Question Number 176437 by cherokeesay last updated on 19/Sep/22 | ||
Answered by HeferH last updated on 19/Sep/22 | ||
$$\:{by}\:{similar}\:{triangles}: \\ $$$$\:{b}\:={AD}\:=\:\frac{\mathrm{8}\sqrt{\mathrm{5}}}{\mathrm{2}\:+\:\mathrm{2}\sqrt{\mathrm{5}}}\:\:=\:\frac{\left(\mathrm{4}\sqrt{\mathrm{5}}\right)}{\left(\mathrm{1}\:+\:\sqrt{\mathrm{5}}\right)} \\ $$$$\:{h}\:=\:\mathrm{4} \\ $$$$\:{A}_{\bigtriangleup{ABD}} \:=\:\frac{{bh}}{\mathrm{2}}\:=\:\frac{\mathrm{8}\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{5}}\:+\:\mathrm{1}}\:=\:\mathrm{10}\:−\:\mathrm{2}\sqrt{\mathrm{5}\:} \\ $$$$\: \\ $$ | ||
Commented by cherokeesay last updated on 19/Sep/22 | ||
$${nice},\:{thank}\:{you}\:! \\ $$ | ||