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Distance between Two Men

Date: 04/24/2003 at 10:44:18
From: Luis
Subject: Triangle

Two men, starting at the same point, walk in opposite directions for 4 
meters, then left, and walk another 3 meters. What is the distance 
between them?

sqrt(3*3 + 4*4) *2 

This is a Mensa question. I guessed by drawing out two triangles with 
a line and came up with 10. I wish to know the formula and not make a 
guess. I know the formula is given here but do not know how to arrive 
at the result. The formula needs to be explained so I will understand. 
Thank you.


Date: 04/25/2003 at 13:10:54
From: Doctor Link
Subject: Re: Triangle

Hi Luis, thanks for writing in. Great question! I have one of those 
Mensa books, and they are definitely a lot of fun.

This problem can make use of the distance formula, which is as 
follows:

                    2         2
    d = sqrt[(x2-x1) + (y2-y1) ]

where d is distance, and x1, x2, y1 and y2 are all values on a graph
that make 2 points. In the question you sent the 2 points are:

                   |
                   |       P(4,3)
                   |
           _ _ _ _ | _ _ _ _
                   |
                   |
           Q(-4,-3)|
                   |

Where P(x1,y1) and Q(x2,y2) describe where x1, x2, y1 and y2 are.

In other words 

     x1 is 4
    x2 is -4
     y1 is 3 
    y2 is -3 

I don't know if you noticed, but the distance formula is very closely
related to Pythagoras' theorem (concerning right triangles), which
states:

     2   2    2
    a + b  = c 

where a, b, and c are the sides of a right triangle, and where c is
the hypotenuse (the longest side).

If we take the square root of the whole equation, we get:

             2    2
   c = sqrt(a  + b )

which is another way of writing it, and is very important in
understanding why the distance formula works the way it does.

Now if we draw out the graph again, this time connecting P and Q with
a line and making a triangle, you will notice that we have created a
right triangle.

                     P  
                    /|  
               |   / |  
               |  /  |  
               | /   |  
           ____|/____| a
               /     |
            c /|     |
             / |    _|
            /__|___|_|
           Q     b

Do you see how (x2 - x1), or the change in x, is really the same thing 
as a?

Do you also see how (y2-y1), or the change in y, is really the same
thing as b? 

And finally do you see how the distance d is the same thing as c?

So the distance from P to Q (which we now see is identical to the 
length of c in this case), is just:
 
              2   2
    c = sqrt(a + b ) (based on Pythagoras' theorem)

or rewriting it, substituting in equivalent values...

                    2         2
    d = sqrt[(x2-x1) + (y2-y1) ]

which is the distance formula I showed you above.

Computing the anwer this way would yield:

                       2            2
    d = sqrt[((-4)-(4))) + ((-3)-(3)) ]

which leads to

                 2      2
    d = sqrt[(-8) + (-6) ]

which simplifies to
                 
    d = sqrt[64 + 36]

which simpifies further to

    d = 10

The explanation for the method that you sent, which was:

    sqrt(3*3 + 4*4)*2  

lies in the fact that instead of using the distance formula, you
just used Pythagoras' theorem twice. Let me show you what I mean: 

                     P  
                    /|  
               |   / |  
               |c1/  |b1  
               | /   |  
           _a2_|/____| 
          |    /  a1   
          |   /|     
        b2|  / |     
          | /c2|     
          |/     
          Q

where a1 = a2, b1 = b2 and c1 = c2 

Computing the distance from P to Q is just a matter of computing c1 or
c2 and then doubling it:

                2     2
    c1 = sqrt(a1  + b1 )

               2    2
    c1 = sqrt(3  + 4 )

    c1 = sqrt(25)

    c1 = 5

and since we know that the two triangles are identical we can just
double this answer to get our final answer of 10.

Does that explain why the formula works?

If you are still stuck or you need help with any other problem then
feel free to write in again.

- Doctor Link, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Triangles and Other Polygons

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