Hmmm. If the legs are a and b, and the hypotenuse is c, then we have
r = (a+b-c)/2 = ab/(a+b+c)
Since A = ab/2, that means
r = 2A / (a+b+c)
with r=2, that means
A = (a+b+c)
But, since a+b+c = (a+b-c)+2c, we have
A = 2r+2c
A = 4+2c
A circle of radius 2 inscribed in a right triangle. Find the simplest expression possible for the triangle's area as a function of its hypotenuse.
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