what is the perimeter of a triangle with vertices (0,0),(2,4),and (5,1)

Question

Thank you i really need help to solve this for my study guide i dont want to get it wrong on my exam

 

Answer 1

Triangle ABC with the vertices 

A = (0,0), B = (2,4), C = (5,1)

The sides will be:

AB, BC, and AC

Calculate the total length of the sides

 

AB =(2-0,4-0)  = (2,4)  Yielding into a length of (2² +4²)^1/2   =4.47 

BC =(5-2, 1-4) =(3,-3)                     length    (3² + -3²)^1/2  =4.24

AC = (5-0,1-0) =(5,1)                      length    (5² +1²)^1/2    =5.1

 

Perimeter is 4.47 + 4.24 + 5.1 = 13.81

Answer 2

The perimeter will be the sum of the distances between each point in the triangle

 

Point A = (0, 0)

Point B = (2, 4)

Point C = (5, 1)

 

The distance between two points A=(x1, y1) & B=(x2, y2) is equal to

 

√( (x2 - x1)² + (y2 - y1)²  )

 

 

So, the distance between point A & B is

 

d1 = √( (2 - 0)² + (4 - 0)² )

 

     = √( (2)2 + (4)2 ) = √( 4 + 16 ) = √(20)

 

Distance Between A & C

 

d2 = √( (5 - 0)2 + (1 - 0)2 )

 

    = √( (5)2 + (1)2 ) = √( 25 + 1) = √( 26 ) 

 

Distance Between B & C 

 

d3 = √( (5 - 2)2 + (1 - 4)2 )

 

     = √( (3)2 + (-3)2 ) = √( 9 + 9 ) = √( 18 )

 

 

Now, you just have to add the distances to find the perimeter (P)

 

P = d1 + d2 + d3

 

P = √( 20 ) + √( 26 ) + √( 18 )

 

 

Plug the numbers into the calculator to solve.