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The Pythagorean Theorem

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  1. Apr 8, 2015 #1
    Anyone want to take a crack at it? My class has been discussing it: Find the length of AB:http://t4.rbxcdn.com/84e25f3830d66e6bbaeaba48e35c0781
     
  2. jcsd
  3. Apr 8, 2015 #2
  4. Apr 8, 2015 #3
    Will you please elaborate?
     
  5. Apr 8, 2015 #4
  6. Apr 8, 2015 #5
    Ok, great, but how did you get there?
     
  7. Apr 8, 2015 #6
    I am probably wrong, but I got sqrt(34).
     
  8. Apr 8, 2015 #7
    Find the coordinates of point A and use the distance formula to get AB. There are pleasant cancellations.
     
  9. Apr 8, 2015 #8
    I just found the hypotenuse of a triangle with legs 4+sqrt(2) and 4-sqrt(2).
     
  10. Apr 8, 2015 #9
    I did, and I still got sqrt(34)... One question though, the square at the top left corner.... If split vertically, to get 4 triangles, wouldn't the legs of one of those triangles be one?
     
  11. Apr 8, 2015 #10
    No, each leg would be sqrt(2) .
     
  12. Apr 8, 2015 #11
    Alright, I see my mistake, I checked my work again. Thanks.
     
  13. Apr 8, 2015 #12
    Yep, no problem. :smile:
     
  14. Apr 9, 2015 #13
    First, we choose the axes so as to simplify our calculation. Let us drop a perpendicular from A to the base and consider the the y-axis.

    The location of A is (0, 4 +√²) and B is (4 - √2, 0)
    Distance is √x²+y²

    As we know, (a - b)² + (a + b)² = 2( a² + b²)

    Which here is 2( 16 +2) = 36, the square root of it is 6.

    I'm sorry if I skipped some steps. It's very difficult to type in mathematical notation.
     
  15. Apr 11, 2015 #14

    epenguin

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    Was it given that the figures with sides 2 and 4 are squares?

    If so, 6.
     
    Last edited: Apr 11, 2015
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