how do you find the perimeter of a parallelogram
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Answer: D
Step-by-step explanation:To find the perimeter of this parallelogram, you would have to find the distance between the coordinates.
I would use the distance formula for this question.
Distance Formula - [tex]d=\sqrt{(x_2 -x_1)^{2}+(y_2 -y_1)^{2}}[/tex]
Ok, let's find the length of side AB.
Coordinates of A = (-5, -7) This will be our second x and y coordinates [[tex]x_2,y2[/tex]]
Coordinates of B = (2, -3) These will be our first x and y coordinates [[tex]x_1,y1[/tex]]
Now Substitute Into The Equation -
[tex]d=\sqrt{(-5-2)^{2} +(-7-(-3))^{2} }[/tex]
Solve/Simplify -
[tex]\sqrt{(-5-2)^2+(-7-(-3))^2} \\ \\ \sqrt{(-5-2)^2+(-7+3)^2} \\\\\sqrt{(-7)^2+(-4)^2}\\\\ \sqrt{49+16} \\\\ \sqrt{65}[/tex]
Now let's solve for side CB
Point C = (-1, 2) This will be our second x and y coordinates [[tex]x_2,y2[/tex]]
Point B = (2, -3) These will be our first x and y coordinates [[tex]x_1,y1[/tex]]
Substitute and Solve -
[tex]\sqrt{(-1-2)^2+(2-(-3))^2}\\\\sqrt{(-1-2)^2+(2+3)^2}\\\\\ \sqrt{(-3)^2+(5)^2}\\\\ \sqrt{9+25}\\\\\sqrt{34}[/tex]
Now solve for the other two sides -
Because this shape is a rectangle, the parallel lines are the same length.
Now, find the perimeter-
[tex]\sqrt{65}+ \sqrt{65}+ \sqrt{34} +\sqrt{34}[/tex]
Hoped this helped!~