In a rectangle ABCD,if AB + BC = 47 and the d | vedclass

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(A) Let $AB = x$ and $BC = y$. Given $x + y = 47$ and $x^2 + y^2 = AC^2 = 37^2 = 1369$.We know that $(x + y)^2 = x^2 + y^2 + 2xy$.Substituting the val

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$AB = 35, BC = 12$

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(A) Let $AB = x$ and $BC = y$. Given $x + y = 47$ and $x^2 + y^2 = AC^2 = 37^2 = 1369$.We know that $(x + y)^2 = x^2 + y^2 + 2xy$.Substituting the values: $47^2 = 1369 + 2xy$.$2209 = 1369 + 2xy$,which gives $2xy = 840$,so $xy = 420$.Now,we have $x + y = 47$ and $xy = 420$. These are the roots of the quadratic equation $t^2 - 47t + 420 = 0$.Solving the quadratic equation: $t^2 - 35t - 12t + 420 = 0$.$t(t - 35) - 12(t - 35) = 0$,so $(t - 35)(t - 12) = 0$.The roots are $t = 35$ and $t = 12$.Since $AB > BC$,we have $AB = 35$ and $BC = 12$.

In a rectangle $ABCD$,if $AB + BC = 47$ and the diagonal $AC = 37$,find the lengths of $AB$ and $BC$,given that $AB > BC$.

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