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Heron's Formula, Cartesian Coordinate Plane


Date: 11/01/2001 at 07:36:16
From: Jacob Kohl
Subject: Geometry

If a triangle has sides 5, 6, and the square root of thirteen, 
what is the area of the triangle?

Thank you for your help.


Date: 11/01/2001 at 12:18:50
From: Doctor Greenie
Subject: Re: Geometry

Hi, Jacob -

Cool problem! Thanks for sending it to us.

When I am given the lengths of the three sides of a triangle and am 
asked to find the area, I have the following thoughts:

(1) Maybe the lengths of the sides satisfy the Pythagorean theorem, 
indicating the triangle is a right triangle; then finding its area is 
easy.

(2) If the triangle is not a right triangle, then maybe I can draw it 
on a Cartesian coordinate plane in such a way that its area is easy to 
determine.

(3) If (1) and (2) don't get me anywhere, there is always Heron's 
formula:

  Area = sqrt[(s)(s-a)(s-b)(s-c)]

where a, b, and c are the lengths of the sides and s is the semi-
perimeter, s = (a+b+c)/2

For the particular problem you give....

(1) The given lengths do not satisfy the Pythagorean theorem, so the 
triangle is not a right triangle.

(2) I looked at this for a while and didn't see anything right away.  
So I went on to Heron's formula, with the idea of coming back to look 
at this approach some more.

(3) The use of Heron's formula looks formidable, with one side 
having length sqrt(13) - but the arithmetic - thanks to 
"(a+b)(a-b) = a^2-b^2" - turns out not to be at all bad. Here it is:

With a = 5, b = 6, and c = sqrt(13), we have s = (11+sqrt(13))/2.  
Then we have

  A = sqrt[(s)(s-a)(s-b)(s-c)]

            (11+sqrt(13))   (11-sqrt(13))   sqrt(13)+1   sqrt(13)-1
  A = sqrt( ------------- * ------------- * ---------- * ---------- )
                  2               2              2            2

            121-13   13-1
  A = sqrt( ------ * ---- )
               4       4

            108*12
  A = sqrt( ------ )
              16

  A = sqrt( 27*3 )

  A = sqrt( 81)

  A = 9

Now let's go back and look at approach (2). I had noticed that the 
side with length sqrt(13) could be represented by the hypotenuse of a 
right triangle with sides of length 2 and 3; so I drew a right 
triangle on a Cartesian coordinate plane with vertices A(0,0), B(2,0), 
and C(0,3):

                                        |
                                        C
                                        |
                                        *
                                        * *
                                        *   *
                                        *     *
                                        *       *
                                        *         *
                                        *           *
              --------------------------*-*-*-*-*-*-*-*-------------
                                      A |               B
                                        |

Then I thought about how I could use BC, with length sqrt(13), as one 
of the sides of the triangle in the problem. With a little playing 
around, I discovered that, if I chose point D(-4,0), I would have 
AD = 4, which with AC = 3 means CD = 5; then DB = DA+AB = 4+2 = 6.  
So then triangle BCD is a triangle with sides 5, 6, and sqrt(13), as 
required:

                                        |
                                        C
                                        |
                                        *
                                    *   * *
                                *       *   *
                            *           *     *
                        *               *       *
                    *                   *         *
                *                       *           *
        ----*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-------------
            D                         A |               B
                                        |

The area of this triangle is easy to calculate, because it has a base 
of 6 and a height of 3: A = one-half base times height = (6*3)/2 = 9.

Thanks again for a nice problem. It is always fun for me to find an 
unusual problem that has a very neat solution.

Write back if you have any questions about any of this.

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

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