Question Number 202276 by professorleiciano last updated on 23/Dec/23
Answered by professorleiciano last updated on 23/Dec/23
$${Area}\left({retangulo}\right)=\mathrm{4}×\mathrm{6}=\mathrm{24}{m}^{\mathrm{2}} \\ $$$${Area}\left({triangulo}\:{I}\right)=\mathrm{3}×\mathrm{6}=\mathrm{18}/\mathrm{2}=\mathrm{9}{m}^{\mathrm{2}} \\ $$$${Area}\left({triangulo}\:{II}\right)=\mathrm{3}×\mathrm{4}=\mathrm{12}/\mathrm{2}=\mathrm{6}{m}^{\mathrm{2}} \\ $$$${Area}\left({triangulo}\:{III}\right)=\mathrm{3}×\mathrm{4}=\mathrm{12}/\mathrm{2}=\mathrm{6}{m}^{\mathrm{2}} \\ $$$${Area}\left({total}\right)=\mathrm{24}{m}^{\mathrm{2}} +\mathrm{9}{m}^{\mathrm{2}} +\mathrm{6}{m}^{\mathrm{2}} +\mathrm{6}{m}^{\mathrm{2}} \\ $$$$=\mathrm{45}{m}^{\mathrm{2}} \\ $$$${Alternativa}\:\left({a}\right) \\ $$
Commented by professorleiciano last updated on 23/Dec/23
Commented by mr W last updated on 24/Dec/23
$${a}\:{better}\:{and}\:{more}\:{general}\:{method} \\ $$$${see}\:{below}. \\ $$
Answered by mr W last updated on 24/Dec/23
Commented by mr W last updated on 24/Dec/23
$${A}_{{k}} =\frac{\left({x}_{{k}+\mathrm{1}} −{x}_{{k}} \right)\left({y}_{{k}} +{y}_{{k}+\mathrm{1}} \right)}{\mathrm{2}} \\ $$$$\begin{array}{|c|c|c|c|c|c|c|c|}{{k}}&\hline{{x}_{{k}} }&\hline{{y}_{{k}} }&\hline{{x}_{{k}+\mathrm{1}} −{x}_{{k}} }&\hline{{y}_{{k}} +{y}_{{k}+\mathrm{1}} }&\hline{{A}_{{k}} }\\{\mathrm{1}}&\hline{\mathrm{1}}&\hline{\mathrm{1}}&\hline{\mathrm{0}}&\hline{\mathrm{8}}&\hline{\mathrm{0}}\\{\mathrm{2}}&\hline{\mathrm{1}}&\hline{\mathrm{7}}&\hline{\mathrm{4}}&\hline{\mathrm{14}}&\hline{\mathrm{28}}\\{\mathrm{3}}&\hline{\mathrm{5}}&\hline{\mathrm{7}}&\hline{\mathrm{2}}&\hline{\mathrm{18}}&\hline{\mathrm{18}}\\{\mathrm{4}}&\hline{\mathrm{7}}&\hline{\mathrm{11}}&\hline{\mathrm{3}}&\hline{\mathrm{22}}&\hline{\mathrm{33}}\\{\mathrm{5}}&\hline{\mathrm{10}}&\hline{\mathrm{11}}&\hline{−\mathrm{5}}&\hline{\mathrm{12}}&\hline{−\mathrm{30}}\\{\mathrm{6}}&\hline{\mathrm{5}}&\hline{\mathrm{1}}&\hline{−\mathrm{4}}&\hline{\mathrm{2}}&\hline{−\mathrm{4}}\\{\mathrm{7}\left(=\mathrm{1}\right)}&\hline{\mathrm{1}}&\hline{\mathrm{1}}&\hline{\diagup}&\hline{\diagup}&\hline{\mathrm{45}}\\\hline\end{array} \\ $$$${Area}\:{of}\:{polygon}\:=\:\Sigma{A}_{{k}} =\mathrm{45}\:\checkmark \\ $$$$\Rightarrow{answer}\:\left({a}\right) \\ $$