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

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Answer 1 · 2017-07-30T10:30:57

Perimeter of the triangle is # 13.46 (2dp)# unit.

Let #A(5,2) , B(2,7),C(1 ,4)# are the three corners of the triangle.

Length # AB = sqrt((x_1-x_2)^2+(y_1-y_2)^2) = sqrt((5-2)^2+(2-7)^2)#

#=sqrt(9+25) = sqrt 34 ~~ 5.83(2dp)#

Length # BC = sqrt((x_2-x_3)^2+(y_2-y_3)^2) = sqrt((2-1)^2+(7-4)^2)#

#=sqrt(1+9) = sqrt 10 ~~ 3.16 (2dp)#

Length # CA = sqrt((x_3-x_1)^2+(y_3-y_1)^2) = sqrt((1-5)^2+(4-2)^2)#

#=sqrt(16+4) = sqrt 20 ~~ 4.47 (2dp)#

Perimeter of the triangle is #P= AB + BC + CA ~~ (5.83+3.16+4.47)# or

# P ~~ 13.46 (2dp)# unit. [Ans]