Question Number 205160 by cortano12 last updated on 11/Mar/24
Answered by A5T last updated on 11/Mar/24
Commented by A5T last updated on 11/Mar/24
$${AG}=\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{2}×\mathrm{5}×\mathrm{5}{cos}\left(\mathrm{135}°\right)}=\mathrm{5}\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}={GE} \\ $$$${BG}=\sqrt{\mathrm{75}+\mathrm{25}\sqrt{\mathrm{2}}−\mathrm{50}\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}{cos}\left(\mathrm{112}.\mathrm{5}\right)}={BE}=\mathrm{5}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$$${Then}\:{apply}\:{Heron}'{s}\:{formula} \\ $$
Answered by mr W last updated on 11/Mar/24
Commented by mr W last updated on 11/Mar/24
$${a}=\mathrm{2}×\frac{\mathrm{5}}{\:\sqrt{\mathrm{2}}}+\mathrm{5}=\mathrm{5}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$$${A}_{{blue}} =\frac{{a}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{45}°}{\mathrm{2}}=\frac{\mathrm{25}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{25}\left(\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}\right)}{\mathrm{4}} \\ $$