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  1. #1
    Member Veronica1999's Avatar
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    triangle

    A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths 3,4,and 5. What is the area of the triangle?

    Not sure how to start the problem...

    Could I get some hints please?
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  2. #2
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    You must draw a diagram. Inscribe a triangle in a circle.
    Draw three radial segments, one from each vertex to the center,
    Now you have three sub-triangles. The sum of their areas is the area you want.

    The radius of the circle is r=\frac{6}{\pi}. How and why?
    The angle subtending the arc of length 5 measures \theta_5=\frac{5\pi}{6}. Again how and why?.

    If a~\&~b are the length of two sides of a triangle and \theta is the measure of the angle between then the area to that triangle is \frac{1}{2}a\cdot b\sin(\theta).
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  3. #3
    Member Veronica1999's Avatar
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    I think I got it. It took me a while though....

    I understand the radius is 6/pi because 2pir = 12 r= 6/pi.

    Then I understood the angle subtending the arc of length 5 is 5pi/6 because 2pi X 5/12 = 5pi/6
    arc of length 4 2pi X 4/12 = 2pi/3
    arc of length 3 2pi X 3/12 = pi/2

    I fully understand why 1/2absin(theta) gets the area therefore,

    1/2 X 6/pi X 6/pi X sin 5pi/6 + 1/2 X 6/pi X 6/pi X sin 2pi/3 + 1/2 X 6/pi X 6/pi X sin pi/2 = 9/pi squared X (3 + root3)


    Thanks!!!
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  4. #4
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    Quote Originally Posted by Veronica1999 View Post
    ....... = 9/pi squared X (3 + root3)
    Agree.
    AS a general case:
    r = radius; a,b,c = 3 given arc lengths with u,v,w = corresponding central angles

    Area = r^2[SIN(u) + SIN(v) + SIN(w)] / 2, where:

    r = (a + b + c) / (2pi)
    u = 360a / (a + b + c)
    v = 360b / (a + b + c)
    w = 360c / (a + b + c)
    Last edited by Wilmer; February 12th 2011 at 09:24 AM. Reason: general case
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