Question Number 200575 by sonukgindia last updated on 20/Nov/23
Answered by AST last updated on 20/Nov/23
$$\frac{{ra}}{\mathrm{2}}=\frac{{bx}}{\mathrm{2}}\Rightarrow{ra}={bx};{x}=\sqrt{{r}^{\mathrm{2}} −{a}^{\mathrm{2}} };{b}={r}−{a} \\ $$$${ra}={bx}\Rightarrow{ra}=\left({r}−{a}\right)\left(\sqrt{{r}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\mathrm{2}{r}={a}\underset{−} {+}{a}\sqrt{\mathrm{5}+\mathrm{4}\sqrt{\mathrm{2}}}\Rightarrow{a}+\mathrm{2}{b}=\underset{−} {+}{a}\sqrt{\mathrm{5}+\mathrm{4}\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow{a}\left(\mathrm{1}\overset{−} {+}\sqrt{\mathrm{5}+\mathrm{4}\sqrt{\mathrm{2}}}\right)=−\mathrm{2}{b}\Rightarrow\frac{{a}}{{b}}=\frac{−\mathrm{2}}{\mathrm{1}\overset{−} {+}\sqrt{\mathrm{5}+\mathrm{4}\sqrt{\mathrm{2}}}} \\ $$$$\Rightarrow\frac{{a}}{{b}}\approx\mathrm{0}.\mathrm{883203} \\ $$
Commented by mr W last updated on 21/Nov/23
$${how}\:{did}\:{you}\:{get}? \\ $$$$\Rightarrow\mathrm{2}{r}={a}\underset{−} {+}{a}\sqrt{\mathrm{5}+\mathrm{4}\sqrt{\mathrm{2}}} \\ $$
Commented by AST last updated on 21/Nov/23
$$\frac{{r}^{\mathrm{2}} {a}^{\mathrm{2}} }{\left({r}−{a}\right)^{\mathrm{2}} }={r}^{\mathrm{2}} −{a}^{\mathrm{2}} \Rightarrow{r}^{\mathrm{2}} {a}^{\mathrm{2}} =\left({r}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)\left({r}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}{ar}\right) \\ $$$$\Rightarrow{r}^{\mathrm{4}} −\mathrm{2}{ar}^{\mathrm{3}} −{a}^{\mathrm{2}} {r}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{3}} {r}−{a}^{\mathrm{4}} =\mathrm{0} \\ $$$${So},{we}\:{can}\:{solve}\:{the}\:{polynomial}\:{in}\:{r},\:{I}\:{used}\: \\ $$$${wolfram}.\:{r}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}\underset{−} {+}{a}\sqrt{\mathrm{5}+\mathrm{4}\sqrt{\mathrm{2}}}\right) \\ $$
Commented by mr W last updated on 21/Nov/23
$${thanks}\:{sir}!\:{i}\:{just}\:{wanted}\:{to}\:{know}\:{if} \\ $$$${you}\:{had}\:{a}\:{way}\:{to}\:{avoid}\:{solving}\: \\ $$$${quadratic}\:{equation}. \\ $$