In rectangle ABCD,AC = 29 and AB + BC = 41. F | vedclass

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(A) Let $AB = x$ and $BC = y$. In rectangle $ABCD$,$	riangle ABC$ is a right-angled triangle with $\angle B = 90^\circ$. By the Pythagorean theorem,$

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(A) Let $AB = x$ and $BC = y$. In rectangle $ABCD$,$	riangle ABC$ is a right-angled triangle with $\angle B = 90^\circ$. By the Pythagorean theorem,$AB^2 + BC^2 = AC^2$. Given $x^2 + y^2 = 29^2 = 841$. Also given $x + y = 41$. We know that $(x + y)^2 = x^2 + y^2 + 2xy$. Substituting the values: $41^2 = 841 + 2xy$. $1681 = 841 + 2xy$. $2xy = 1681 - 841 = 840$. $xy = 420$. Now,we have $x + y = 41$ and $xy = 420$. These are the roots of the quadratic equation $t^2 - (x+y)t + xy = 0$. $t^2 - 41t + 420 = 0$. Factoring the equation: $(t - 21)(t - 20) = 0$. So,$t = 21$ or $t = 20$. Since $AB > BC$,we have $AB = 21$ and $BC = 20$. Thus,$AB = 21$.

In rectangle $ABCD$,$AC = 29$ and $AB + BC = 41$. Find $AB$ (given $AB > BC$).

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