Embedding in Circles

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We come now to the heart of this Microworld.  We saw that given the ordered sequence of side lengths: a, b, c, and d,  if they satisfy the condition  stated earlier that  s > a, s > b, s > c, and s > d  (where s is the  semi-perimeter)  then many quadrilaterals may be constructed with these side lengths forming a cycle around the perimeter.  Among these, only  those with shapes that can be embedded in circles have the maximal area.

 

We will now calculate that area.  This will give insight into Heron's Formula.

 

          The picture is this:

 

 

  and our job is to calculate the area.  Given that   (radians),  we want to calculate 

 

 

  so we use the relation between T and P:

 

         

         Now , and

                    ,since T + P  =   

 

         So the area is:   .   If we call the area A, then

 

          also, 

         

            From this last equation, it follows that

 

             and   

 

             from the expression for   above.  Thus,

or

(4.1)

 

               Rewriting this as:

 

               we see that this difference of squares is equal to:

 

       

            and this is equal to:

 

 

            and finally,  this is equal to:

 

 

 

             Recalling that the semi-perimeter   

             we see that:

 

              or

 

              This last formula for the area of a quadrilateral that can be embedded in a circle generalizes Heron's Formula.  In fact,  it is more symmetric than Heron's Formula.  We may now easily derive Heron's formula from this one.

 

           Consider the case of a quadrilateral determined by four numbers a,b,c, and d  where d = 0.  This means that the last length is 0,  and so this "quadrilateral" is actually a triangle.  Since every triangle may be embedded in a circle, it is clear that every triangle may be approximated as closely as we like by a quadrilateral that is  embedded in the same circle.  In fact, let us suppose that the triangle has sides a, b, and c.   Then we can keep a and b fixed, and imagine that those approximating quadrilaterals have sides c' and d' where c' approaches c and d' approaches 0.  Then the areas of  those quadrilaterals must approach the area of the triangle.  But the limiting value of the areas of the quadrilaterals (as c' approaches c  and d' approaches 0) may be shown from the Calculus (using  the fact that the function defining area is continuous) to be equal to what is obtained by substituting c'=c and d'=0 into the formula.

 

               When we do that, we get Heron's formula! 

     

               That is almost the whole story.  We finish this microworld with a challenge and an amusing exercise on the next page.

 

               Exercise:  Using the formula for the squared area (4.1) above, show

                                  that if a triangle has sides a, b, and c then its area is:

 

 

 

 

            This formula for the area is surprising because it appears as though the order of a, b, and c matters.   Also, it is not obvious that the expression under the radical must be non-negative.  But if the formula is true, then it must, and the order cannot matter.   Any rearrangement (permutation) of a,b, and c should produce the same area.   You may test this formula on the first exploration page, in the section  on areas of triangles using the calculator there.

 

            On the next page, you will construct an arbitrary quadrilateral again, and you will see that if you connect the midpoints of the sides, you get something very interesting!