Question Number 93986 by $@ty@m123 last updated on 16/May/20 | ||
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Answered by mr W last updated on 16/May/20 | ||
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Commented by mr W last updated on 16/May/20 | ||
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$$\left[{AEF}\right]=\frac{\mathrm{3}}{\mathrm{5}}\left[{ABF}\right]=\frac{\mathrm{3}}{\mathrm{5}}×\frac{\mathrm{1}}{\mathrm{6}}\left[{ABC}\right] \\ $$$$\left[{BED}\right]=\frac{\mathrm{2}}{\mathrm{5}}\left[{BAD}\right]=\frac{\mathrm{2}}{\mathrm{5}}×\frac{\mathrm{4}}{\mathrm{7}}\left[{ABC}\right] \\ $$$$\left[{EFCD}\right]=\left[{ABC}\right]−\left[{AEF}\right]−\left[{BED}\right] \\ $$$$=\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{5}}×\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{2}}{\mathrm{5}}×\frac{\mathrm{4}}{\mathrm{7}}\right)\left[{ABC}\right] \\ $$$$=\frac{\mathrm{47}}{\mathrm{70}}\left[{ABC}\right] \\ $$$$\Rightarrow\frac{\left[{EFCD}\right]}{\left[{ABC}\right]}=\frac{\mathrm{47}}{\mathrm{70}} \\ $$ | ||
Commented by $@ty@m123 last updated on 17/May/20 | ||
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$${Thanks}\:{a}\:{lot}. \\ $$ | ||
Answered by $@ty@m123 last updated on 17/May/20 | ||
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Commented by $@ty@m123 last updated on 17/May/20 | ||
Got an alternative solution from my friend. | ||