What is the perimeter of a triangle with corners at #(1 ,5 )#, #(6 , 2 )#, and #(2 ,7 )#?

1 Answer

Perimeter #P=sqrt(41)+sqrt(5)+sqrt(34)#
Perimeter #P=14.47014411" "#units

Explanation:

let the points be #A(1, 5)#, #B(6, 2)#, #C(2, 7)#

compute lengths of sides a, b , c

#a=sqrt((x_B-x_C)^2+(y_B-y_C)^2)#
#a=sqrt((6-2)^2+(2-7)^2)#
#a=sqrt((4)^2+(-5)^2)#
#a=sqrt(16+25)#
#a=sqrt(41)#

#b=sqrt((x_A-x_C)^2+(y_A-y_C)^2)#
#b=sqrt((1-2)^2+(5-7)^2)#
#b=sqrt((-1)^2+(-2)^2)#
#b=sqrt((1+4)#
#b=sqrt(5)#

#c=sqrt((x_A-x_B)^2+(y_A-y_B)^2)#
#c=sqrt((1-6)^2+(5-2)^2)#
#c=sqrt((-5)^2+(3)^2)#
#c=sqrt(25+9)#
#c=sqrt(34)#

Perimeter #P=sqrt(41)+sqrt(5)+sqrt(34)#
Perimeter #P=14.47014411" "#units