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Question Number 205861 by mr W last updated on 01/Apr/24

Answered by A5T last updated on 01/Apr/24

9×9=(2r−x)x⇒2r−x=((81)/x)  21×21=(2r−16−x)(16+x)=(((81)/x)−16)(16+x)  ⇒x=2⇒r=21.25

9×9=(2rx)x2rx=81x21×21=(2r16x)(16+x)=(81x16)(16+x)x=2r=21.25

Commented by mr W last updated on 01/Apr/24

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Answered by mr W last updated on 01/Apr/24

(√(R^2 −(12+9)^2 ))+16=(√(R^2 −9^2 ))  R^2 −21^2 +16^2 +32(√(R^2 −21^2 ))=R^2 −9^2   4(√(R^2 −21^2 ))=13  R=(√(21^2 +(((13)/4))^2 ))=((85)/4)=21.25

R2(12+9)2+16=R292R2212+162+32R2212=R2924R2212=13R=212+(134)2=854=21.25

Commented by Skabetix last updated on 02/Apr/24

  Hello sir, how did you arrive at the first equation? Which theorems did you use? Congratulations anyway.

Hello sir, how did you arrive at the first equation? Which theorems did you use? Congratulations anyway.

Commented by mr W last updated on 02/Apr/24

Commented by mr W last updated on 02/Apr/24

AB=(√(R^2 −(12+9)^2 ))  AC=(√(R^2 −9^2 ))  AB+16=AC  (√(R^2 −(12+9)^2 ))+16=(√(R^2 −9^2 ))

AB=R2(12+9)2AC=R292AB+16=ACR2(12+9)2+16=R292

Answered by mr W last updated on 01/Apr/24

Commented by mr W last updated on 01/Apr/24

R=((AC)/(2 sin (π−θ)))      =((√(16^2 +(12+9+9)^2 ))/(2×((16)/( (√(16^2 +12^2 ))))))=((85)/4)=21.15

R=AC2sin(πθ)=162+(12+9+9)22×16162+122=854=21.15

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