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Question-64544




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

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