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
$$\mathrm{I}\:\mathrm{got}\:\:\mathrm{1}.\mathrm{5} \\ $$
Commented by mr W last updated on 22/Jun/24
Commented by mr W last updated on 22/Jun/24
$$\frac{\mathrm{1}}{\:\sqrt{{r}}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{6}}}\:\Rightarrow{r}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 22/Jun/24
$${or} \\ $$$$\mathrm{6}^{\mathrm{2}} +\left(\mathrm{6}−{r}\right)^{\mathrm{2}} =\left(\mathrm{6}+{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$
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
$$\mathrm{Sir}\:\mathrm{please}\:\mathrm{question}\:\mathrm{208741} \\ $$
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
$$\left(\mathrm{6}+{r}\right)^{\mathrm{2}} =\mathrm{6}^{\mathrm{2}} +\left(\mathrm{6}−{r}\right)^{\mathrm{2}} \\ $$$$\mathrm{36}+\mathrm{12}{r}+{r}^{\mathrm{2}} =\mathrm{36}+\mathrm{36}−\mathrm{12}{r}+{r}^{\mathrm{2}} \\ $$$$\mathrm{24}{r}=\mathrm{36}\rightarrow{r}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${luas}=\pi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{4}}\pi \\ $$
Commented by Tawa11 last updated on 22/Jun/24
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$