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Math Help - help on circles

  1. #1
    Member helloying's Avatar
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    help on circles

    help me solve part (b)
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  2. #2
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    Area

    Hello -

    Area of APBC = 2 x Area of triangle OBC + area of sector OAPB, found in part (a)

    Area of triangle OBC = 0.5 x base x height, where base = OC and height = perpendicular distance from B to OP.

    Height = OB x sin (angle BOP), and angle BOP = 2 x angle BCP (angle at centre = twice angle at circumference.

    I hope you can do it now.

    Grandad
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  3. #3
    Member helloying's Avatar
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    Quote Originally Posted by Grandad View Post
    Hello -

    Area of APBC = 2 x Area of triangle OBC + area of sector OAPB, found in part (a)

    Area of triangle OBC = 0.5 x base x height, where base = OC and height = perpendicular distance from B to OP.

    Height = OB x sin (angle BOP), and angle BOP = 2 x angle BCP (angle at centre = twice angle at circumference.

    I hope you can do it now.

    Grandad
    Hello Grandad

    Thanks I got it now. However i got stuck in part (c) . Can u teach me please?
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  4. #4
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    Area

    Hi -

    In triangle OBT we know OB = 6, angle BOT = \frac{\pi}{3} and angle OBT = 90^o. So you can work out the length of BT and hence the area of the triangle.

    Then subtract the area of the sector OAPB (found in b) from area of quadrilateral OABT.

    Grandad
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