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Question-181153




Question Number 181153 by cortano1 last updated on 22/Nov/22
Answered by Acem last updated on 22/Nov/22
Commented by Acem last updated on 22/Nov/22
I got the question as wrong, hope my English′s   getting better.
$${I}\:{got}\:{the}\:{question}\:{as}\:{wrong},\:{hope}\:{my}\:{English}'{s} \\ $$$$\:{getting}\:{better}. \\ $$
Answered by mr W last updated on 22/Nov/22
Commented by Acem last updated on 22/Nov/22
Aww thank you Sir very much for the 2nd   scientific term too!
$${Aww}\:{thank}\:{you}\:{Sir}\:{very}\:{much}\:{for}\:{the}\:\mathrm{2}{nd} \\ $$$$\:{scientific}\:{term}\:{too}! \\ $$
Commented by mr W last updated on 22/Nov/22
AD=DC=R=((BC)/2)=((√(1^2 +3^2 ))/2)=((√(10))/2)  β=(π/4)−α  ED=R−r  AE=(√2)r  cos β=((((√2)r)^2 +R^2 −(R−r)^2 )/(2(√2)rR))=cos ((π/4)−α)  ((r+2R)/(2(√2)R))=(1/( (√2)))(cos α+sin α)=(1/( (√2)))((1/( (√(10))))+(3/( (√(10)))))  ⇒r=((R(4(√(10))−10))/( 5))         =(((√(10))(4(√(10))−10))/(10))         =4−(√(10))≈0.838
$${AD}={DC}={R}=\frac{{BC}}{\mathrm{2}}=\frac{\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }}{\mathrm{2}}=\frac{\sqrt{\mathrm{10}}}{\mathrm{2}} \\ $$$$\beta=\frac{\pi}{\mathrm{4}}−\alpha \\ $$$${ED}={R}−{r} \\ $$$${AE}=\sqrt{\mathrm{2}}{r} \\ $$$$\mathrm{cos}\:\beta=\frac{\left(\sqrt{\mathrm{2}}{r}\right)^{\mathrm{2}} +{R}^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}{rR}}=\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}−\alpha\right) \\ $$$$\frac{{r}+\mathrm{2}{R}}{\mathrm{2}\sqrt{\mathrm{2}}{R}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{cos}\:\alpha+\mathrm{sin}\:\alpha\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}}+\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}}\right) \\ $$$$\Rightarrow{r}=\frac{{R}\left(\mathrm{4}\sqrt{\mathrm{10}}−\mathrm{10}\right)}{\:\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{10}}\left(\mathrm{4}\sqrt{\mathrm{10}}−\mathrm{10}\right)}{\mathrm{10}} \\ $$$$\:\:\:\:\:\:\:=\mathrm{4}−\sqrt{\mathrm{10}}\approx\mathrm{0}.\mathrm{838} \\ $$
Commented by Acem last updated on 22/Nov/22
@Mr. W, Hello Sir!  Circumcircle means the circle that passing   by vertices?
$$@{Mr}.\:{W},\:{Hello}\:{Sir}! \\ $$$${Circumcircle}\:{means}\:{the}\:{circle}\:{that}\:{passing} \\ $$$$\:{by}\:{vertices}? \\ $$
Commented by mr W last updated on 22/Nov/22
yes, sir.
$${yes},\:{sir}. \\ $$
Commented by mr W last updated on 22/Nov/22
the circle you have in your diagram  is the incircle.
$${the}\:{circle}\:{you}\:{have}\:{in}\:{your}\:{diagram} \\ $$$${is}\:{the}\:{incircle}.\: \\ $$

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