Source: M. Teitelbaum, "Simpler solution," *Mathematics Teacher*, October 1987

**Hint:** The solution path can be either complicated or straight-forward....which approach are you trying?

For the straight-forward approach, draw the picture (i.e. circle incribed in a right triangle), then stand back and look at it....what geometrical "laws" do you see in action?

The graphic (and title) is a hint as well.

**Solution Commentary:**
For the more complicated approach, consult Problem #27 in the *Mathematics Teacher" (December 1986).*

Teitelbaum suggests one focus on the circle breaking the triangle into three examples of pairs of tangents to a circle from an external point. Remember, these pairs of tangents are congruent. Is this enough to get you on a good track....you should get an answer of a radius = 27 cm.

Solution proposed by D.K. (Fife, WA): *
Suppose right triangle ABC has right angle B. Use the perimeter of 324 cm, Pythagoreans theorem and trial and error to find AB = a = 81, BC = b = 108 and we know the AC = c = 135. Suppose x, y, and z are points of tangency to circle with center i on AC, AB, and BC respectively. As we know points of tangency are perpendicular to the radius, they can be used as heights of triangles.*

*
Since all radii are congruent, xi = yi = zi = r (radius).
Divide triangle ABC into three smaller triangles: AiB, BiC and AiC. Then area of triangle ABC = sum of areas of triangles AiB, BiC, and AiC. Or,*