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Question Number 200457 by cortano12 last updated on 19/Nov/23

Commented by cortano12 last updated on 19/Nov/23

$$\cancel{\mathit{x}}$$

Answered by mr W last updated on 19/Nov/23

$$R=biggerradius$$$$r=smalleradius$$$$wehaveR=\sqrt{2}r$$$$\frac{x}{2r}=\frac{1+x}{2R}$$$$\Rightarrow x=\frac{1+x}{\sqrt{2}}\Rightarrow x=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1✓$$

Commented by cortano12 last updated on 19/Nov/23

$$\mathrm{very}\mathrm{simple}$$

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