Question Number 201853 by sonukgindia last updated on 14/Dec/23
Answered by AST last updated on 14/Dec/23
$${Let}\:{green}\:{segment}={g};{orange}\:{segment}={r} \\ $$$${and}\:{blue}\:{segment}={b};{side}\:{of}\:{square}={s} \\ $$$$\frac{{b}}{{r}}=\frac{\mathrm{7}−\mathrm{5}}{\mathrm{7}−\mathrm{3}}\Rightarrow{r}=\mathrm{2}{b};\frac{{g}}{{g}+{s}}=\frac{\mathrm{2}}{\mathrm{4}}\Rightarrow{g}={s} \\ $$$$\frac{\mathrm{2}{b}}{\mathrm{2}{b}+{s}}=\frac{\mathrm{7}−\mathrm{3}}{\mathrm{13}−\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{5}}\Rightarrow\mathrm{5}{r}=\mathrm{2}{r}+\mathrm{2}{s}\Rightarrow{r}=\frac{\mathrm{2}{s}}{\mathrm{3}} \\ $$$${r}^{\mathrm{2}} +\left({s}+{g}\right)^{\mathrm{2}} =\mathrm{16}\Rightarrow\frac{\mathrm{4}{s}^{\mathrm{2}} }{\mathrm{9}}+\mathrm{4}{s}^{\mathrm{2}} =\mathrm{16}\Rightarrow\mathrm{40}{s}^{\mathrm{2}} =\mathrm{16}×\mathrm{9} \\ $$$$\Rightarrow{s}^{\mathrm{2}} =\frac{\mathrm{4}×\mathrm{9}}{\mathrm{10}}=\mathrm{3}.\mathrm{6} \\ $$