By using the <em>perimeter</em> formula for an <em>orthogonally oriented</em> rectangle set on a <em>Cartesian</em> plane, we find that the perimeter of the figure is 68/3 units.
<h3>How to determine the perimeter of orthogonally oriented rectangle</h3>
In this question we have a rectangle oriented with respect to the two <em>orthogonal</em> axes of a <em>Cartesian</em> plane. In this case, the vertices of the figure are of the form:
A(x, y) = (a, b), B(x, y) = (c, b), C(x, y) = (a, d), D(x, y) = (c, d)
And the perimeter of this rectangle is equal to this:
p = 2 · |a - c| + 2 · |b - d|
If we know that a = - 5, b = 2, c = 2, d = - 7/3, then the perimeter of the rectangle is:
p = 2 · |- 5 - 2| + 2 · |2 - (- 7/3)|
p = 14 + 26/3
p = 68/3
By using the <em>perimeter</em> formula for an <em>orthogonally oriented</em> rectangle set on a <em>Cartesian</em> plane, we find that the perimeter of the figure is 68/3 units.
To learn more on perimeters: brainly.com/question/6465134
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