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Math Help - The problem of shaded region's area

  1. #1
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    The problem of shaded region's area

    The problem of shaded region's area-maths-1.jpg


    I don't know how to do this question and the correct answer is A.
    Can anybody tell me how to proof that the Answer A is correct?
    And how to find 82π​?
    Thanks for answering.
    Last edited by william24; July 14th 2012 at 08:52 PM.
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  2. #2
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    Re: The problem of shaded region's area

    HAVE NAMED THE RECTANGLE AS ABCD
    A IS THE VERTEX AT LEFT HAND CORNER, B AT RIGHT HAND CORNER, C AND D ARE THE VERTICES ON THE DIAMETER OF THE SEMI-CIRCLE
    Let the center of the circle be at O
    Radius of circle=OA
    NOW OA=SQRT(100+64)=SQRT(164)=RADIUS
    AREA OF SEMICIRCLE-1/2(PI*OA*OA)=82PI
    SHADED AREA=AREA OF SEMICIRCLE-AREA OF RECTANGLE
    =82*PI-160
    Last edited by hemvanezi; July 14th 2012 at 09:57 PM.
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  3. #3
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    Re: The problem of shaded region's area

    Quote Originally Posted by william24 View Post
    Click image for larger version. 

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    I don't know how to do this question and the correct answer is A.
    Can anybody tell me how to proof that the Answer A is correct?
    And how to find 82π​?
    Thanks for answering.
    To evaluate the area of a circle or semicircle, you need to know what the radius is.

    Draw a segment from the centre of the circle (on the diameter) to one of the corners of the rectangle that touches the semicircle. This is the radius of the circle and semicircle. It also creates a right angle triangle, so you can use Pythagoras to evaluate the length of the radius. Then you can evaluate the area of the semicircle and subtract the area of the rectangle.
    Thanks from william24
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