Question-212949

Question Number 212949 by Spillover last updated on 27/Oct/24

Answered by som(math1967) last updated on 27/Oct/24

Commented by som(math1967) last updated on 27/Oct/24

AB=BC=CD=AD=x DE=r(√2)⇒OD=r(√2)+r+R BD=2(r(√2)+r+R) ∴ x=((BD)/( (√2)))=(√2)(r(√2)+r+R) ((FG)/(BG))=tan 22(1/2) BG=R×cot22(1/2)=R((√2)+1) AB=x=AG+GB=R+R((√2)+1) ∴(√2)(r(√2)+r+R)=R(2+(√2)) ⇒r(2+(√2))=R(2+(√2)−(√2)) (r/R)=(2/((2+(√2))))=((2−(√2))/1)

$${AB}={BC}={CD}={AD}={x} \\ $$$$\:{DE}={r}\sqrt{\mathrm{2}}\Rightarrow{OD}={r}\sqrt{\mathrm{2}}+{r}+{R} \\ $$$${BD}=\mathrm{2}\left({r}\sqrt{\mathrm{2}}+{r}+{R}\right) \\ $$$$\therefore\:{x}=\frac{{BD}}{\:\sqrt{\mathrm{2}}}=\sqrt{\mathrm{2}}\left({r}\sqrt{\mathrm{2}}+{r}+{R}\right) \\ $$$$\:\frac{{FG}}{{BG}}=\mathrm{tan}\:\mathrm{22}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${BG}={R}×{cot}\mathrm{22}\frac{\mathrm{1}}{\mathrm{2}}={R}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$$${AB}={x}={AG}+{GB}={R}+{R}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$$$\therefore\sqrt{\mathrm{2}}\left({r}\sqrt{\mathrm{2}}+{r}+{R}\right)={R}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right) \\ $$$$\:\Rightarrow{r}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)={R}\left(\mathrm{2}+\sqrt{\mathrm{2}}−\sqrt{\mathrm{2}}\right) \\ $$$$\:\frac{{r}}{{R}}=\frac{\mathrm{2}}{\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{1}} \\ $$

Answered by ajfour last updated on 27/Oct/24

r(1+(√2))+R=R(1+(√2))=(a/( (√2))) (r/R)=((√2)/(1+(√2)))=2−(√2) ≈ 0.586 R=a(1−(1/( (√2))))≈0.393a r=(3−2(√2))a ≈0.172a

$${r}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)+{R}={R}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)=\frac{{a}}{\:\sqrt{\mathrm{2}}} \\ $$$$\frac{{r}}{{R}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{1}+\sqrt{\mathrm{2}}}=\mathrm{2}−\sqrt{\mathrm{2}}\:\:\approx\:\mathrm{0}.\mathrm{586} \\ $$$${R}={a}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\approx\mathrm{0}.\mathrm{393}{a}\:\:\: \\ $$$${r}=\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right){a}\:\approx\mathrm{0}.\mathrm{172}{a} \\ $$

Answered by A5T last updated on 27/Oct/24

Let side be s⇒diagonal=s(√2) (s^2 /2)=(R/2)(2s+s(√2))⇒R=(s/(2+(√2)))⇒(s/R)=2+(√2) r(√2)+r+R=((s(√2))/2)⇒^(/R) (r/R)(1+(√2))=(((√2)s)/(2R))−1 ⇒(r/R)=((√2)/(1+(√2)))=((((√2))((√2)−1))/( ((√2)+1)((√2)−1)))=2−(√2)

$${Let}\:{side}\:{be}\:{s}\Rightarrow{diagonal}={s}\sqrt{\mathrm{2}} \\ $$$$\frac{{s}^{\mathrm{2}} }{\mathrm{2}}=\frac{{R}}{\mathrm{2}}\left(\mathrm{2}{s}+{s}\sqrt{\mathrm{2}}\right)\Rightarrow{R}=\frac{{s}}{\mathrm{2}+\sqrt{\mathrm{2}}}\Rightarrow\frac{{s}}{{R}}=\mathrm{2}+\sqrt{\mathrm{2}} \\ $$$${r}\sqrt{\mathrm{2}}+{r}+{R}=\frac{{s}\sqrt{\mathrm{2}}}{\mathrm{2}}\overset{/{R}} {\Rightarrow}\frac{{r}}{{R}}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)=\frac{\sqrt{\mathrm{2}}{s}}{\mathrm{2}{R}}−\mathrm{1} \\ $$$$\Rightarrow\frac{{r}}{{R}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{1}+\sqrt{\mathrm{2}}}=\frac{\left(\sqrt{\mathrm{2}}\right)\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}=\mathrm{2}−\sqrt{\mathrm{2}} \\ $$

Answered by Spillover last updated on 27/Oct/24