What is the perimeter of a triangle with corners at #(1 ,9 )#, #(8, 2 )#, and #(9 ,4 )#?

1 Answer
Apr 8, 2016

Perimeter#=7sqrt2+sqrt5+sqrt89#, units

Explanation:

Distance between any two points #(x_1,y_1) and (x_2,y_2)#, is given by the formula
#d=sqrt((x_2-x_1)^2 + (y_2-y_1)^2)#

Using this formula distance #d_1# between two corners of triangle #(1,9) and (8,2)# is

#d_1=sqrt((8-1)^2 + (2-9)^2)#
or #d_1=sqrt((7)^2 + (-7)^2)#
or #d_1=sqrt(49 + 49)#
or #d_1=sqrt(2xx49)#
or #d_1=7sqrt2#

Similarly distance #d_2# between points #(8,2) and (9,4)# is
#d_2=sqrt((9-8)^2 + (4-2)^2)#
or #d_2=sqrt((1)^2 + (2)^2)#
or #d_2=sqrt(1 + 4)#
or #d_2=sqrt5#

And distance #d_3# between points #(9,4) and (1,9)# is
#d_3=sqrt((1-9)^2 + (9-4)^2)#
or #d_3=sqrt((-8)^2 + (5)^2)#
or #d_3=sqrt(64 +25)#
or #d_3=sqrt89#

Now perimeter of the triangle is sum of all three sides.
Perimeter#=d_1+d_2+d_3#
Perimeter#=7sqrt2+sqrt5+sqrt89#
or Perimeter#=7sqrt2+sqrt5+sqrt89#