What is the perimeter of a triangle with corners at #(7 ,6 )#, #(4 ,5 )#, and #(3 ,1 )#?

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Answer 1 · 2017-08-10T08:43:33

#Perimeter=sqrt10+sqrt17+sqrt41#

Equation for finding distance between #2# coordinates #=# #sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

Let,
#(7,6)=#point #A#

#(4,5)=#point #B#

#(3,1)=#point #C#

Distance between #AB#

#x_1=7,x_2=4,y_1=6,y_2=5#

#AB=sqrt((4-7)^2+(5-6)^2)#

#=>sqrt((-3)^2+(-1)^2#

#=>sqrt(9+1)#

#=>sqrt10#

#AB=sqrt10#

Distance between #BC#

#x_1=4,x_2=3,y_1=5,y_2=1#

#BC=sqrt((3-4)^2+(1-5)^2)#

#=>sqrt((-1)^2+(-4)^2#

#=>sqrt(1+16)#

#sqrt17#

#BC=sqrt17#

Distance between #AC#

#x_1=7,x_2=3,y_1=6,y_2=1#

#AC=sqrt((3-7)^2+(1-6)^2)#

#=>sqrt((-4)^2+(-5)^2#

#=>sqrt(16+25)#

#=>sqrt41#

#AC=sqrt41#

Perimeter of a triangle is the sum of its #3# sides, that #AB+BC+AC.#

#Perimeter=sqrt10+sqrt17+sqrt41#