Question Number 182135 by Acem last updated on 04/Dec/22 | ||
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Commented by Acem last updated on 04/Dec/22 | ||
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$${Sum}\:{of}\:{their}\:{areas}\:\uparrow \\ $$ | ||
Answered by mr W last updated on 04/Dec/22 | ||
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Commented by mr W last updated on 04/Dec/22 | ||
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$$\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{7}×\mathrm{2}+\mathrm{5}\left({x}−\mathrm{4}\right)+{xy}+\mathrm{3}\left(\mathrm{3}+{y}\right)\right\}+\mathrm{35}=\left(\mathrm{3}+{x}\right)\left(\mathrm{5}+{y}\right) \\ $$$$\Rightarrow\mathrm{5}{x}+\mathrm{3}{y}+{xy}=\mathrm{43} \\ $$$${sum}\:{of}\:{areas}\:{of}\:{triangles}: \\ $$$$\left(\mathrm{3}+{x}\right)\left(\mathrm{5}+{y}\right)−\mathrm{35} \\ $$$$=\mathrm{5}{x}+\mathrm{3}{y}+{xy}−\mathrm{20} \\ $$$$=\mathrm{43}−\mathrm{20}=\mathrm{23}\:\checkmark \\ $$ | ||
Commented by Acem last updated on 04/Dec/22 | ||
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$${Thanks}\:{Sir}! \\ $$ | ||
Answered by Acem last updated on 04/Dec/22 | ||
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Commented by Acem last updated on 04/Dec/22 | ||
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$$;\:{S}^{\:\ast} =\:\mathrm{4}×\mathrm{3}\:,\:{S}:\:{Quadrilateral} \\ $$ | ||