What is the perimeter of a triangle with corners at #(1 ,4 )#, #(3 , 2 )#, and #(9 ,7 )#?

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Answer 1 · 2016-08-30T16:47:21

The perimeter of #Delta=2sqrt2+sqrt61+sqrt73~~2.828+7.810+8.544=19.182#.

Let the given corners be #A(1,4), B(3,2), and, C(9,7)#.

Then, using The Distance Formula,

#AB=sqrt{(1-3)^2+(4-2)^2}=sqrt8=2sqrt2#.

#BC=sqrt{(3-9)^2+(2-7)^2}=sqrt61#.

#CA=sqrt{(9-1)^2+(7-4)^2}=sqrt73#.

Hence, the perimeter of #Delta ABC#

#=AB+BC+CA=2sqrt2+sqrt61+sqrt73#.

#~~2.828+7.810+8.544=19.182#.