What is the perimeter of a triangle with corners at #(3 ,6 )#, #(1 ,8 )#, and #(8 ,7 )#?

1 Answer
Apr 26, 2018

#7sqrt(2)+sqrt(26)#

Explanation:

#sqrt((x_2-x_1)^2+(y_2-y_1)^2)# is the distance formula. If you simply plug in all the points one by one by replace #x_1# and #x_2# as well as #y_1# and #y_2# with three pairs of points you can solve each single one.

So the first pair between (3,6) and (1,8) is #sqrt((1-3)^2+(8-6)^2)# is #sqrt(8)#.

Second is the pair of points (1,8) and (8,7). Plug this in once again for #sqrt((8-1)^2+(7-8)^2)# and simply to #sqrt(50)#.

Finally for the third pair of points, (8,7) and (3,6) you plug it in for #sqrt((3-8)^2+(6-7)^2)#. Simply to #sqrt(26)#.

Now simply the radicals as much as you can. #sqrt(8)# can go to #2sqrt(2)#, #sqrt(50)# can simplify to #5sqrt(2)#, and #sqrt(26)# cannot be simplified.

Now you just add like terms and thats it so in the end you get #2sqrt(2)+5sqrt(2)+sqrt(26)#. The answer is #7sqrt(2)+sqrt(26)#