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Question-216180




Question Number 216180 by mr W last updated on 29/Jan/25
Answered by AntonCWX last updated on 29/Jan/25
Commented by AntonCWX last updated on 29/Jan/25
FG^2 =FC×FB  6^2 =4(CB+4)⇒CB=5  CE=CB=5    GE=a  EH=b  ED=2r−a    GE×ED=CE×EH  a(2r−a)=5b  2r−a=((5b)/a)  2r=((5b+a^2 )/a)  r=((5b+a^2 )/(2a))    From ΔFCE,  FE^2 =FC^2 +CE^2   FE^2 =4^2 +5^2 ⇒FE^2 =41    From ΔFGE,  FE^2 =FG^2 +GE^2   41=6^2 +a^2   a^2 =5⇒a=(√5)    r=((5b+5)/( 2(√5)))    From ΔCBH,  BH^2 =BC^2 +CH^2   (2r)^2 =5^2 +(5+b)^2   4(((5b+5)/(2(√5))))^2 =25+25+10b+b^2   ((25b^2 +50b+25)/( 5))=50+10b+b^2   25b^2 +50b+25=250+50b+5b^2   20b^2 =225⇒b=(√((45)/4))=((3(√5))/2)    r=((5(((3(√5))/2))+5)/(2(√5)))=((2(√5)+15)/4)
FG2=FC×FB62=4(CB+4)CB=5CE=CB=5GE=aEH=bED=2raGE×ED=CE×EHa(2ra)=5b2ra=5ba2r=5b+a2ar=5b+a22aFromΔFCE,FE2=FC2+CE2FE2=42+52FE2=41FromΔFGE,FE2=FG2+GE241=62+a2a2=5a=5r=5b+525FromΔCBH,BH2=BC2+CH2(2r)2=52+(5+b)24(5b+525)2=25+25+10b+b225b2+50b+255=50+10b+b225b2+50b+25=250+50b+5b220b2=225b=454=352r=5(352)+525=25+154
Commented by mr W last updated on 29/Jan/25
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Answered by A5T last updated on 29/Jan/25
Commented by AntonCWX last updated on 30/Jan/25
I think you mistakenly typed in the 1 there...  ((2(√5)+15)/(41))  Remove the 1 and its done.
Ithinkyoumistakenlytypedinthe1there25+1541Removethe1anditsdone.
Commented by A5T last updated on 30/Jan/25
y=ED=EF ⇒ 6^2 =4(4+y)⇒y=5  6^2 +AD^2 =4^2 +5^2 ⇒AD=(√5)⇒CD=(√(41))  (AD×CE)+(AC×DE)=CD×EA   ⇒EA=((4(√5)+30)/( (√(41))))=z  2r((√5))(2r−(√5))+5^2 (2r)=(4r^2 −EA^2 )(√5)+EA^2 (2r−(√5))  ⇒r(EA^2 −20)=EA^2 (√5)⇒r=((EA^2 (√5))/(EA^2 −20))  ⇒r=((49(√5)+60)/(8+12(√5)))=((2(√5)+15)/4)≈4.868
y=ED=EF62=4(4+y)y=562+AD2=42+52AD=5CD=41(AD×CE)+(AC×DE)=CD×EAEA=45+3041=z2r(5)(2r5)+52(2r)=(4r2EA2)5+EA2(2r5)r(EA220)=EA25r=EA25EA220r=495+608+125=25+1544.868
Commented by A5T last updated on 30/Jan/25
It was a typo, thanks.
Itwasatypo,thanks.
Commented by mr W last updated on 29/Jan/25
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Answered by mr W last updated on 29/Jan/25
Commented by mr W last updated on 29/Jan/25
4(4+x)=6^2   ⇒x=5  6^2 +y^2 =4^2 +5^2   ⇒y=(√5)  (R−(√5))^2 =((x/2))^2 +(x−(√(R^2 −((x/2))^2 )))^2   10+R(√5)=5(√(R^2 −2.5^2 ))  R^2 −(√5)R−((205)/(16))=0  ⇒R=(((√5)+(√(5+((205)/4))))/2)=((2(√5)+15)/4)≈4.868
4(4+x)=62x=562+y2=42+52y=5(R5)2=(x2)2+(xR2(x2)2)210+R5=5R22.52R25R20516=0R=5+5+20542=25+1544.868
Answered by A5T last updated on 29/Jan/25
Commented by A5T last updated on 29/Jan/25
4×(4+EF)=6^2 ⇒EF=ED=5  6^2 +AD^2 =CE^2 +ED^2 ⇒AD=(√5)  AD×DH=ED×DG⇒(√5)(2r−(√5))=5DG...(i)  In △EFG; GF^2 =EF^2 +EG^2   ⇒4r^2 =25+(5+DG)^2 ⇒4r^2 =DG^2 +10DG...(ii)  (i) in (ii)  ⇒4r^2 =25+(5+((2r−(√5))/( (√5))))^2 ⇒4r^2 =((125+(2r+4(√5))^2 )/5)  ⇒16r^2 −16(√5)r−205=0  ⇒r=((16(√5)+_− (√(16^2 (5)−4(16)(−205))))/(32))=((16(√5)+_− 120)/(32))>0  ⇒r=((2(√5)+15)/4)
4×(4+EF)=62EF=ED=562+AD2=CE2+ED2AD=5AD×DH=ED×DG5(2r5)=5DG(i)InEFG;GF2=EF2+EG24r2=25+(5+DG)24r2=DG2+10DG(ii)(i)in(ii)4r2=25+(5+2r55)24r2=125+(2r+45)2516r2165r205=0r=165+162(5)4(16)(205)32=165+12032>0r=25+154
Commented by AntonCWX last updated on 31/Jan/25
Same idea as mine  ⋛
Sameideaasmine

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