Question Number 182910 by mnjuly1970 last updated on 16/Dec/22 | ||
Answered by mr W last updated on 16/Dec/22 | ||
$${a}={side}\:{length}\:{of}\:{square} \\ $$$${R}={radius}\:{of}\:{circle} \\ $$$${R}^{\mathrm{2}} =\left({R}−{a}\right)^{\mathrm{2}} +\left({R}−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{5}{a}^{\mathrm{2}} −\mathrm{12}{Ra}+\mathrm{4}{R}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\frac{{a}}{{R}}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{2}}{\mathrm{5}}×\mathrm{5}=\mathrm{2}\:{cm} \\ $$$${area}\:{of}\:{square}\:={a}^{\mathrm{2}} =\mathrm{4}\:{cm}^{\mathrm{2}} \\ $$ | ||
Commented by mnjuly1970 last updated on 16/Dec/22 | ||
$${grateful}\:{sir}\:{W} \\ $$ | ||