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

Commented by Tawa11 last updated on 22/Jun/24

Area of red circle.

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

Commented by Tawa11 last updated on 22/Jun/24

I got 1.5

$$\mathrm{I}\:\mathrm{got}\:\:\mathrm{1}.\mathrm{5} \\ $$

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

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

(1/( (√r)))=(2/( (√6))) ⇒r=(3/2)

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

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

or 6^2 +(6−r)^2 =(6+r)^2 ⇒r=(3/2)

$${or} \\ $$$$\mathrm{6}^{\mathrm{2}} +\left(\mathrm{6}−{r}\right)^{\mathrm{2}} =\left(\mathrm{6}+{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Commented by Tawa11 last updated on 22/Jun/24

I used this. Based on what I learnt here

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

Commented by Tawa11 last updated on 22/Jun/24

Thanks sir

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

Commented by Tawa11 last updated on 22/Jun/24

Sir please question 208741

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

Commented by Tawa11 last updated on 22/Jun/24

Please prove this theorem sir.

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

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→r=(3/2) luas=π((3/2))^2 =(9/4)π

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

Commented by Tawa11 last updated on 22/Jun/24

Thanks sir.

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

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