If A(0, 0),B(2, 0),C(2, 2),and D(0, 2),then s | vedclass

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(A) To determine the type of quadrilateral $ABCD$,we calculate the lengths of its sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 -

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(A) To determine the type of quadrilateral $ABCD$,we calculate the lengths of its sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.$1$. Length of $AB = \sqrt{(2 - 0)^2 + (0 - 0)^2} = \sqrt{2^2 + 0^2} = 2$.$2$. Length of $BC = \sqrt{(2 - 2)^2 + (2 - 0)^2} = \sqrt{0^2 + 2^2} = 2$.$3$. Length of $CD = \sqrt{(0 - 2)^2 + (2 - 2)^2} = \sqrt{(-2)^2 + 0^2} = 2$.$4$. Length of $DA = \sqrt{(0 - 0)^2 + (0 - 2)^2} = \sqrt{0^2 + (-2)^2} = 2$.Since all sides are equal $(AB = BC = CD = DA = 2)$,the quadrilateral is either a square or a rhombus.Next,we check the diagonals $AC$ and $BD$:Diagonal $AC = \sqrt{(2 - 0)^2 + (2 - 0)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.Diagonal $BD = \sqrt{(0 - 2)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.Since all sides are equal and the diagonals are equal,$\square ABCD$ is a square.

If $A(0, 0)$,$B(2, 0)$,$C(2, 2)$,and $D(0, 2)$,then $\square ABCD$ is a $\ldots \ldots \ldots$

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