Editing Incircle and excircles (section)

====Relation to area of the triangle====
{{Redirect|Inradius|the three-dimensional equivalent|Inscribed sphere}}
The radius of the incircle is related to the [[area]] of the triangle.<ref>Coxeter, H.S.M. "Introduction to Geometry'' 2nd ed. Wiley, 1961.''</ref> The ratio of the area of the incircle to the area of the triangle is less than or equal to <math>\pi \big/ 3\sqrt3</math>,
with equality holding only for [[equilateral triangle]]s.<ref>Minda, D., and Phelps, S., "Triangles, ellipses, and cubic polynomials", ''[[American Mathematical Monthly]]'' 115, October 2008, 679-689: Theorem 4.1.</ref>

Suppose <math>\triangle ABC</math> has an incircle with radius <math>r</math> and center <math>I</math>. Let <math>a</math> be the length of <math>\overline{BC}</math>, <math>b</math> the length of <math>\overline{AC}</math>, and <math>c</math> the length of <math>\overline{AB}</math>.

Now, the incircle is tangent to <math>\overline{AB}</math> at some point <math>T_C</math>, and so <math>\angle AT_CI</math> is right. Thus, the radius <math>T_CI</math> is an [[altitude (triangle)|altitude]] of <math>\triangle IAB</math>.

Therefore, <math>\triangle IAB</math> has base length <math>c</math> and height <math>r</math>, and so has area <math>\tfrac12 cr</math>.

Similarly, <math>\triangle IAC</math> has area <math>\tfrac12 br</math> and <math>\triangle IBC</math> has area <math>\tfrac12 ar</math>.

Since these three triangles decompose <math>\triangle ABC</math>, we see that the area <math>\Delta \text{ of} \triangle ABC</math> is:
:<math display=block>\Delta = \tfrac12 (a + b + c)r = sr,</math>

{{spaces|4}} and {{spaces|4}}<math>r = \frac{\Delta}{s},</math>

where <math>\Delta</math> is the area of <math>\triangle ABC</math> and <math>s = \tfrac12(a + b + c)</math> is its [[semiperimeter]].

For an alternative formula, consider <math>\triangle IT_CA</math>. This is a right-angled triangle with one side equal to <math>r</math> and the other side equal to <math>r \cot \tfrac{A}{2}</math>. The same is true for <math>\triangle IB'A</math>. The large triangle is composed of six such triangles and the total area is:{{Citation needed|date=May 2020}}
:<math display=block>\Delta = r^2 \left(\cot\tfrac{A}{2} + \cot\tfrac{B}{2} + \cot\tfrac{C}{2}\right).</math>