Problem

Using the inclusive definition, so a square also qualifies as a rhombus, verify that the ordered vertices \(W(0,3)\), \(X(3,0)\), \(Y(6,3)\), \(Z(3,6)\) form rhombus \(WXYZ\). Give \(\text{all}~\text{four}~\text{squared}~\text{side}~\text{lengths}\), compare them, and state the classification.

Source coordinate plane: Quadrilateral WXYZ has vertices W(0,3), X(3,0), Y(6,3), and Z(3,6); no side lengths or rhombus conclusion are shown.

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squared side lengths = WX²=18, XY²=18, YZ²=18, ZW²=18; equality = all four are equal and positive; conclusion = WXYZ is a rhombus (indeed, it is also a square)

squared side lengths = WX²=18, XY²=18, YZ²=18, ZW²=18; equality = all four equal; conclusion = WXYZ is not a rhombus because it is also a square

squared side lengths = WX²=18, XY²=18, YZ²=9, ZW²=9; equality = not all four are equal; conclusion = WXYZ is not a rhombus

squared side lengths = WX²=18, XY²=18, YZ²=18, ZW²=18; equality = all four equal; conclusion = WXYZ is not a rhombus because only diagonal lengths determine a rhombus

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Using the inclusive definition, so a square also qualifies as a rhombus, verify that the ordered vertices \(W(0,3)\), \(X(3,0)\), \(Y(6,3)\), \(Z(3,6)\) form rhombus \(WXYZ\). Give \(\text{all}~\text{four}~\text{squared}~\text{side}~\text{lengths}\), compare them, and state the classification.

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