Question Number 201562 by Mingma last updated on 09/Dec/23
Answered by mr W last updated on 09/Dec/23
$${r}={radius}\:{of}\:{quarter}\:{circle} \\ $$$${a}={side}\:{length}\:{of}\:{square} \\ $$
Commented by mr W last updated on 09/Dec/23
Commented by mr W last updated on 09/Dec/23
$${r}^{\mathrm{2}} =\left(\frac{{a}}{\:\sqrt{\mathrm{2}}}+\frac{{a}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\left(\frac{{a}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} =\frac{\mathrm{5}{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{\mathrm{2}{r}^{\mathrm{2}} }{\mathrm{5}}=\mathrm{0}.\mathrm{4}={area}\:{of}\:{square} \\ $$
Commented by Mingma last updated on 09/Dec/23
Perfect ��
Answered by cherokeesay last updated on 09/Dec/23
Commented by Mingma last updated on 09/Dec/23
Perfect ��
Answered by ajfour last updated on 09/Dec/23
Commented by ajfour last updated on 09/Dec/23
$$\left({a}+\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${a}^{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{10}}=\mathrm{0}.\mathrm{4} \\ $$