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Question Number 212878 by Spillover last updated on 25/Oct/24

Answered by som(math1967) last updated on 26/Oct/24

Commented by som(math1967) last updated on 26/Oct/24

$$letradofgreencircle=R$$$$purplecircle=r$$$$\therefore a^2=2R\times 2r=4Rr$$$$\therefore a=2\sqrt{Rr}$$$$again{b}^2={(R+r)}^2-{(R-r)}^2$$$$b=\sqrt{4Rr}=2\sqrt{Rr}$$$$\therefore a:b=1:1$$

Answered by Spillover last updated on 26/Oct/24

Answered by Spillover last updated on 26/Oct/24

Answered by Spillover last updated on 26/Oct/24

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