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Square root of 3?

  1. Aug 21, 2005 #1
    An old man gives you a set square (http://www.buchhandlung-umbach.de/pbs/geodreick.gif) and then asks you to draw a line of exactly the length [tex] \sqrt{3} [/tex].

    How would you do it?
  2. jcsd
  3. Aug 21, 2005 #2
    In white: Draw a right triangle of lengths 1 and 2. The hypotenuse will have length [itex]\sqrt{3}[/itex].
    Last edited: Aug 21, 2005
  4. Aug 21, 2005 #3
    first you what units are we using?

    second, as far as i can see there are two ways:
    first you can straight away construct a 30-60-90 triangle with the hypotenuse (2) and the other leg of 1 (the one which is adjacent to the 60 degree) and then the other leg is sqrt(3).
    the second approach is you build an isosceles right triangle with the legs of 1, and then on the hypotenuse you bulid another 1 leg (when the degree between sqrt(2) and the third leg is 90), and then the second hypotenuse is sqrt(3).

    this was way too easy, so perhaps i totally misinterpratated your question.
  5. Aug 21, 2005 #4
    @jimmy: Your length would be [tex]\sqrt{5}[/tex]

    @ Loop:
    I thought about the second approach. You're right.

    Next question: What was the name of the old man :biggrin:
  6. Aug 21, 2005 #5
    Man, talk about a blind spot! I have no explanation for what I was thinking when I wrote that.
  7. Aug 28, 2005 #6
    think quite easy !

    1- first you draw a squareroot of 2 by using P=1 AND b =1

    then measure the length of the squareroot of 2, take it as p , and take 1 as b, then you can join the end of the these two sides.

    correct !

    hope so !
  8. Aug 28, 2005 #7


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    the word "exactly" means none of these answers are correct, nor is there any correct answer. this is impossible with drawing instruments, but only in the mind, or in the fantasy world of perfect mathematics can this be done exactly.
  9. Aug 29, 2005 #8
    I figure the old man must be Pythagoras.

    To get the answer “Exactly” we can’t be drawing so many lines as has been suggested.
    Observing the “Square” provided by the old man we see all three sides are neatly inscribe with halfway points. Thus we shall define it as 2 units by 2 units with a H of 2*sqrt{2} units.
    After inscribing an exact right angle of 1 by 2, we place a mark on the long length of exactly sqrt{2} by using the midpoint of the H side of the provided square.
    Now, again only using the square provided, we see it is long enough to draw a straight line from that marked point to the end of the short line. Giving us an exact sqrt{3} .

    But we will never be able to provide an exact measure of that line based on any fractional measure of the units we have selected, ie that measure is irrational.
    Note: The old man would appreciate it if you kept this part a secret – he fears the idea that a number can be irrational could cause a public panic! :wink:
    Last edited: Aug 29, 2005
  10. Aug 29, 2005 #9
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