Question Number 220016 by fantastic last updated on 04/May/25
Answered by efronzo1 last updated on 04/May/25
$$\:\frac{\mathrm{PO}}{\mathrm{sin}\:\mathrm{60}°}\:=\:\frac{\mathrm{8}}{\mathrm{sin}\:\mathrm{75}°}\:=\:\frac{\mathrm{QO}}{\mathrm{sin}\:\mathrm{45}°} \\ $$$$\:\mathrm{PO}\:=\:\frac{\mathrm{16}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{6}}\:+\sqrt{\mathrm{2}}}\:=\mathrm{4}\left(\mathrm{3}\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{6}}\:\right) \\ $$$$\:\mathrm{area}\:\Delta\mathrm{POQ}\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\:\mathrm{8}.\:\mathrm{4}\sqrt{\mathrm{2}}\:\left(\mathrm{3}−\sqrt{\mathrm{3}}\right) \\ $$$$\:=\:\mathrm{16}\sqrt{\mathrm{2}}\:\left(\mathrm{3}−\sqrt{\mathrm{3}}\:\right) \\ $$