The perimeter of rectangle ABCD is 28. If AC | vedclass

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(A) Let the sides of the rectangle be $AB = x$ and $BC = y$.The perimeter of a rectangle is given by $2(x + y) = 28$,which simplifies to $x + y = 14$.

Vedclass · Accepted Answer

$6$ and $8$

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(A) Let the sides of the rectangle be $AB = x$ and $BC = y$.The perimeter of a rectangle is given by $2(x + y) = 28$,which simplifies to $x + y = 14$.In a rectangle,the diagonal $AC$ forms a right-angled triangle $ABC$ with sides $AB$ and $BC$.By the Pythagorean theorem,$x^2 + y^2 = AC^2 = 10^2 = 100$.We have the system of equations:$1) x + y = 14$$2) x^2 + y^2 = 100$From $(1)$,$y = 14 - x$. Substituting this into $(2)$:$x^2 + (14 - x)^2 = 100$$x^2 + 196 - 28x + x^2 = 100$$2x^2 - 28x + 96 = 0$$x^2 - 14x + 48 = 0$$(x - 6)(x - 8) = 0$So,$x = 6$ or $x = 8$.If $x = 6$,then $y = 14 - 6 = 8$. Since $AB If $x = 8$,then $y = 14 - 8 = 6$. This contradicts $AB Therefore,the sides are $AB = 6$ and $BC = 8$.

The perimeter of rectangle $ABCD$ is $28$. If $AC = 10$ and $AB < BC$,find the measures of its sides.

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