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    (Original post by Student403)
    Yeah there are lots of ways. I used C12 at first (integration) to just brute force it.
    I just threw in coordinate geometry/equation of a circle. Elegance is not my forte
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    (Original post by 13 1 20 8 42)
    I just threw in coordinate geometry/equation of a circle. Elegance is not my forte
    To be honest I'd have done the same on another day

    Anyway if it's stupid but works, it ain't stupid
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    (Original post by HFancy1997)
    4(25-(Pie*5^(2)/4)=21.46

    dunno how you get the right answer but I got this
    Edit: I thought the bottom bit is shaded nvm im wrong

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    Edit
    3(25-(pie*5^(2)/4)+
    (15-(25/2)*(ArcTan(4/3))

    =19.5(3sf)

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    (Original post by Student403)


    Find the area of the shaded bit
    Nice question, loved doing it because it was quite challenging

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    (Original post by HFancy1997)
    Nice question, loved doing it because it was quite challenging

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    Good job
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    (Original post by Student403)
    Good job
    Got the exact answer of
    90-\frac{75}{4}\pi -\frac{5}{36}\pi \cdot tan^{-1}(\frac{1}{2})

    Finding that corner bit proved to be slightly annoying but straight forward once I started to consider only one half of the overall rectangle. I'd be surprised if anyone from GCSE get the right answer
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    (Original post by RDKGames)
    Got the exact answer of
    90-\frac{75}{4}\pi -\frac{5}{36}\pi \cdot tan^{-1}(\frac{1}{2})

    Finding that corner bit proved to be slightly annoying but straight forward once I started to consider only one half of the overall rectangle. I'd be surprised if anyone from GCSE get the right answer
    could you explain how you worked it out?
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    (Original post by RDKGames)
    I'd be surprised if anyone from GCSE get the right answer
    Same
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    (Original post by Student403)
    Same
    Mind if I explain how to work it out? Or do you want to keep it low-key?
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    (Original post by Hashtosh302)
    could you explain how you worked it out?
    You're the guy who made the AMA thread about wanting to go to Oxbridge Now's your chance to prove yourself

    (Original post by RDKGames)
    Mind if I explain how to work it out? Or do you want to keep it low-key?
    I reckon if he's serious about Oxbridge, he should get it himself lol
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    (Original post by RDKGames)
    Mind if I explain how to work it out? Or do you want to keep it low-key?
    Yeah could you expalin it
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    Got \displaystyle  90+\frac{25}{2} \arcsin \left (\frac{3}{5 \right )} -25\pi .
    Is it even possible by merely using GCSE techniques.
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    (Original post by B_9710)
    Got \displaystyle  90+\frac{25}{2} \arcsin \left (\frac{3}{5 \right )} -25\pi .
    Is it even possible by merely using GCSE techniques.
    Yes, I restricted myself to only GCSE knowledge and did it, but the process is simply too long for the corner area for most GCSE student's to bother.
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    If anyone wants a more rigorous treatment of the answer, this question was asked on math.stackexchange a week ago.

    Definitely not a GCSE level problem though, since you need to integrate
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    (Original post by thatllbeme)
    If anyone wants a more rigorous treatment of the answer, this question was asked on math.stackexchange a week ago.

    Definitely not a GCSE level problem though, since you need to integrate
    you dont need to integrate, as I didnt but yeah I agree its not gcse problem unless someone shows me a method without A level methodss

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    (Original post by thatllbeme)
    If anyone wants a more rigorous treatment of the answer, this question was asked on math.stackexchange a week ago.

    Definitely not a GCSE level problem though, since you need to integrate
    Wrong. You don't need to integrate. That is one way of doing it (and makes it quite easy)

    I said it's GCSE level because it's possible to do it with entirely GCSE taught techniques
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    (Original post by thatllbeme)
    If anyone wants a more rigorous treatment of the answer, this question was asked on math.stackexchange a week ago.

    Definitely not a GCSE level problem though, since you need to integrate
    The congruent triangles argument there is technically GCSE. And nice..
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    (Original post by HFancy1997)
    you dont need to integrate, as I didnt but yeah I agree its not gcse problem unless someone shows me a method without A level methodss

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    look at that stack exchange page, there is an argument using only GCSE knowledge
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    (Original post by thatllbeme)
    If anyone wants a more rigorous treatment of the answer, this question was asked on math.stackexchange a week ago.

    Definitely not a GCSE level problem though, since you need to integrate
    (Original post by HFancy1997)
    you dont need to integrate, as I didnt but yeah I agree its not gcse problem unless someone shows me a method without A level methodss

    Posted from TSR Mobile
    This problem is not aimed at GCSE students by any means, but it can indeed be solved using only GCSE knowledge.
    Take a look at this diagram I've spent some time coming up with:
    Name:  DIAG2.PNG
Views: 104
Size:  12.6 KB

    Firstly, there is a crap load of symmetry in here. That means the yellow highlighted areas are exactly the same. Furthermore, the two portions of the circle within that right-angled triangle add up to create an identical circle that you see in the diagram, so literally you can get most of the area by simply doing the area of the triangle take the area of the circle. You can work out the red area by simply doing area of a square take area of a quarter circle. GCSE stuff. The red area is the same at every corner and can be expressed as a sum of two areas; A+B or whatever you wish, then either A or B will have to be equal to the yellow area. Pay attention to the angle, it's simple trigonometry, you can find the angle using inverse tan. You can then shift focus on the circle on the right where I've labelled the angles that GCSE student's should know. The area of the right angled triangle XYZ can easily be worked out, and it is made up from the areas of the isosceles triangle within the circle, the sector of the area, AND the most important of all, the black shaded area which can be used to work out the yellow highlighted corner because we know what they add up to! GCSE student's should be able to find the area of the isosceles triangle using sine, and area of the sector, at which point it's just add/subtract some areas to get what you want.

    Really, nothing too complicated here. Once you find the yellow shaded area you just take it away from whatever you need to in order to get the shaded bits only. I can understand that A-Level methods can prove to be quicker and less tedious but this problem is easily accessible to GCSE student's once they have an idea of what to do and when throughout their process.
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    Thanks for the question, it was really interesting! I'm GCSE student and I managed to get it
 
 
 
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