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    ok can you help me with this question:
    "in the diagram the chord AB of the circle with centre O is such that AOB = 1.4 radians. given that the shaded area is 4cm^2 find the radius of the circle correct to 2 decimal places."

    would this be the right idea? 1/2r^2•-1/2bcsin•=4

    • is theta (the angle)

    i have attached a picture of it below
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    (Original post by greenie787)
    ok can you help me with this question:
    "in the diagram the chord AB of the circkw with centre O is such that AOB = 1.4 radians. given that the shaded area is 4cm^2 find the radius of the circle correct to 2 decimal places."

    would this be the right idea? 1/2r^2•-1/2bcsin•=4

    • is theta (the angle)

    i have attached a picture of it below
    Use
    area of the sector = r^2 x angle
    and

    area of the triagle = r^2 x sin angle

    to form simulatius equation (they minus 2 make that shaded area) the r is the only variable!!!!


    is that any help???
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    (Original post by Harveer)
    Use
    area of the sector = r^2 x angle
    and

    area of the triagle = r^2 x sin angle

    to form simulatius equation (they minus 2 make that shaded area) the r is the only variable!!!!


    is that any help???
    i thought it was 1/2 bc sin angle to find the area of the triangle?
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    That diagram cheered me up a bit
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    I get the shaded area to be:

    As = 0.5*q*R^2 - R(sin(q/2)*cos(q/2))

    where q is theta
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    (Original post by greenie787)
    i thought it was 1/2 bc sin angle to find the area of the triangle?
    well, boh of them r used...but it fairly depends !!!
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    Area of the sector is

    1/2.r^2.[theta]

    Area of the triangle is

    1/2.r^2.sin[theta]

    The shaded segment is their difference:

    1/2.r^2.([theta] - sin[theta])

    These formulae only work when theta is in radians, btw.

    You know this area is 4cm^2, and theta is 1.4, so it all becomes

    4 = 1/2.r^2.(1.4 - sin(1.4))

    which you can solve for r. My arithmetic is wretched but I got 4.39cm (2dp)

    Hope this helps.

    Jon
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    thanks every 1 i should be able to work it out now.
 
 
 
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