Question Number 151973 by ajfour last updated on 24/Aug/21
Commented by ajfour last updated on 24/Aug/21
$${OB}={OC}=\sqrt{{OA}}\:\:;\:{OM}=\mathrm{1} \\ $$$${yellow}\:{area}=\mathrm{1}/\mathrm{4}\:,\:{then}\:{find} \\ $$$${OB}={OC}\:={x}\:\:\:{or}\:\:{OA}={x}^{\mathrm{2}} . \\ $$
Answered by ajfour last updated on 24/Aug/21
Commented by ajfour last updated on 25/Aug/21
$$\left({s}+\sqrt{{s}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right)\left({a}+{b}\right)=\mathrm{2}{c}+\mathrm{2}\sqrt{{s}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} ={s}^{\mathrm{2}} −\mathrm{1}+\left(\mathrm{1}+\sqrt{{b}^{\mathrm{2}} +{s}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$\left({s}^{\mathrm{2}} −\mathrm{1}\right)\left({s}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$${To}\:{someone}\:{who}\:{cares},\:{for}\:{this} \\ $$$${Find}\:{s}\:{from}\:{above}\:\mathrm{3}\:{eqs}. \\ $$