Question Number 179159 by mr W last updated on 25/Oct/22
Commented by mr W last updated on 25/Oct/22
$${find}\:{the}\:{area}\:{of}\:{ABCD}. \\ $$
Answered by mr W last updated on 25/Oct/22
Commented by cherokeesay last updated on 26/Oct/22
$${so}\:{beautiful}.\:{thank}\:{you} \\ $$
Commented by mr W last updated on 25/Oct/22
![[ABCD]=(((7×9−4×4)×(√3))/(2×2))=((47(√3))/4)](https://www.tinkutara.com/question/Q179169.png)
$$\left[{ABCD}\right]=\frac{\left(\mathrm{7}×\mathrm{9}−\mathrm{4}×\mathrm{4}\right)×\sqrt{\mathrm{3}}}{\mathrm{2}×\mathrm{2}}=\frac{\mathrm{47}\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$
Commented by cortano1 last updated on 26/Oct/22
$$\:\mathrm{L}_{\mathrm{ABCD}} =\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{7}.\mathrm{9}.\mathrm{sin}\:\mathrm{60}°−\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{4}.\mathrm{4}.\mathrm{sin}\:\mathrm{60}° \\ $$$$\: \\ $$