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Thread: Thales [semi circles, not the other one]

  1. #1 Thales [semi circles, not the other one] 
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    Is there a formula to predict the changing area when using Thales theorem?
    ie. inside a semi-circle you place a right angled triangle with the edge on the curve; now if you move the right-angle edge around the outer curve the size of that triangle will change, but within the limits of the original semi-circle. Is there a formula for this change?
    Thanks,


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  3. #2  
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    Here I just wrote it
    Area A- area B= constant
    Area inside the triangle is a constant
    Area,A+areaB+Area of triangle=area of the,semicircle
    AreaA+AreaB+(1/2)base*height=1/2(pi)(r^2)


    Last edited by fiveworlds; August 25th, 2013 at 08:16 AM.
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  4. #3  
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    No, but I found an easier example.
    http://en.wikipedia.org/wiki/File:Thales%27_Theorem_Simple.svg

    I
    f you move B closer to C the area of the triangle is tiny, if you move it away the area is much larger. Is there a way to calculate this change if you know the size of the circle?
    Thanks,
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  5. #4  
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    No
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  6. #5  
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    It is actually no ph the best we can do is work in cartesian co-ordinates but even then.. You could accurately say that the rate of change of one area is proportionate to the rate of change of the others aftet that.... I dont know you might be better with the cartesian than I
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  7. #6  
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    Not really mostly its just approximate. pythagoras works for the right angled triangles but the shapes on either side...
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  8. #7  
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    We could use the arc length fomula or whatever with that but its always approximate. always
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  9. #8  
    KJW
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    I get for the area of the triangle, where is the radius of the circle, and is the angle at the centre of the circle from the base of the semi-circle to the third point of the triangle.
    anticorncob28 likes this.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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  10. #9  
    KJW
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    Quote Originally Posted by PhDemon View Post
    Do you have any clue what fiveworlds is on about?
    No.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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  11. #10  
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    You should know this in mathematics. A+B=C In reality LiquidA+LiquidB!=liquid C all in ml since c is c returns false then the proof is wrong
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  12. #11  
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    Why irrelevant?? No reactions have occured no gasses produced its accurate
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  13. #12  
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    I already told him the awnser the pythagoran theorem will calculate the change in area of the triangle. Where hyp^2 is my diameter we can find the two sides easy. Then just find the area of the triangle
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  14. #13  
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    Not really 1 I said that we,dont have a way to really do this (true) 2 I,said how it could be done and warned it,was completely inaccurate 3 I proved exactly why it was completely inaccurate
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  15. #14  
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    Quote Originally Posted by fiveworlds View Post
    Here I just wrote it
    Area A- area B= constant
    Area inside the triangle is a constant
    Area,A+areaB+Area of triangle=area of the,semicircle
    AreaA+AreaB+(1/2)base*height=1/2(pi)(r^2)
    You haven't defined A;
    you haven't defined B;
    from the construction of your claim that a triangle is constant infers an inductive proof, yet you skip ahead and post a deductive formula without answering the question.

    You've been warned several times to tighten up your contributions to the math forum. Giving you month off this time, since a few days obviously hasn't focused your attention.
    KALSTER, tk421, PhDemon and 1 others like this.
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  16. #15  
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    Was just about to do something similar. Next time though it will be permanent.
    Disclaimer: I do not declare myself to be an expert on ANY subject. If I state something as fact that is obviously wrong, please don't hesitate to correct me. I welcome such corrections in an attempt to be as truthful and accurate as possible.

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  17. #16  
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    Let's see. Suppose you have a point T along the circumference of the semi-circle in which you create the right triangle. Given the angle measure of T (from 0 to pi) on the semi-circle, you are asking for the area of the triangle, is this correct? In that case it's pretty simple. Suppose the arc measure up to T is theta. Then there is a theorem in geometry stating that the arc measure is twice the inscribed angle, so one angle in the triangle is theta/2. Now you can use basic trigonometry to solve for the base and height, since you know the hypotenuse of the triangle is twice the radius (say, r). You get base = 2r*sin(theta/2) and height = 2r*cos(theta/2). Then area = 1/2*(2r)^2*sin(theta/2)*cos(theta/2) = 2r^2sin(theta/2)cos(theta/2). Using the double-angle identity sin(x)cos(x) = 1/2*sin(2x), you get the area = r^2*sin(theta). So the answer below is right. If you want the formula for the "change" in the area, as far as I think that's the derivative of this function which would simply be r^2cos(x) if r is a constant.
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  18. #17  
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    Quote Originally Posted by anticorncob28 View Post
    Let's see. Suppose you have a point T along the circumference of the semi-circle in which you create the right triangle. Given the angle measure of T (from 0 to pi) on the semi-circle, you are asking for the area of the triangle, is this correct? In that case it's pretty simple. Suppose the arc measure up to T is theta. Then there is a theorem in geometry stating that the arc measure is twice the inscribed angle, so one angle in the triangle is theta/2. Now you can use basic trigonometry to solve for the base and height, since you know the hypotenuse of the triangle is twice the radius (say, r). You get base = 2r*sin(theta/2) and height = 2r*cos(theta/2). Then area = 1/2*(2r)^2*sin(theta/2)*cos(theta/2) = 2r^2sin(theta/2)cos(theta/2). Using the double-angle identity sin(x)cos(x) = 1/2*sin(2x), you get the area = r^2*sin(theta). So the answer below is right. If you want the formula for the "change" in the area, as far as I think that's the derivative of this function which would simply be r^2cos(x) if r is a constant.
    That's precisely the method I used to obtain the answer posted in #11.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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  19. #18  
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    That's precisely the method I used to obtain the answer posted in #11.
    I thought so. I'm not too sure if we really answered the question because the answer was very straight forward, and unless this person is particularly bad at maths, why he didn't figure it out himself. Maybe he just didn't know/think of the inscribed angle theorem.
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  20. #19  
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    You haven't defined A;
    you haven't defined B;
    from the construction of your claim that a triangle is constant infers an inductive proof, yet you skip ahead and post a deductive formula without answering the question.
    Judging from the equation "Area,A+areaB+Area of triangle=area of the,semicircle" I think he means that when you place a triangle inside a semicircle the way Thales did, you get two regions that are not inside the triangle, and fiveworlds simply labeled them as areas A and B, but he did not give any method for working out the values or an answer to this question. And the claim the triangle area is constant is false, unless we were only worrying about a specific case. All in all, I do agree with the month ban.
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  21. #20  
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    Quote Originally Posted by anticorncob28 View Post
    That's precisely the method I used to obtain the answer posted in #11.
    I thought so. I'm not too sure if we really answered the question because the answer was very straight forward, and unless this person is particularly bad at maths, why he didn't figure it out himself. Maybe he just didn't know/think of the inscribed angle theorem.
    Thales' theorem (the title of this thread) is just a special case of inscribed angle theorem. My concern as to whether or not we answered the question (whether our method is appropriate) is whether we have used mathematics that is more advanced than the result we are trying to obtain (e.g. the inscribed angle theorem is more advanced than Thales' theorem). On the other hand, the fact that the answer contains indicates that trigonometry is necessary.
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  22. #21  
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    Quote Originally Posted by KJW View Post
    Quote Originally Posted by anticorncob28 View Post
    That's precisely the method I used to obtain the answer posted in #11.
    I thought so. I'm not too sure if we really answered the question because the answer was very straight forward, and unless this person is particularly bad at maths, why he didn't figure it out himself. Maybe he just didn't know/think of the inscribed angle theorem.
    Thales' theorem (the title of this thread) is just a special case of inscribed angle theorem. My concern as to whether or not we answered the question (whether our method is appropriate) is whether we have used mathematics that is more advanced than the result we are trying to obtain (e.g. the inscribed angle theorem is more advanced than Thales' theorem). On the other hand, the fact that the answer contains indicates that trigonometry is necessary.
    It turns out that my intuition was right and that the method I used to obtain the answer was far more complicated than it needed to be. In fact, the area of a triangle is half the base times the height. The base is and the height is , giving as the area. I'm guessing that Thales' theorem misdirected me (and anticorncob28) to consider the triangle as a right triangle with the lengths of the right-angled sides being sought.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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  23. #22  
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    This problem can also be approached using co-ordinate geometry. Let the semicircle be described by the Cartesian equation . A general point on the semicircle is given by where is the angle the line joining the point and the origin makes with the positive x-axis. Then the area of the triangle with vertices , and is given by formula as:


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