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To the nearest tenth, find the perimeter of triangle ABC with vertices A(3, 2), B(-2, 3), and C(2, 6)

Respuesta :

Answer:

14.2 units

Step-by-step explanation:

To find the distance between the points, you use the formula

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

 Therefore, we need to get the distances between points A and B, between points B and C and between points A and C.

Length [tex]AB=\sqrt {(3--2)^{2}+(2-3)^{2}}=\sqrt 26\\BC=\sqrt{(2--2)^{2}+(6-3)^{2}}=5\\AC=\sqrt {(2-3)^{2}+(6-2)^{2}}=\sqrt 17\\[/tex]

Perimeter=[tex]\sqrt 26+5+\sqrt 17[/tex]=14.2 units (to the nearest tenth)