Whomever please

Find the perimeter of the triangle that has vertices at the points R(2,1), S (2,5), and T(4,
1). Round the answer to the nearest hundredth (2 decimal places).

A) 35.78
B) 32
C) 21.54
D)10.47

Respuesta :

The perimeter of the triangle is 10.47 units

The distance between two points is given by:

[tex]Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

Given the points R(2,1), S (2,5), and T(4,  1). Hence:

[tex]RS=\sqrt{(5-1)^2+(2-2)^2}=4\ units[/tex]

[tex]RT=\sqrt{(1-1)^2+(4-2)^2}=2\ units\\\\ST=\sqrt{(4-2)^2+(1-5)^2}=4.47\ units[/tex]

The perimeter of the triangle = 4 + 2 + 4.47 = 10.47 units

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