Question Number 195996 by sonukgindia last updated on 15/Aug/23
Answered by Rasheed.Sindhi last updated on 15/Aug/23
$${Let}\:{radius}\:{of}\:{blue}\:{quarter}={r} \\ $$$${and}\:{of}\:{green}={R} \\ $$$${R}={r}+\mathrm{1} \\ $$$${Diagonal}={R}+{r} \\ $$$$\left({R}+{r}\right)^{\mathrm{2}} =\left({R}+\mathrm{1}\right)^{\mathrm{2}} +\left({r}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{Rr}={R}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{R}+{r}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{r} \\ $$$$\mathrm{2}{Rr}=\mathrm{2}+\mathrm{2}{R}+\mathrm{2}{r} \\ $$$${Rr}=\mathrm{1}+{R}+{r} \\ $$$${Rr}={R}+{R} \\ $$$${r}=\mathrm{2} \\ $$$${R}={r}+\mathrm{1}=\mathrm{2}+\mathrm{1}=\mathrm{3} \\ $$$${Perimeter}=\mathrm{2}\left(\:\left({r}+\mathrm{1}\right)+\left({R}+\mathrm{1}\right)\:\right) \\ $$$$\:\:\:\:\:=\mathrm{2}\left({R}+{r}+\mathrm{2}\right) \\ $$$$\:\:\:\:\:=\mathrm{2}\left(\mathrm{3}+\mathrm{2}+\mathrm{2}\right)=\mathrm{14}\checkmark \\ $$
Answered by cherokeesay last updated on 15/Aug/23
