Question Number 200092 by ajfour last updated on 13/Nov/23
Commented by ajfour last updated on 13/Nov/23
$${If}\:{semicircle}\:{has}\:{radius}\:\mathrm{1},\:{find} \\ $$$${radii}\:{of}\:{blue}\:{and}\:{green}\:{circles}. \\ $$$${One}\:{end}\:{of}\:{each}\:{semicircle}\:{is}\:{at} \\ $$$${center}\:{of}\:{other}. \\ $$
Answered by mr W last updated on 13/Nov/23
Commented by mr W last updated on 13/Nov/23
$$\sqrt{\left(\mathrm{1}+{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }=\mathrm{1}+\sqrt{\left(\mathrm{1}−{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$\sqrt{\mathrm{1}+\mathrm{2}{a}}=\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{2}{a}} \\ $$$$\mathrm{4}{a}−\mathrm{1}=\mathrm{2}\sqrt{\mathrm{1}−\mathrm{2}{a}} \\ $$$$\mathrm{16}{a}^{\mathrm{2}} =\mathrm{3} \\ $$$$\Rightarrow{a}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\:\checkmark \\ $$$$\sqrt{\left(\mathrm{1}−{b}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{1}−\mathrm{2}{b}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{b}=\frac{\mathrm{3}}{\mathrm{8}}\:\checkmark \\ $$
Commented by ajfour last updated on 13/Nov/23
$${excellent}\:{workings}!\:{thank}\:{you}\:{sir}. \\ $$