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Math Help - Need the derivation of area formula of a segment

  1. #1
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    Need the derivation of area formula of a segment

    This is not in my syllabus. But on my geometry box i saw the formula.

    Radius(Pie Theta/360 - Sin Theta/2)

    We have area of segment in our syllabus but that consists of getting area of sector then subtracting the triangular area.
    I tried to do some stuffs using that., using trignometry.

    I first though Sin Theta gets us the Height of the triangle from the angle Theta. Like in 60 Degree it would be Root 3/2 but then i concluded its wrong.

    Can anyone derive and explain the formula to me?
    I am curious to know about it.
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  2. #2
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    Have a look at this: Circular Segment -- from Wolfram MathWorld
    Is that anything close to what you are asking?
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  3. #3
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    i dont think so. Because the last formula is different. Another thing, i dont know integration yet.
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  4. #4
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    Quote Originally Posted by shaurya View Post
    i dont think so. Because the last formula is different. Another thing, i dont know integration yet.
    Well then, you must do a better job in describing the problem.
    The description that you gave leads to the area of a sector or Circular Segments
    So reread the question and post a clarification.

    BTW: it is pi for \pi. We like apple pie.
    Last edited by Plato; February 20th 2009 at 08:49 AM.
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  5. #5
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    The thing there written is correct and the diagram too but its outta my reach bcuz half of thing i havent yet learned.
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  6. #6
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    The area of a "sector of a circle", the region in a circle of radius R between two radii making angle \theta, is given by \frac{\pi r^2\theta}{360} (when the angle is measured in degrees. The formula is simpler in radians). To see that, recall that a full circle is 360 degrees and has area \pi r^2 while angle \theta is \frac{\theta}{360} of the whole circle. Multiply \pi r^2 by that fraction.

    To get the segment, you need to subtract off the area of that triangular part between the center of circle and the chord. That is an isosceles triangle with side length R and and angle \theta. If you drop a perpendicular you get two right triangles with hypotenuse R and angle \theta/2.

    Now call the "near side" of that right triangle (part of the radius) "h" and the "opposite side" (half of the chord) "b".

    sin(\theta/2)= "opposite side/hyptotenuse= b/R so b= R sin(\theta/2).

    cos(\theta/2)= "near side/hypotenuse= h/R so h= R cos(\theta/2)

    The base of the entire isoscelese triangle is 2b so the area is (1/2)(2b)(h)= bh= R^2 sin(\theta/2)cos(\theta/2)

    Now we need to use a trigonometric identity: for any angle \phi, sin(2\phi)= 2 sin(\phi)cos(\phi). Replacing \phi by \theta/2, sin(\theta)= 2 sin(\theta/2)cos(\theta/2) so sin(\theta/2)cos(\theta/2)= \frac{1}{2}sin(\theta).

    That identity makes the area of the triangle \frac{R^2}{2}sin(\theta)

    Finally, then, the area of the sector is \frac{\pi R^2\theta}{360}- \frac{R^2}{2} sin(\theta)= R^2\left(\frac{\pi\theta}{360}- \frac{1}{2}sin(\theta)\right).

    Be very careful about fractions when writing "in line"! What you wrote would be interpreted as sin(\frac{\theta}{2}) but what you want is \frac{sin(\theta)}{2}.
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