See complete Problem 82
Area of the Contact Triangle, Inradius, Circumradius. Level: High School, SAT Prep, College geometry
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Monday, May 19, 2008
Elearn Geometry Problem 82
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S1 = Tr.IDE + Tr.IEF + Tr.IFD
ReplyDelete= [r^2*S/bc + r^2*S/ca + r^2*S/ab] by previous problem.
Thus, S1= S*r^2[1/bc + 1/ca + 1/ab]
and S = abc/4R where a,b & c have their usual meaning.
Hence. S1 = (abcr^2/4R)[1/bc + 1/ca + 1/ab]
= (r^2/4R)(a+b+c)
= (r/2R)[r*(a+b+c)/2]
We know that S = Inradius*Semisum of sides
= [r*(a+b+c)/2]
Hence, S1 = (r/2R)* S or S1/S = r/2R
Ajit: ajitathle@gmail.com
Alternative solution to problem 82.
ReplyDeleteIt was proved in problem 081 that S(DIE) = (a.r*2)/(4R). Similarly, S(DIF) = (b.r*2)/(4R) and S(EIF) = (c.r*2)/(4R). The area S1 of tr. DEF is the sum of those three areas, so S1 = (a+b+c).r*2/(4R).
We also know that the area of tr. ABC is given by S = (a+b+c).r/2. Hence
S1/S = ((a+b+c).r*2).2)/(4R.(a+b+c).r) = r/2R
Reference my proof of Problem 81,
ReplyDeleteS(DEF) = ar^2/4R + br^2/4R + cr^2/4R
Hence S1 = (r/2R)(ra/2 + rb/2 + rc/2) = (r/2R).S
Sumith Peiris
Moratuwa
Sri Lanka