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Question-195012

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Question Number 195012 by sonukgindia last updated on 22/Jul/23
Answered by a.lgnaoui last updated on 23/Jul/23
Commented by a.lgnaoui last updated on 23/Jul/23
Answered by mr W last updated on 23/Jul/23
Commented by mr W last updated on 23/Jul/23
r=radius of small circle R=radius of semi circle=2r θ=60° a=(√7)+(√3) OD=(r/(tan (θ/2))) AD=2r−(r/(tan (θ/2))) DB=2r+(r/(tan (θ/2))) AC^2 =(2r−(r/(tan (θ/2))))^2 +a^2 +2a(2r−(r/(tan (θ/2))))cos θ BC^2 =(2r+(r/(tan (θ/2))))^2 +a^2 −2a(2r+(r/(tan (θ/2))))cos θ AC^2 +BC^2 =AB^2 (2r−(r/(tan (θ/2))))^2 +a^2 +2a(2r−(r/(tan (θ/2))))cos θ+(2r+(r/(tan (θ/2))))^2 +a^2 −2a(2r+(r/(tan (θ/2))))cos θ=(4r)^2 r^2 (4−(1/(tan^2 (θ/2))))=a^2 ⇒r^2 =(a^2 /(4−tan^2 (θ/2))) ⇒r=(a/( (√(4−tan^2 (θ/2))))) area of small circle =πr^2 =((πa^2 )/(4−tan^2 (θ/2)))=((π((√7)+(√3))^2 )/(4−((1/( (√3))))^2 ))=((6π(5+(√(21))))/(11))
$${r}={radius}\:{of}\:{small}\:{circle} \\ $$$${R}={radius}\:{of}\:{semi}\:{circle}=\mathrm{2}{r} \\ $$$$\theta=\mathrm{60}° \\ $$$${a}=\sqrt{\mathrm{7}}+\sqrt{\mathrm{3}} \\ $$$${OD}=\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$${AD}=\mathrm{2}{r}−\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$${DB}=\mathrm{2}{r}+\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$${AC}^{\mathrm{2}} =\left(\mathrm{2}{r}−\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} +\mathrm{2}{a}\left(\mathrm{2}{r}−\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}\right)\mathrm{cos}\:\theta \\ $$$${BC}^{\mathrm{2}} =\left(\mathrm{2}{r}+\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}{a}\left(\mathrm{2}{r}+\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}\right)\mathrm{cos}\:\theta \\ $$$${AC}^{\mathrm{2}} +{BC}^{\mathrm{2}} ={AB}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}{r}−\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} +\mathrm{2}{a}\left(\mathrm{2}{r}−\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}\right)\mathrm{cos}\:\theta+\left(\mathrm{2}{r}+\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}{a}\left(\mathrm{2}{r}+\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}\right)\mathrm{cos}\:\theta=\left(\mathrm{4}{r}\right)^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} \left(\mathrm{4}−\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}\right)={a}^{\mathrm{2}} \\ $$$$\Rightarrow{r}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{\mathrm{4}−\mathrm{tan}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}\:\Rightarrow{r}=\frac{{a}}{\:\sqrt{\mathrm{4}−\mathrm{tan}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}} \\ $$$${area}\:{of}\:{small}\:{circle}\:=\pi{r}^{\mathrm{2}} \\ $$$$\:\:\:\:\:=\frac{\pi{a}^{\mathrm{2}} }{\mathrm{4}−\mathrm{tan}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}=\frac{\pi\left(\sqrt{\mathrm{7}}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }{\mathrm{4}−\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }=\frac{\mathrm{6}\pi\left(\mathrm{5}+\sqrt{\mathrm{21}}\right)}{\mathrm{11}} \\ $$

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