Question Number 153432 by naka3546 last updated on 07/Sep/21
Commented by talminator2856791 last updated on 07/Sep/21
$$\:\mathrm{how}? \\ $$
Answered by talminator2856791 last updated on 07/Sep/21
$$\: \\ $$$$\:\mathrm{17}\:=\:\sqrt{\mathrm{13}}{x}\centerdot\mathrm{sin}\left({X}+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)\right)\:\left(\mathrm{not}\:\mathrm{needed}\right)\: \\ $$$$\:\mathrm{17}\:=\:\mathrm{3}{x}\centerdot\mathrm{sin}\left({X}\right)+\mathrm{2}{x}\centerdot\mathrm{cos}\left({X}\right) \\ $$$$\:\mathrm{26}\:=\:\mathrm{5}{x}\centerdot\mathrm{cos}\left({X}\right)+\mathrm{2}{x}\centerdot\mathrm{sin}\left({X}\right) \\ $$$$\: \\ $$$$\:{x}\:=\:\mathrm{5} \\ $$$$\: \\ $$$$\:\:\:\:\:\:\mathrm{shaded}\:\mathrm{area}\: \\ $$$$\:=\:\mathrm{17}×\mathrm{26}−\mathrm{7}\left(\mathrm{5}×\mathrm{5}\right) \\ $$$$\:=\:\mathrm{267} \\ $$$$\: \\ $$
Answered by mr W last updated on 07/Sep/21
Commented by mr W last updated on 07/Sep/21
$${r}={red},\:{g}={green} \\ $$$$\mathrm{26}=\mathrm{2}{r}+\mathrm{5}{g} \\ $$$$\mathrm{17}=\mathrm{3}{r}+\mathrm{2}{g} \\ $$$$\Rightarrow{g}=\mathrm{4},\:{r}=\mathrm{3} \\ $$$${side}\:{length}\:{of}\:{small}\:{squares} \\ $$$$=\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }=\mathrm{5} \\ $$$${shaded}\:{area}=\mathrm{26}×\mathrm{17}−\mathrm{7}×\mathrm{5}^{\mathrm{2}} =\mathrm{267} \\ $$