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Question Number 180299 by mnjuly1970 last updated on 10/Nov/22

Answered by som(math1967) last updated on 10/Nov/22

Commented by som(math1967) last updated on 10/Nov/22

let O is centre of semicircle AO= (((a+b)/2) −r) [semicircle and circle touch internally] BO=b−((a+b)/2)=((b−a)/2) (((a+b)/2)−r)^2 =r^2 +(((b−a)/2))^2 (((a+b)^2 )/4) −r(a+b)+r^2 =r^2 +(((b−a)^2 )/4) ⇒(((a+b)^2 −(b−a)^2 )/4)=r(a+b) ⇒((4ab)/4)=r(a+b) ⇒ (1/r)=((a+b)/(ab)) ∴ (1/r)=(1/a)+(1/b)

letOiscentreofsemicircleAO=(a+b2−r)[semicircleandcircletouchinternally]BO=b−a+b2=b−a2(a+b2−r)2=r2+(b−a2)2(a+b)24−r(a+b)+r2=r2+(b−a)24⇒(a+b)2−(b−a)24=r(a+b)⇒4ab4=r(a+b)⇒1r=a+bab∴1r=1a+1b

Commented by mnjuly1970 last updated on 10/Nov/22

thanks alot sir ..excellent solution...

thanksalotsir..excellentsolution...

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