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Question Number 137893 by bemath last updated on 07/Apr/21

Answered by benjo_mathlover last updated on 08/Apr/21

$$byalgebra$$$$letF(0,0),D(1,0),A(0,1),C(1,1)$$$$B(\frac{1}{2},1),E(\frac{1}{4},0)$$$$eqoflineAE:x+\frac{1}{4}y=\frac{1}{4};$$$$eqoflineFB:y=2x$$$$intersectatpointG(\frac{1}{6},\frac{1}{3})$$$$eqoflineEC:4x-3y=1$$$$eqoflineBD:2x+y=2$$$$intersectatpointH(\frac{7}{10},\frac{3}{5})$$$$shadowofareais$$$$\frac{1}{2}\mid \left|\begin{array}{c}\frac{1}{4}0\\ \frac{7}{10}\frac{3}{5}\end{array}\right|+\left|\begin{array}{c}\frac{7}{10}\frac{3}{5}\\ \frac{1}{2}1\end{array}\right|+\left|\begin{array}{c}\frac{1}{2}1\\ \frac{1}{6}\frac{1}{3}\end{array}\right|+\left|\begin{array}{c}\frac{1}{6}\frac{1}{3}\\ \frac{1}{4}0\end{array}\right|\mid$$$$=\frac{1}{2}\mid (\frac{3}{20}+\frac{4}{10}+0-\frac{1}{12})\mid$$$$=\frac{1}{2}(\frac{11}{20}-\frac{1}{12})=\frac{7}{30}$$

Answered by mr W last updated on 08/Apr/21

$$\frac{EG}{GA}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}\Rightarrow EG=\frac{1}{2}GA=\frac{1}{3}AE$$$$\Rightarrow \mathrm{\Delta }EBG=\frac{1}{3}\mathrm{\Delta }EBA=\frac{1}{6}\mathrm{\Delta }EAC=\frac{[ACDF[}{12}$$$$\frac{EH}{HC}=\frac{\frac{3}{4}}{\frac{1}{2}}=\frac{3}{2}\Rightarrow EH=\frac{3}{2}HC=\frac{3}{5}EC$$$$\Rightarrow \mathrm{\Delta }EBH=\frac{3}{5}\mathrm{\Delta }EBC=\frac{3}{10}\mathrm{\Delta }EAC=\frac{3\left[ACDF\right]}{20}$$$$$$$$shadedarea=\mathrm{\Delta }EBG+\mathrm{\Delta }EBH$$$$=(\frac{1}{12}+\frac{3}{20})\left[ACDF\right]=\frac{7}{30}\times \left[ACDF\right]$$$$=\frac{7}{30}{cm}^2$$

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