A circle touches all the sides of square ABCD | vedclass

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(A) For a quadrilateral circumscribing a circle,the sum of opposite sides is equal,i.e.,$AB + CD = BC + AD$.Given that $\overline{AB}$ is the largest

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$\overline{CD}$

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(A) For a quadrilateral circumscribing a circle,the sum of opposite sides is equal,i.e.,$AB + CD = BC + AD$.Given that $\overline{AB}$ is the largest side,let $AB = x$.Since the circle is tangent to all sides,the lengths of the tangents from the vertices to the circle are equal. Let the lengths of the tangents from vertices $A, B, C, D$ be $a, b, c, d$ respectively.Then $AB = a + b$,$BC = b + c$,$CD = c + d$,and $AD = a + d$.Since $AB$ is the largest side,$a+b$ is the maximum value.In a tangential quadrilateral,the side opposite to the largest side is generally the smallest side. Thus,$\overline{CD}$ is the smallest side.

$A$ circle touches all the sides of $\square ABCD$. If the largest side of $\square ABCD$ is $\overline{AB}$,then the smallest side is:

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