What is the perimeter of a triangle with the vertices located at (1,3), (2,6), and (0,4), rounded to the nearest hundredth?
A. 7.40 units
B. 9.07 units
C. 8.49 units
D. 7.07 units

Question

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Respuesta :

carlosego carlosego · Answer 1 · 2019-04-08T21:55:25+02:00

Answer: OPTION A

Step-by-step explanation:

You can plot the points given in the problem, as you can see in the figure attached.

Apply the formula for calculate the distance between two points, which is:

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

Distance AB:

[tex]d_{AB}=\sqrt{(2-1)^2+(6-3)^2}=3.162units[/tex]

Distance AC:

[tex]d_{AC}=\sqrt{(0-1)^2+(4-3)^2}=1.414units[/tex]

Distance BC:

[tex]d_{BC}=\sqrt{(2-0)^2+(6-4)^2}=2.828units[/tex]

The perimeter is:

[tex]P=3.162units+1.414units+2.828units\\P=7.40units[/tex]

alinakincsem alinakincsem · Answer 2 · 2019-04-08T22:01:39+02:00

Answer:

The correct answer option is A. 7.40 units.

Step-by-step explanation:

We are given the following coordinates of the vertices of a triangle and we are to find its perimeter:

(1,3), (2,6), and (0,4)

Finding the lengths of all three sides:

AB = [tex] \sqrt { ( 1 - 2 ) ^ 2 + ( 3 - 6 ) ^ 2 } = \sqrt{10} =3.16[/tex]

BC = [tex] \sqrt { ( 2 - 0 ) ^ 2 + ( 6 - 4 ) ^ 2 } = \sqrt{8} =2.83[/tex]

CA = [tex] \sqrt { ( 0 - 1 ) ^ 2 + ( 4 - 3 ) ^ 2 } = \sqrt{2} =1.41[/tex]

Perimeter of the triangle = [tex]3.16+2.83+1.41[/tex] = 7.40 units