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Need help with this DIFFICULT QUESTION :/ watch

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    is this A Level
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    Polar integration will help. But I'm assuming this is AS?
    Actually looking at it polar would be of no use really.
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    Notice that it says OB and OC are arcs of circles.
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    AOB = Equilateral Triangle.

    The angle = pie/3 as a result.

    Find the area of the semi circle to the left.

    Use pie/3 as an angle to find both sectors.

    Add these up, then take away from the area of the whole circle.
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    (Original post by Bigbosshead)
    AOB = Equilateral Triangle.

    The angle = pie/3 as a result.

    Find the area of the semi circle to the left.

    Use pie/3 as an angle to find both sectors.

    Add these up, then take away from the area of the whole circle.
    But how do you know AOB is Equilateral Triangle?
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    Solomon paper L
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    I was thinking the same until i realised that:

    AO = OB

    AB = AO

    therefore AO = AB = OB
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    (Original post by Adorable98)
    But how do you know AOB is Equilateral Triangle?
    You are told that arc OB is an arc of a circle with radius r and centre A.

    So AO and AB must be equal since they have to be radii of that circle with centre A.

    And AO = OB since they're both radii of the main circle.

    So AO = AB = OB.
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    Same question was posted 3 days ago:
    http://www.thestudentroom.co.uk/show...729&highlight=

    should be useful^
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    (Original post by B_9710)
    Polar integration will help. But I'm assuming this is AS?
    Actually looking at it polar would be of no use really.
    I mean, you could...

    Spoiler:
    Show
    Observe that the circle formed by extending the arc OB has equation \rho = 2r\sin \theta in polar coordinates (\rho, \theta). Given that the equation of the main circle is \rho = r, it is easily seen that the first intersection of the two circles occurs at B = (r,\pi /6) (in polars). Hence the area of the segment defined by sector AOB is given by:

    \dfrac{1}{2}\displaystyle\int_0^  {\pi /6} (2r\sin \theta)^2 d\theta = r^2 \displaystyle\int_0^{\pi /6} [1-\cos 2 \theta] d\theta = r^2 [ \pi / 6 - \sqrt 3 /4]

    Noting also that the area of sector OBC is \dfrac{1}{2}r^2 \cdot \dfrac{\pi}{3} and that the segment defined by sector OCD is equal in area to the one found above by symmetry, we see the shaded region is given by:

    r^2\dfrac{\pi}{6} - 2r^2 [ \pi / 6 - \sqrt 3 /4] = \dfrac{1}{6}r^2[3\sqrt{3} - \pi]


    But this is slightly more involved than simply working out the segment directly (since, in this case, the arc is all circular and nice) and therefore excessive.
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    (Original post by notnek)
    You are told that arc OB is an arc of a circle with radius r and centre A.

    So AO and AB must be equal since they have to be radii of that circle with centre A.

    And AO = OB since they're both radii of the main circle.

    So AO = AB = OB.
    (Original post by Bigbosshead)
    I was thinking the same until i realised that:

    AO = OB

    AB = AO

    therefore AO = AB = OB
    (Original post by Bigbosshead)
    AOB = Equilateral Triangle.

    The angle = pie/3 as a result.

    Find the area of the semi circle to the left.

    Use pie/3 as an angle to find both sectors.

    Add these up, then take away from the area of the whole circle.
    (Original post by B_9710)
    Notice that it says OB and OC are arcs of circles.
    (Original post by SM45367)
    Same question was posted 3 days ago:
    http://www.thestudentroom.co.uk/show...729&highlight=

    should be useful^
    Thank you so much!!!
 
 
 
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