The perimeter of the bean bag toss board is 12 units
The points are given as:
C (3,2) D (3,6) E (5,2) F (5,6)
Calculate the distance between adjacent points using:
[tex]d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/tex]
So, we have:
[tex]CD = \sqrt{(3 - 3)^2 + (2 - 6)^2} = 4[/tex]
[tex]DF = \sqrt{(3 - 5)^2 + (6 - 6)^2} = 2[/tex]
[tex]FE = \sqrt{(5 - 5)^2 + (2 - 6)^2} = 4[/tex]
[tex]CE = \sqrt{(3 - 5)^2 + (2 - 2)^2} = 2[/tex]
The perimeter is then calculated as:
P = CD + DF + FE + CE
This gives
P = 4 + 2 + 4 + 2
Evaluate
P =12
Hence, the perimeter of the bean bag toss board is 12 units
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