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The Segments of a Polygon

  1. Aug 25, 2011 #1
    1. The problem statement, all variables and given/known data
    1. In an equilateral triangle ABC, a line segment is drawn from each vertex to a point of the opposite side so that the segment divides the side in the ratio 1:2, creating another triangle DEF.
    triangle.jpg
    a. What is the ratio of the area of the two equilateral triangles?
    b. Check the ratio of the areas for different ratio of side (1:n, pick your own value of n)
    c. By analyzing the results above, conjecture a relationship between the ratios of the sides and the ratio of the areas of the triangles
    d. Prove this conjecture analytically
    e. Does this conjecture hold for non-equilateral triangles? Explain

    2. Do a similar construction in a square where each side is divided into the ratio of 1:2.
    a. Compare the area of the inner square to the area of the original square
    b. How do the areas compare if each side is divided into the ratio 1:n?
    c. Prove the conjecture

    3. If segments were constructed in a similar manner in other regular polygons, would similar relationship exist? Investigate the relationship in another regular polygon




    2. Relevant equations
    Not sure


    3. The attempt at a solution
    I have already stuck at first question. I guess I use equation:
    Area of triangle = 1/2. a . b . sin θ

    But I can't find the side of smaller triangle DEF in terms of side of triangle ABC. Please help

    Thanks
     
  2. jcsd
  3. Aug 26, 2011 #2
    Perhaps you can use coordinate geometry to get the coordinates of the points. If you let the side length be a, the two lower points could be (say) (0,0) and (a,0). Find the coordinates of the upper point, then find the points on the sides which divide them into the given ratio. You can find the coordinates of the sides of the inner triangle by calculating the equation of the lines joining the dividing with their opposite vertices.

    EDIT: This seems like an IB Math assessment problem to me!
     
  4. Aug 26, 2011 #3

    tiny-tim

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    hi songoku! :smile:
    my inclination would be to use the cosine and sine rules to find the angles, and then the area, of the smallest triangles, and then to find the are of the small equilateral triangle by subtraction

    (btw, why do your very small diagrams take up so much space? :redface:)
     
  5. Aug 26, 2011 #4
    That seems to be a much more efficient approach!
     
  6. Aug 26, 2011 #5
    Yes, it is; and now it is my homework

    hi tiny - tim! :smile:
    let line BE meets side AC at Q. I can find the angle AQE but I still can't find the area of AEQ, if it is the smallest triangle you mean. I do not know how to find EQ or AE

    not sure :redface:
     
  7. Aug 26, 2011 #6

    tiny-tim

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    angle AEQ = 60° :wink:
     
  8. Aug 27, 2011 #7
    I get it. Thanks a lot :smile:
     
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