Question Number 196225 by mr W last updated on 20/Aug/23
Commented by mr W last updated on 20/Aug/23
$${find}\:{the}\:{area}\:{of}\:{red}\:{part}\:{in}\:{the} \\ $$$${parallelogram}. \\ $$
Answered by liuxinnan last updated on 20/Aug/23
$${s}_{\mathrm{1}} +{s}_{\mathrm{2}} +{s}_{\mathrm{5}} +{s}_{\mathrm{6}} =\frac{\mathrm{1}}{\mathrm{2}}{S} \\ $$$${s}_{\mathrm{2}} +{s}_{\mathrm{3}} +{s}_{\mathrm{4}} +{s}_{\mathrm{6}} +{s}_{\mathrm{7}} =\frac{\mathrm{1}}{\mathrm{2}}{S} \\ $$$${s}_{\mathrm{7}} =\mathrm{79}+\mathrm{10}−\mathrm{8}−\mathrm{72}=\mathrm{9} \\ $$
Commented by liuxinnan last updated on 20/Aug/23
Commented by mr W last updated on 20/Aug/23
$${great}! \\ $$
Answered by mr W last updated on 20/Aug/23
Commented by mr W last updated on 20/Aug/23
$$\mathrm{10}+{E}+\mathrm{79}+{F}=\frac{{parallelogram}}{\mathrm{2}} \\ $$$$\mathrm{8}+{E}+\mathrm{72}+{F}+{red}=\frac{{parallelogram}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{8}+{E}+\mathrm{72}+{F}+{red}=\mathrm{10}+{E}+\mathrm{79}+{F} \\ $$$$\Rightarrow{red}=\mathrm{10}+\mathrm{79}−\mathrm{8}−\mathrm{72}=\mathrm{9}\:\checkmark \\ $$