Question Number 202581 by Frix last updated on 30/Dec/23
Commented by Frix last updated on 30/Dec/23
$$\mathrm{An}\:\mathrm{easy}\:\mathrm{one}:\:\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rectangle}. \\ $$
Answered by mr W last updated on 30/Dec/23
Commented by mr W last updated on 30/Dec/23
$${R}={radius}\:{of}\:{circumcircle}\:{of}\:\Delta{ABC} \\ $$$$\frac{\sqrt{\mathrm{5}}{a}}{\mathrm{sin}\:\mathrm{135}°}=\mathrm{2}{R}=\mathrm{10} \\ $$$$\Rightarrow{a}=\frac{\sqrt{\mathrm{10}}{R}}{\:\mathrm{5}}=\sqrt{\mathrm{10}} \\ $$$${area}\:{of}\:{rectangle}\:=\mathrm{2}{a}^{\mathrm{2}} =\mathrm{20} \\ $$
Commented by Frix last updated on 30/Dec/23
��
Answered by ajfour last updated on 30/Dec/23
Commented by ajfour last updated on 30/Dec/23
$${r}^{\mathrm{2}} =\left({a}+\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +{k}^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} =\left({a}+{k}\right)^{\mathrm{2}} +\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{k}=\frac{{a}}{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} =\frac{\mathrm{5}{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${Area}\:=\:\mathrm{2}{a}^{\mathrm{2}} =\frac{\mathrm{4}{r}^{\mathrm{2}} }{\mathrm{5}}=\frac{\mathrm{4}}{\mathrm{5}}×\mathrm{25}=\mathrm{20} \\ $$
Commented by mr W last updated on 30/Dec/23
$${right}!\:{i}\:{got}\:{the}\:{same}. \\ $$
Commented by Frix last updated on 30/Dec/23
��