Area of a triangle: Incenter (Part 6)

Incenter_750Source: http://mathworld.wolfram.com/Incircle.html

The incenter I of a triangle \triangle ABC is defined by the intersection of the angle bisectors of its three angles. A circle can be inscribed within \triangle ABC, as shown in the picture.

This incircle provides a different way of finding the area of \triangle ABC commonly needed for high school math contests like the AMC 10/12. Suppose that the sides a, b, and c are known and the inradius r is also known. Then \triangle ABI is a right triangle with base c and height r. So

\hbox{Area of ~} \triangle ABI = \displaystyle \frac{1}{2} cr

Similarly,

\hbox{Area of ~} \triangle ACI = \displaystyle \frac{1}{2} br

\hbox{Area of ~} \triangle BCI = \displaystyle \frac{1}{2} ar

Since the area of \triangle ABC is the sum of the areas of these three smaller triangles, we conclude that

\hbox{Area of ~} \triangle ABC = \displaystyle \frac{1}{2} r (a+b+c),

or

\hbox{Area of ~} \triangle ABC = rs,

where s = (a+b+c)/2 is the semiperimeter of \triangle ABC.

green lineThis also permits the computation of r itself. By Heron’s formula, we know that

\hbox{Area of ~} \triangle ABC = \sqrt{s(s-a)(s-b)(s-c)}

Equating these two expressions for the area of \triangle ABC, we can solve for the inradius r:

r = \displaystyle \sqrt{ \frac{(s-a)(s-b)(s-c)}{s} }

For much more about the inradius and incircle, I’ll refer to the MathWorld website.

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