Question Number 64544 by LPM last updated on 19/Jul/19 | ||
Commented by LPM last updated on 19/Jul/19 | ||
$${area}\:{of}\:\:\Box{ABCD} \\ $$ | ||
Answered by MJS last updated on 19/Jul/19 | ||
$$\mid{CD}\mid=\mathrm{4sin}\:\mathrm{60}°\:=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\mathrm{area}\:\left({ABCD}\right)=\left(\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} =\mathrm{12} \\ $$ | ||
Commented by LPM last updated on 19/Jul/19 | ||
$$\mathrm{good} \\ $$ | ||