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Question Number 187961 by normans last updated on 24/Feb/23

Answered by HeferH last updated on 24/Feb/23

Commented by HeferH last updated on 24/Feb/23

$$i.{(R-3)}^2=9+\frac{R^2}{4}$$$$R^2+9-6R=9+\frac{R^2}{4}$$$$4R^2-24R=R^3$$$$3R^2=24R$$$$R=8$$$$ii.{(R-r)}^2-r^2={(\frac{R}{\sqrt{2}}+r)}^2-{(\frac{R}{\sqrt{2}}-r)}^2$$$$R^2-2Rr=\frac{4Rr}{2}\sqrt{2}$$$$R-2r=2r\sqrt{2}$$$$R=2r+2r\sqrt{2}$$$$r=\frac{R}{2+2\sqrt{2}}$$$$r=\frac{8}{2+2\sqrt{2}}=\frac{4}{1+\sqrt{2}}=\frac{4(\sqrt{2}-1)}{2-1}=4(\sqrt{2}-1)$$

Commented by normans last updated on 25/Feb/23

$$nice$$

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