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Topic: RE:Radius of the incircle puzzle
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Jim Swift

Posts: 38
Registered: 12/6/04
RE:Radius of the incircle puzzle
Posted: Mar 9, 1994 8:19 AM
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>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(s-a)(s-b)(s-c))
Area = root(9*4*3*2) = 6 root(6)
Now divide the triangle into three smaller triangles, by joining the
in-centre to each vertex.
The height of each triangle is r, the in-radius
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





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