What is the perimeter of a triangle with corners at #(3 ,4 )#, #(2 ,7 )#, and #(4 ,5 )#?

1 Answer
Jun 8, 2017

#P=sqrt10 + 3sqrt2#

Explanation:

The perimeter of an object is the sum of the lengths of its sides. To find the lengths of the sides in coordinates, we use a formula that is based off of Pythagoras's Theorem:
#sqrt((x""_1 - x""_2)^2+(y""_1 - y""_2)^2)#, where
#(x""_1,y""_1),(x""_2,y""_2)# are the coordinates. With this formula you will get the exact distance from point A to point B. Since we have three coordinates, we need the exact distance from Point A-B, A-C and B-C. Our formula will now look like:

#sqrt((x""_1 - x""_2)^2+(y""_1 - y""_2)^2) + sqrt((x""_1 - x""_3)^2+(y""_1 - y""_3)^2) + sqrt((x""_2 - x""_3)^2+(y""_2 - y""_3)^2)# Where #(x""_1,y""_1),(x""_2,y""_2), (x""_3,y""_3)# are the coordinates of
#(3,4),(2,7),(4,5)#. Now we substitute the numbers into the formula, and solve.

#sqrt((3 - 2)^2+(4 -7)^2) + sqrt((3 - 4)^2+(4 - 5)^2) + sqrt((2 - 4)^2+(7 - 5)^2)#
#=sqrt((1)^2+(-3)^2) + sqrt((-1)^2+(-1)^2) + sqrt((-2)^2+(2)^2)#
#=sqrt(1+9) + sqrt(1+1) + sqrt(4+4)#
#=sqrt10 + sqrt2 + sqrt8#.

The last part involves the simplification and addition of square roots (#sqrt#), and if you want further information on how to do that then ask the question and I, or other, will answer. As it is I will just speed through the explanations. Thus,

#=sqrt10 + sqrt2 + sqrt8#

#=sqrt10+sqrt2+2sqrt2#

#=sqrt10+3sqrt2#

There you have the perimeter of a triangle with coordinates (3,4),(2,7) and (4,5).

I hope I helped!