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Rectangle [tex]\( WXYZ \)[/tex] has consecutive vertices [tex]\( W(-9, -3), X(-9, 5), Y(-2, 5) \)[/tex], and [tex]\( Z(-2, -3) \)[/tex].

1. Find the perimeter of rectangle [tex]\( WXYZ \)[/tex].
[tex]\(\square\)[/tex] units

2. Find the area of rectangle [tex]\( WXYZ \)[/tex].
[tex]\(\square\)[/tex] square units

Answer :

First, notice that the vertices of rectangle [tex]$WXYZ$[/tex] are given by
[tex]$$
W(-9,-3),\quad X(-9,5),\quad Y(-2,5),\quad Z(-2,-3).
$$[/tex]

Since the vertices are consecutive, we can determine the lengths of two adjacent sides.

1. The side [tex]$WX$[/tex] is vertical (because the [tex]$x$[/tex]-coordinates are the same). Its length is the difference between the [tex]$y$[/tex]-coordinates:
[tex]$$
WX = 5 - (-3) = 8.
$$[/tex]

2. The side [tex]$XY$[/tex] is horizontal (because the [tex]$y$[/tex]-coordinates are the same). Its length is the difference between the [tex]$x$[/tex]-coordinates:
[tex]$$
XY = -2 - (-9) = 7.
$$[/tex]

Now, the perimeter [tex]$P$[/tex] of a rectangle is given by
[tex]$$
P = 2 \times (\text{length} + \text{width}),
$$[/tex]
so we have
[tex]$$
P = 2 \times (8 + 7) = 2 \times 15 = 30\text{ units}.
$$[/tex]

Next, the area [tex]$A$[/tex] of the rectangle is
[tex]$$
A = \text{length} \times \text{width} = 8 \times 7 = 56\text{ square units}.
$$[/tex]

Thus, the perimeter of rectangle [tex]$WXYZ$[/tex] is [tex]$30$[/tex] units and the area is [tex]$56$[/tex] square units.

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