Trapezoid [tex]$WXYZ$[/tex] has vertices [tex]$W(-1,2)$[/tex], [tex]$X(2,2)$[/tex], [tex]$Y(3,-1)$[/tex], and [tex]$Z(-3,-1)$[/tex].

Is the trapezoid an isosceles trapezoid?

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Trapezoid [tex]$WXYZ$[/tex] [tex]$\square$[/tex] Choose an isosceles trapezoid.
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Answer :

To determine whether trapezoid [tex]\( WXYZ \)[/tex] is an isosceles trapezoid, we need to verify if the non-parallel sides (legs) are of equal length. Let's proceed step by step:

First, we calculate the distances between the vertices using the distance formula:
[tex]\[ \text{distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]

### Calculate the lengths of the sides

Length of [tex]\( WX \)[/tex]:
[tex]\[ W(-1,2) \text{ and } X(2,2) \][/tex]
[tex]\[ \text{distance} = \sqrt{(2 - (-1))^2 + (2 - 2)^2} \][/tex]
[tex]\[ = \sqrt{(3)^2 + (0)^2} \][/tex]
[tex]\[ = \sqrt{9} \][/tex]
[tex]\[ = 3 \][/tex]

Length of [tex]\( XY \)[/tex]:
[tex]\[ X(2,2) \text{ and } Y(3,-1) \][/tex]
[tex]\[ \text{distance} = \sqrt{(3 - 2)^2 + (-1 - 2)^2} \][/tex]
[tex]\[ = \sqrt{(1)^2 + (-3)^2} \][/tex]
[tex]\[ = \sqrt{1 + 9} \][/tex]
[tex]\[ = \sqrt{10} \][/tex]

Length of [tex]\( YZ \)[/tex]:
[tex]\[ Y(3,-1) \text{ and } Z(-3,-1) \][/tex]
[tex]\[ \text{distance} = \sqrt{(-3 - 3)^2 + (-1 - (-1))^2} \][/tex]
[tex]\[ = \sqrt{(-6)^2 + (0)^2} \][/tex]
[tex]\[ = \sqrt{36} \][/tex]
[tex]\[ = 6 \][/tex]

Length of [tex]\( ZW \)[/tex]:
[tex]\[ Z(-3,-1) \text{ and } W(-1,2) \][/tex]
[tex]\[ \text{distance} = \sqrt{(-1 - (-3))^2 + (2 - (-1))^2} \][/tex]
[tex]\[ = \sqrt{(2)^2 + (3)^2} \][/tex]
[tex]\[ = \sqrt{4 + 9} \][/tex]
[tex]\[ = \sqrt{13} \][/tex]

### Compare the lengths of the legs

For trapezoid [tex]\( WXYZ \)[/tex], the non-parallel sides (legs) are [tex]\( XY \)[/tex] and [tex]\( ZW \)[/tex].

Lengths of the legs:
[tex]\[ XY = \sqrt{10} \][/tex]
[tex]\[ ZW = \sqrt{13} \][/tex]

Since [tex]\(\sqrt{10} \neq \sqrt{13}\)[/tex], the legs of the trapezoid [tex]\( WXYZ \)[/tex] are not equal in length.

Therefore, trapezoid [tex]\( WXYZ \)[/tex] is not an isosceles trapezoid.