Answer:
Option C is correct.
Step-by-step explanation:
Given: Δ ABC on coordinate plane.
From Coordinate plane, Coordinates are A( 1 , 1 ) , B( 3 , 5 ) and C( 5 , 1 )
We need to find Approximate perimeter of the triangle.
we use the distance formula of two point [tex](x_1,y_1)\:and\:(x_2,y_2)[/tex]
[tex]Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
So,
[tex]AB=\sqrt{(3-1)^2+(5-1)^2}=\sqrt{(2)^2+(4)^2}=\sqrt{4+16}=\sqrt{20}=4.47\:(approx.)[/tex]
[tex]CB=\sqrt{(5-3)^2+(1-5)^2}=\sqrt{(2)^2+(-4)^2}=\sqrt{4+16}=\sqrt{20}=4.47\:(approx.)[/tex]
[tex]AC=\sqrt{(5-1)^2+(1-1)^2}=\sqrt{(4)^2+(0)^2}=\sqrt{16}=4[/tex]
Thus, Perimeter = AB + AC + CB = 4.47 + 4.47 + 4 = 12.94 units
Therefore, Option C is correct.
