If the vertices of square ABCD are A(1, 3), B | vedclass

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(C) To identify the quadrilateral,we calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.Side $A

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rectangle

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(C) To identify the quadrilateral,we calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.Side $AB = \sqrt{(4 - 1)^2 + (3 - 3)^2} = \sqrt{3^2 + 0^2} = 3$.Side $BC = \sqrt{(4 - 4)^2 + (5 - 3)^2} = \sqrt{0^2 + 2^2} = 2$.Side $CD = \sqrt{(1 - 4)^2 + (5 - 5)^2} = \sqrt{(-3)^2 + 0^2} = 3$.Side $DA = \sqrt{(1 - 1)^2 + (3 - 5)^2} = \sqrt{0^2 + (-2)^2} = 2$.Since opposite sides are equal ($AB = CD = 3$ and $BC = DA = 2$) and adjacent sides are not equal,it is a parallelogram.Now,check the diagonals: $AC = \sqrt{(4 - 1)^2 + (5 - 3)^2} = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$.$BD = \sqrt{(1 - 4)^2 + (5 - 3)^2} = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$.Since the opposite sides are equal and the diagonals are equal,$\square ABCD$ is a rectangle.

If the vertices of $\square ABCD$ are $A(1, 3), B(4, 3), C(4, 5)$ and $D(1, 5)$,then $\square ABCD$ is a $\ldots$

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