Answer:
And we can find the distance between JK like this:
[tex] d_{JK} =\sqrt{(3-3)^2 +(2+1)^2}= 3[/tex]
And we can do something similar for KL and we got:
[tex] d_{KL} =\sqrt{(-5-3)^2 +(2-2)^2}= 8[/tex]
And we can do something similar for LM and we got:
[tex] d_{LM} =\sqrt{(-5+5)^2 +(-1-2)^2}= 3[/tex]
And we can do something similar for MJ and we got:
[tex] d_{MJ} =\sqrt{(-5-3)^2 +(-1+1)^2}= 8[/tex]
And the perimeter would be:
[tex] P = d_{JK} + d_{KL} +d_{LM} +d_{MJ} = 3+8+3+8 =22[/tex]
Step-by-step explanation:
For this case we can use the formula for the euclidean distance given by:
[tex] d = \sqrt{(y_2 -y_1)^2 +(x_2 -x_1)^2}[/tex]
And we can find the distance between JK like this:
[tex] d_{JK} =\sqrt{(3-3)^2 +(2+1)^2}= 3[/tex]
And we can do something similar for KL and we got:
[tex] d_{KL} =\sqrt{(-5-3)^2 +(2-2)^2}= 8[/tex]
And we can do something similar for LM and we got:
[tex] d_{LM} =\sqrt{(-5+5)^2 +(-1-2)^2}= 3[/tex]
And we can do something similar for MJ and we got:
[tex] d_{MJ} =\sqrt{(-5-3)^2 +(-1+1)^2}= 8[/tex]
And the perimeter would be:
[tex] P = d_{JK} + d_{KL} +d_{LM} +d_{MJ} = 3+8+3+8 =22[/tex]
The perimeter of the rectangle with the given vertices is: 22 units.
Recall:
Thus, find the length (JK) and width (KL) of the rectangle.
Width of rectangle = distance between J(-1,3) and K(2,3):
[tex]JK = \sqrt{(3 - 3)^2 + (2 - (-1))^2} \\\\JK = \sqrt{(0)^2 + (3)^2} \\\\JK = \sqrt{9} \\\\JK = 3[/tex]
Length of rectangle = distance between K(2,3) and L(2,-5)
[tex]KL = \sqrt{(-5 - 3)^2 + (2 - 2)^2} \\\\KL = \sqrt{(-8)^2 + (0)^2} \\\\KL = \sqrt{64} \\\\KL = 8[/tex]
Perimeter of the rectangle = 2(length x width)
Perimeter of the rectangle = 2(8 + 3)
Perimeter = 2(11)
Perimeter = 22 units
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