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RE:Radius of the incircle puzzle
Posted:
Mar 9, 1994 8:19 AM


>Problem: > Given a triangle with sides 5, 6, and 7, what is the radius of the incircle? > >jgus@tenet.edu >Judie Gustafson An approach through area will work. If 2s = perimeter of the triangle then 2s = 18 and s = 9 From Hero's formula, the area is root (s(sa)(sb)(sc)) Area = root(9*4*3*2) = 6 root(6) Now divide the triangle into three smaller triangles, by joining the incentre to each vertex. The height of each triangle is r, the inradius so the area is .5(ar + br + cr) = .5*r*2s = rs so 9r = 6*root(6) so r = 2*root(6)/3 Jim Swift Burnaby South 2000 S. S. jswift@cln.etc.bc.ca



