See complete Problem 82

Area of the Contact Triangle, Inradius, Circumradius. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Elearn Geometry Problem 82

Labels:
area,
circumcircle,
circumradius,
contact,
incircle,
inradius,
triangle

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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