Question and Answers Forum
Question Number 175637 by mnjuly1970 last updated on 04/Sep/22
Answered by behi834171 last updated on 04/Sep/22
![I. r_a =p.tg(A/2)=p.((√(((p−b)(p−c))/(bc)))/( (√((p(p−a))/(bc)))))=(√((p(p−b)(p−c))/((p−a))))= =(S/(p−a)) and so on.... ⇒Σr_a =S.Σ((1/(p−a)))=S.((Σ(p−b)(p−c))/(Π(p−a)))= =(p/(4S)).[2ab+2bc+2ca−a^2 −b^2 −c^2 ]= =((4(r^2 +4R.r))/(4r))=r+4R .■ [(p−b)(p−c)=(1/4)(a+c−b)(a+b−c)= =(1/4)[a^2 +ab−ac+ac+bc−c^2 −ab−b^2 +bc]= =(1/4)[a^2 −(b−c)^2 ] (and so on) +(1/4)[c^2 −(b−a)^2 ]+ +(1/4)[b^2 −(a−c)^2 ]= =(1/4)[Σa^2 −(2Σa^2 −2Σab)]=(1/4)[2Σab−Σa^2 ]] II. d^2 =R^2 −2R.r ,d≥0⇒R≥2r . [d=distance from center of: incircle to center of outcircle]](Q175648.png)
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Question and Answers Forum | ||
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Question Number 175637 by mnjuly1970 last updated on 04/Sep/22 | ||
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Answered by behi834171 last updated on 04/Sep/22 | ||
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