Question Number 93540 by M±th+et+s last updated on 13/May/20 | ||
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Commented by M±th+et+s last updated on 13/May/20 | ||
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$${correct}\:{sir}\: \\ $$ | ||
Commented by M±th+et+s last updated on 13/May/20 | ||
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$${The}\:{digonal}\:{BD}\:{of}\:{the}\:{square}\:{ABCD} \\ $$$${cuts}\:{the}\:\mathrm{45}°−{triangle}\:{into}\:\mathrm{2}\:{parts}. \\ $$$${Area}\:{of}\:{red}\::\:{Area}\:{of}\:{blue}=? \\ $$ | ||
Commented by Shakhzod last updated on 13/May/20 | ||
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$${What}\:{should}\:{we}\:{do}\:{it}? \\ $$ | ||
Commented by M±th+et+s last updated on 13/May/20 | ||
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$${find}\:{the}\:{ratio}\:{of}\:{two}\:{parts} \\ $$ | ||
Commented by mr W last updated on 13/May/20 | ||
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$${without}\:{calculation}: \\ $$$$\frac{{red}}{{blue}}=\mathrm{1} \\ $$ | ||
Commented by Ari last updated on 13/May/20 | ||
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$$\mathrm{why}\:\mathrm{can}\:\mathrm{samone}\:\mathrm{explayn}\:\mathrm{this}? \\ $$ | ||
Commented by mr W last updated on 14/May/20 | ||
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$${just}\:{look}\:{at}\:{the}\:{situation}\:{when}\:{the}\: \\ $$$${red}+{blue}\:{areas}\:{form}\:{the}\:{triangle} \\ $$$$\Delta{ABC}. \\ $$ | ||