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Finding area of octagon

  1. Aug 10, 2012 #1
    1. The problem statement, all variables and given/known data
    The question says that there is a square that has a side of length 1cm. It then says 4 equal triangles are cut, one off each corner. The resulting shape is a regular octagon. What is its area?


    2. Relevant equations
    I believe the sides of the triangles to both be (1-x)/2 assuming we let each side of the octagon be length x. This means that length x is represented by (triangle side length)^2 + (triangle side length)^2 = x^2.



    3. The attempt at a solution
    I essentially let the above equations equal and I get...
    (0.5 - x^2/2)^2 + (0.5 - x^2/2)^2 = x^2
    Using quadratic after simplifying the above...
    X= rt2 - 1
    From there I let that equal x again. I find the total area of all 4 triangles and get 1- area of four triangles which leaves me with 0.828cm^2

    Is this correct? I'm not convinced it is actually correct. Any guidance?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 10, 2012 #2

    PeterO

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    Homework Helper

    I got that answer calculating a different way, so it is probably correct.
     
  4. Aug 11, 2012 #3

    LCKurtz

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    Gold Member

    That's correct, but why give a decimal approximation when you have an exact answer:$$
    1-\frac 2 {(\sqrt 2 + 2)^2}$$
     
  5. Aug 11, 2012 #4
    A better simplification would be 2(√2 - 1)
     
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