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(D) Let the square have side length $2 \ cm$. $A$ cut is made from one corner to a point on an adjacent side,creating a right-angled triangle and a qu

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$4\sqrt{2} + 8$

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(D) Let the square have side length $2 \ cm$. $A$ cut is made from one corner to a point on an adjacent side,creating a right-angled triangle and a quadrilateral (trapezium).Let the cut start from a corner and end at a distance $x$ from the corner on one side and $y$ from the corner on the other side. However,based on the figure,the cut is a straight line segment of length $L$ connecting a point at distance $x$ from the corner on one side to the corner itself on the other side. Let the triangle have sides $x$,$2$,and $L$,where $L = \sqrt{x^2 + 2^2}$.The perimeter of the triangle is $P_1 = x + 2 + L$.The remaining figure is a quadrilateral with sides $2$,$2$,$(2-x)$,and $L$. The perimeter of the quadrilateral is $P_2 = 2 + 2 + (2-x) + L = 6 - x + L$.The sum of the perimeters is $S = P_1 + P_2 = (x + 2 + L) + (6 - x + L) = 8 + 2L$.To maximize $S$,we must maximize $L$. The maximum length of a straight line segment $L$ that can be cut from a corner of a square of side $2$ is the diagonal of the square.Thus,$L = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.The maximum sum of the perimeters is $S_{max} = 8 + 2(2\sqrt{2}) = 8 + 4\sqrt{2} \ cm$.

Consider a square of side $2 \ cm$. It is cut from one of its corners as shown in the adjacent figure. The maximum value of the sum of the perimeters of the two plane figures thus formed is

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