Area of Inscribed Circle

Date: 12/01/98 at 13:24:52
From: Anna 
Subject: Trigonometry

The question is: Find the area of the circle inscribed in a triangle 
a, b, c.  The hint is: Draw lines to the center of the circle and use 
Heron's law.

I would appreciate it if you could help me!
Anna

Date: 12/01/98 at 15:41:37
From: Doctor Floor
Subject: Re: Trigonometry

Hi Anna,

Thanks for your question!

Let us consider triangle ABC with sidelengths BC = a, AC = b, AB = c 
and center of the inscribed circle (incenter) I. Drop perpendicular 
altitudes from I to Ia, Ib and Ic on the sides of ABC. The lengths of 
these altitudes are equal to the radius r of the incircle. This radius 
we will have to calculate in order to find the area of the inscribed 
circle (= incircle).

           C
          / \
         /   \
        /     \
      Ib       Ia
      /         \
     /     I     \
    /             \
   A-------Ic------B


First, we know that area(ABC) = area(AIB) + area(BIC) + area(CIA).

We can calculate area(AIB) = 0.5 * AB * IIc = 0.5cr. And in the same 
way we find area(BIC) = 0.5ar and area(CIA) = 0.5br.

The conclusion is that area(ABC) = 0.5(a+b+c)r = sr, where s is the 
semiperimeter of ABC.

Then, from Heron's formula, we also know that:

   area(ABC) = sqrt(s(s-a)(s-b)(s-c))

Combining the two formulas for the area of ABC we can derive:

     sr = sqrt(s(s-a)(s-b)(s-c))
      r = sqrt(s(s-a)(s-b)(s-c)) / s
      r = sqrt((s-a)(s-b)(s-c) / s )

And now we can finish off the area of the incircle:

    area(incircle) = pi * r^2
                   = pi * sqrt((s-a)(s-b)(s-c) / s )^2

                     pi(s-a)(s-b)(s-c)
                   = -----------------
                            s

Quite nice!

Best regards,

- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/