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Question Number 183393 by cherokeesay last updated on 25/Dec/22

Answered by mr W last updated on 25/Dec/22

Commented by mr W last updated on 25/Dec/22

(x/2)=(2/(2+2))  ⇒x=1  R=((1+2+2)/2)=(5/2)  cos θ=(((R−r)^2 +(R−1)^2 −(r+1)^2 )/(2(R−r)(R−1)))  cos θ=(((r+2)^2 −(R−1)^2 −(R−r)^2 )/(2(R−r)(R−1)))  (r+2)^2 −(R−1)^2 −(R−r)^2 =(R−r)^2 +(R−1)^2 −(r+1)^2   9(r−(1/4))−(9/4)=7((3/4)−r)+(9/4)  ⇒r=(3/4)  S=((π×2^2 )/4)=π  S_1 =πr^2   S_2 =((π×1^2 )/2)=(π/2)  (S/(S_1 +S_2 ))=(π/(πr^2 +(π/2)))=(2/(1+2r^2 ))=((16)/(17)) ✓

$$\frac{{x}}{\mathrm{2}}=\frac{\mathrm{2}}{\mathrm{2}+\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{1} \\ $$$${R}=\frac{\mathrm{1}+\mathrm{2}+\mathrm{2}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\theta=\frac{\left({R}−{r}\right)^{\mathrm{2}} +\left({R}−\mathrm{1}\right)^{\mathrm{2}} −\left({r}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}\left({R}−{r}\right)\left({R}−\mathrm{1}\right)} \\ $$$$\mathrm{cos}\:\theta=\frac{\left({r}+\mathrm{2}\right)^{\mathrm{2}} −\left({R}−\mathrm{1}\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} }{\mathrm{2}\left({R}−{r}\right)\left({R}−\mathrm{1}\right)} \\ $$$$\left({r}+\mathrm{2}\right)^{\mathrm{2}} −\left({R}−\mathrm{1}\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} =\left({R}−{r}\right)^{\mathrm{2}} +\left({R}−\mathrm{1}\right)^{\mathrm{2}} −\left({r}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{9}\left({r}−\frac{\mathrm{1}}{\mathrm{4}}\right)−\frac{\mathrm{9}}{\mathrm{4}}=\mathrm{7}\left(\frac{\mathrm{3}}{\mathrm{4}}−{r}\right)+\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${S}=\frac{\pi×\mathrm{2}^{\mathrm{2}} }{\mathrm{4}}=\pi \\ $$$${S}_{\mathrm{1}} =\pi{r}^{\mathrm{2}} \\ $$$${S}_{\mathrm{2}} =\frac{\pi×\mathrm{1}^{\mathrm{2}} }{\mathrm{2}}=\frac{\pi}{\mathrm{2}} \\ $$$$\frac{{S}}{{S}_{\mathrm{1}} +{S}_{\mathrm{2}} }=\frac{\pi}{\pi{r}^{\mathrm{2}} +\frac{\pi}{\mathrm{2}}}=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{2}{r}^{\mathrm{2}} }=\frac{\mathrm{16}}{\mathrm{17}}\:\checkmark \\ $$

Commented by mr W last updated on 25/Dec/22

error is fixed. thanks!

$${error}\:{is}\:{fixed}.\:{thanks}! \\ $$

Commented by cherokeesay last updated on 25/Dec/22

c′est ((16)/(17)), il y a erreur dans vos calculs sir.

$${c}'{est}\:\frac{\mathrm{16}}{\mathrm{17}},\:{il}\:{y}\:{a}\:{erreur}\:{dans}\:{vos}\:{calculs}\:{sir}. \\ $$

Answered by Acem last updated on 25/Dec/22

Commented by Acem last updated on 27/Dec/22

 BD is a half of the chord so ∣BD∣^2 = ∣AD∣. ∣DC∣

$$\:{BD}\:{is}\:{a}\:{half}\:{of}\:{the}\:{chord}\:{so}\:\mid{BD}\mid^{\mathrm{2}} =\:\mid{AD}\mid.\:\mid{DC}\mid \\ $$

Commented by Acem last updated on 25/Dec/22

 Continue...

$$\:{Continue}... \\ $$

Commented by Acem last updated on 26/Dec/22

Commented by Acem last updated on 27/Dec/22

Answered by HeferH last updated on 25/Dec/22

Commented by cherokeesay last updated on 25/Dec/22

il y a erreur dans vos calculs !  c′est ((16)/(17))

$${il}\:{y}\:{a}\:{erreur}\:{dans}\:{vos}\:{calculs}\:! \\ $$$${c}'{est}\:\frac{\mathrm{16}}{\mathrm{17}} \\ $$

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