What is the perimeter of a triangle with corners at #(6 ,4 )#, #(9 ,2 )#, and #(5 ,7 )#?

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Answer 1 · 2017-01-08T22:30:59

#P=sqrt13+sqrt41+sqrt10#

Each side of the triangle can be measured by the distance formula:

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

Let's use labels for corners:

#A(6;4);B(9;2);C(5;7)#

Then

#d_(AB)=sqrt((9-6)^2+(2-4)^2)=sqrt13#

#d_(BC)=sqrt((5-9)^2+(7-2)^2)=sqrt41#

#d_(AC)=sqrt((5-6)^2+(7-4)^2)=sqrt10#

Then the perimeter of the given triangle is:

#P=sqrt13+sqrt41+sqrt10#