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Question Number 161482 by amin96 last updated on 18/Dec/21

Commented by amin96 last updated on 18/Dec/21

$$circlesarea=?$$

Answered by FongXD last updated on 18/Dec/21

Commented by Tawa11 last updated on 18/Dec/21

$$\mathrm{Great}\mathrm{sir}.$$

Answered by FongXD last updated on 18/Dec/21

Commented by amin96 last updated on 18/Dec/21

$$greatsirweldone$$

Commented by mr W last updated on 18/Dec/21

$$nicejob!$$

Answered by mr W last updated on 18/Dec/21

Commented by mr W last updated on 18/Dec/21

$$\tan 2\alpha =\frac{1}{2}\Rightarrow \tan \alpha =\frac{\frac{1}{\sqrt{5}}}{1+\frac{2}{\sqrt{5}}}=\frac{1}{2+\sqrt{5}}$$$$\tan 2\beta =2\Rightarrow \tan \beta =\frac{\frac{2}{\sqrt{5}}}{1+\frac{1}{\sqrt{5}}}=\frac{2}{1+\sqrt{5}}$$$$$$$$GF=8-\frac{r}{\tan \alpha }=8-(2+\sqrt{5})r$$$$DG=8-r$$$$DF=8+r$$$${(8+r)}^2={(8-r)}^2+{(8-(2+\sqrt{5})r)}^2$$$$0=64-16(4+\sqrt{5})r+{(2+\sqrt{5})}^2{r}^2$$$$r=\frac{8(4+\sqrt{5})-16\sqrt{3+\sqrt{5}}}{9+4\sqrt{5}}\approx 0.739880$$$$$$$$HK=8-\frac{R}{\tan \beta }=8-\frac{(1+\sqrt{5})R}{2}$$$$DK=8-R$$$$DH=8+R$$$${(8+R)}^2={(8-R)}^2+{(8-\frac{(1+\sqrt{5})R}{2})}^2$$$$0=64-8(5+\sqrt{5})R+\frac{{(1+\sqrt{5})}^2{R}^2}{4}$$$$R=\frac{8(3-\sqrt{5})}{3+\sqrt{5}}=28-12\sqrt{5}\approx 1.167184$$

Commented by Ar Brandon last updated on 19/Dec/21

$$\mathrm{When}\mathrm{will}\mathrm{I}\mathrm{be}\mathrm{able}\mathrm{to}\mathrm{do}\mathrm{this}?$$

Commented by mr W last updated on 19/Dec/21

$$youarealreadyable!justdosome$$$$exercises.$$

Commented by peter frank last updated on 19/Dec/21

$$\mathrm{may}\mathrm{be}\mathrm{we}\mathrm{should}\mathrm{ask}\mathrm{him}\mathrm{how}$$$$\mathrm{he}\mathrm{managed}?$$

Commented by peter frank last updated on 19/Dec/21

$$\mathrm{To}\mathrm{mrW}.\mathrm{we}\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}$$$$\mathrm{for}\mathrm{being}\mathrm{here}\mathrm{in}\mathrm{this}\mathrm{forum}$$$$\mathrm{we}\mathrm{learn}\mathrm{alot}\mathrm{from}\mathrm{you}(\mathrm{Physics},\mathrm{Maths})$$

Commented by mr W last updated on 19/Dec/21

$$thanksforfeedback!ificouldhelp$$$$yousomehow,ididwithpleasure.$$

Commented by Ar Brandon last updated on 19/Dec/21

$$\mathrm{Alright}\mathrm{Sir}.\mathrm{But}\mathrm{I}\mathrm{need}\mathrm{some}\mathrm{ressources}.$$$$\mathrm{I}\mathrm{have}\mathrm{none}.\mathrm{Documents}\mathrm{on}\mathrm{this}\mathrm{topic}.$$

Commented by mr W last updated on 20/Dec/21

$$trytofindtheradiiofthethreenew$$$$circlesifyouwant:$$

Commented by mr W last updated on 20/Dec/21

Commented by Ar Brandon last updated on 20/Dec/21

$$\mathrm{Sir},\mathrm{I}\mathrm{notice}\mathrm{the}\mathrm{smallest}\mathrm{radius}\mathrm{is}\frac{\mathrm{R}}{2}$$$$\mathrm{But}\mathrm{have}\mathrm{no}\mathrm{idea}\mathrm{on}\mathrm{how}\mathrm{to}\mathrm{solve}\mathrm{the}\mathrm{others}.$$

Commented by Ar Brandon last updated on 20/Dec/21

Commented by Ar Brandon last updated on 20/Dec/21

$$\mathrm{Here}\mathrm{is}\mathrm{my}\mathrm{sketch}.\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{what}\mathrm{to}\mathrm{do}\mathrm{next}.$$

Commented by mr W last updated on 20/Dec/21

Commented by Ar Brandon last updated on 21/Dec/21

$$\mathrm{Sir},\mathrm{how}\mathrm{come}2\beta ?$$

Commented by mr W last updated on 21/Dec/21

Commented by mr W last updated on 21/Dec/21

$$bothredmarkedanglesarethesame.$$

Commented by Ar Brandon last updated on 21/Dec/21

OK Sir. I see! Thanks ��

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