What is the perimeter of a triangle with corners at #(1 ,4 )#, #(6 ,7 )#, and #(4 ,2 )#?

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Answer 1 · 2016-09-17T17:01:36

Perimeter #=sqrt(34)+sqrt(29)+sqrt(13)=3.60555#

#A(1,4)# and #B(6,7)# and #C(4,2)# are the vertices of the triangle.

Compute for the length of the sides first.

Distance AB

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

#d_(AB)=sqrt((1-6)^2+(4-7)^2)#

#d_(AB)=sqrt((-5)^2+(-3)^2)#

#d_(AB)=sqrt(25+9)#

#d_(AB)=sqrt(34)#

Distance BC

#d_(BC)=sqrt((x_B-x_C)^2+(y_B-y_C)^2)#

#d_(BC)=sqrt((6-4)^2+(7-2)^2)#

#d_(BC)=sqrt((2)^2+(5)^2)#

#d_(BC)=sqrt(4+25)#

#d_(BC)=sqrt(29)#

Distance BC

#d_(AC)=sqrt((x_A-x_C)^2+(y_A-y_C)^2)#

#d_(AC)=sqrt((1-4)^2+(4-2)^2)#

#d_(AC)=sqrt((-3)^2+(2)^2)#

#d_(AC)=sqrt(9+4)#

#d_(AC)=sqrt(13)#

Perimeter #=sqrt(34)+sqrt(29)+sqrt(13)=3.60555#

God bless....I hope the explanation is useful.