Question-86282

Question Number 86282 by ajfour last updated on 27/Mar/20

Commented by ajfour last updated on 27/Mar/20

$${Find}\:{s}/{r}\:.\:\:\left({s}\:{being}\:{side}\:{of}\:{square}\right). \\ $$

Commented by Prithwish Sen 1 last updated on 28/Mar/20

$$\boldsymbol{\mathrm{I}}\:\boldsymbol{\mathrm{get}} \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{3}} −\mathrm{6}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{\mathrm{x}}−\mathrm{4}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{where}}\:\boldsymbol{\mathrm{x}}\:=\:\frac{\boldsymbol{\mathrm{s}}}{\boldsymbol{\mathrm{r}}} \\ $$

Commented by ajfour last updated on 28/Mar/20

please explain how you reduced it to the cubic, Sir ? I get x≈4.38298 , your cubic too fetches the same, but i had resorted to trigonometric way..

$${please}\:{explain}\:{how}\:{you}\:{reduced} \\ $$$${it}\:{to}\:{the}\:{cubic},\:{Sir}\:? \\ $$$${I}\:{get}\:{x}\approx\mathrm{4}.\mathrm{38298}\:,\:{your}\:{cubic} \\ $$$${too}\:{fetches}\:{the}\:{same},\:{but}\:{i} \\ $$$${had}\:{resorted}\:{to}\:{trigonometric} \\ $$$${way}.. \\ $$

Commented by Prithwish Sen 1 last updated on 28/Mar/20

b= ((√(s^2 −4sr))/2) ∴ Diagonal AC = AE+EC i.e s(√2)= (√(r^2 +{(√(s^2 −4sr))+(s−r)^2 )) +r(√2) ((s/r)−1)(√2)= (√(1+{(√(((s/r))^2 −4((s/r))))+((s/r)+1)}^2 )) putting (s/r) = x and simplify

$$\boldsymbol{\mathrm{b}}=\:\frac{\sqrt{\boldsymbol{\mathrm{s}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{\mathrm{sr}}}}{\mathrm{2}} \\ $$$$\therefore\:\boldsymbol{\mathrm{Diagonal}}\:\:\boldsymbol{\mathrm{AC}}\:=\:\boldsymbol{\mathrm{AE}}+\boldsymbol{\mathrm{EC}} \\ $$$$\mathrm{i}.\mathrm{e}\:\mathrm{s}\sqrt{\mathrm{2}}=\:\sqrt{\mathrm{r}^{\mathrm{2}} +\left\{\sqrt{\mathrm{s}^{\mathrm{2}} −\mathrm{4sr}}+\left(\mathrm{s}−\mathrm{r}\right)^{\mathrm{2}} \right.}\:+\mathrm{r}\sqrt{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{s}}{\mathrm{r}}−\mathrm{1}\right)\sqrt{\mathrm{2}}=\:\sqrt{\mathrm{1}+\left\{\sqrt{\left(\frac{\mathrm{s}}{\mathrm{r}}\right)^{\mathrm{2}} −\mathrm{4}\left(\frac{\mathrm{s}}{\mathrm{r}}\right)}+\left(\frac{\mathrm{s}}{\mathrm{r}}+\mathrm{1}\right)\right\}^{\mathrm{2}} } \\ $$$$\:\:\:\mathrm{putting}\:\frac{\mathrm{s}}{\mathrm{r}}\:=\:\mathrm{x}\:\mathrm{and}\:\mathrm{simplify} \\ $$

Commented by Prithwish Sen 1 last updated on 28/Mar/20