What is the approximate perimeter of a triangle with vertices (3, 1), (3, 8), and (6, 1)?

1 Answer
thelittlestleafeon
Mar 5, 2018

#P=10+sqrt(58)# or #P=17.62#.

Explanation:

I'm going to name each point. Point A will be (3,1), point B will be (3,8), and point C will be (6,1). We know that the perimeter of the triangle has to be the sum of the length of the three sides. If the three sides are AB, AC, and BC then the perimeter P is #P=AB+AC+BC# .
First we will find the length of AB. The equation for the length of AB (and any line) is #AB=sqrt((x_b-x_a)^2+(y_b-y_a)^2)#.

#x_b-x_a=3-3=0# and #y_b-y_a=8-1=7#, so #AB=sqrt((0)^2+(7)^2)=sqrt(0+49)=sqrt(49)=7#.

Now let's do the same process for sides AC and BC.
#AC=sqrt((x_c-x_a)^2+(y_c-y_a)^2)#.
#x_c-x_a=6-3=3# and #y_c-y_a=1-1=0#, so #AC=sqrt((3)^2+(0)^2)=sqrt(9+0)=sqrt(9)=3#.

#BC=sqrt((x_c-x_b)^2+(y_c-y_b)^2)# .
#x_c-x_b=6-3=3# and #y_c-y_b=1-8=-7#, so #BC=sqrt((3)^2+(-7)^2)=sqrt(9+49)=sqrt(58)#.

Therefore our perimeter is #P=7+3+sqrt(58)=10+sqrt(58)# or, in decimal form, #P=7+3+sqrt(58)=10+7.62=17.62#.

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