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Question Number 208743 by Tawa11 last updated on 22/Jun/24

Commented by Tawa11 last updated on 22/Jun/24

$$\mathrm{Area}\mathrm{of}\mathrm{red}\mathrm{circle}.$$

Commented by Tawa11 last updated on 22/Jun/24

$$\mathrm{I}\mathrm{got}1.5$$

Commented by mr W last updated on 22/Jun/24

Commented by mr W last updated on 22/Jun/24

$$\frac{1}{\sqrt{r}}=\frac{2}{\sqrt{6}}\Rightarrow r=\frac{3}{2}$$

Commented by mr W last updated on 22/Jun/24

$$or$$$$6^2+{(6-r)}^2={(6+r)}^2$$$$\Rightarrow r=\frac{3}{2}$$

Commented by Tawa11 last updated on 22/Jun/24

$$\mathrm{I}\mathrm{used}\mathrm{this}.$$$$\mathrm{Based}\mathrm{on}\mathrm{what}\mathrm{I}\mathrm{learnt}\mathrm{here}$$

Commented by Tawa11 last updated on 22/Jun/24

$$\mathrm{Thanks}\mathrm{sir}$$

Commented by Tawa11 last updated on 22/Jun/24

$$\mathrm{Sir}\mathrm{please}\mathrm{question}208741$$

Commented by Tawa11 last updated on 22/Jun/24

$$\mathrm{Please}\mathrm{prove}\mathrm{this}\mathrm{theorem}\mathrm{sir}.$$

Commented by mr W last updated on 22/Jun/24

https://en.m.wikipedia.org/wiki/Descartes%27_theorem

Answered by Sutrisno last updated on 22/Jun/24

$${(6+r)}^2=6^2+{(6-r)}^2$$$$36+12r+r^2=36+36-12r+r^2$$$$24r=36\to r=\frac{3}{2}$$$$luas=\pi {\left(\frac{3}{2}\right)}^2=\frac{9}{4}\pi$$

Commented by Tawa11 last updated on 22/Jun/24

$$\mathrm{Thanks}\mathrm{sir}.$$

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