work, I still don't understand why the original way I did it was wrong
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    (Original post by AG98)
    I'm probably being extremely dumb here, so sorry for that

    But I still don't get why you can't just get the area of half the circle and take away the areas outlined in red here:

    I understand that you could work out the area of the triangle OCB and take away the unshaded segments, but that just seems like a more complicated way of doing it, and while it does work, I still don't understand why the original way I did it was wrong
    You can take away the areas in red. But what are these areas?

    They're not sectors...
 
 
 
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