Question Number 222673 by ajfour last updated on 04/Jul/25
Commented by ajfour last updated on 04/Jul/25
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Find}\:{small}\:{circle}\:{radius}\:\boldsymbol{{a}}. \\ $$$${Oh}!\:{i}\:{forgot}\:{to}\:{say}\:{almost}-\:{ABCD}\:{is}\:{sq}. \\ $$
Answered by mr W last updated on 04/Jul/25
$$\left(\mathrm{2}+{a}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{3}}−{a}\right)^{\mathrm{2}} +\left(\mathrm{1}+\sqrt{\mathrm{3}}−{a}\right)^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}\right){a}+\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$$\Rightarrow{a}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}\left(\mathrm{9}+\mathrm{5}\sqrt{\mathrm{3}}\right)}\:\approx\mathrm{0}.\mathrm{521} \\ $$
Commented by ajfour last updated on 04/Jul/25
$${excellent}\:{sir}.\:{Precise}! \\ $$