What is the perimeter of a triangle with corners at #(5 ,2 )#, #(9 ,7 )#, and #(1 ,4 )#?

1 Answer
Ben
May 15, 2018

The perimeter is 19.419, or #(sqrt(41)+sqrt(73)+2sqrt(5))#

Explanation:

Let's call each of the corners/vertices #A#, #B#, and #C#.

#A=(5,2)#
#B=(9,7)#
#C=(1,4)#

The Perimeter #P#, can be described as:

#P=bar(AB)+bar(AC)+bar(BC)#

Where the value with the bar over it is the length of the triangle's side.

We can calculate the individual lengths by treating the #x# and #y# displacements as right triangles, and using The Pythagorean Theorem to calculate the length:

If #c=sqrt(a^2+b^2)#, and:

#a=Deltax=(x_2-x_1)#
#b=Deltay=(y_2-y_1)#
#c="side length"#

We can re-write a general equation for any length as:

#c=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

Now, let's do the math for one leg, #bar(AB)#:

#bar(AB)=sqrt((x_B-x_A)^2+(y_B-y_A)^2)#

#bar(AB)=sqrt((9-5)^2+(7-2)^2)=sqrt(4^2+5^2)#

#bar(AB)=sqrt(16+25)#

#bar(AB)=sqrt(41)=6.403#

Repeating the process for #bar(BC)# and #bar(AC)#:

#bar(BC)=sqrt(73)=8.544#

#bar(AC)=sqrt(20)=2sqrt(5)=4.472#

Now we can calculate the perimeter, #P#:

#P=sqrt(41)+sqrt(73)+2sqrt(5)#

#P=6.403+8.544+4.472#

#color(green)(P=19.419#

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