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It is given that ∆ABC ~ ∆DEF and \(\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{1}{5}\). Then \(\frac{\operatorname{ar}(D E F)}{\operatorname{ar}(A B C)}\) is equal to

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It is given that ∆ABC ~ ∆DEF and \(\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{1}{5}\). Then \(\frac{\operatorname{ar}(D E F)}{\operatorname{ar}(A B C)}\) is equal to






 
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