What is the perimeter of a triangle with corners at #(2 ,5 )#, #(9 ,2 )#, and #(3 ,1 )#?

Question

What is the perimeter of a triangle with corners at #(2  ,5  )#, #(9 ,2  )#, and #(3 ,1  )#?

Sophia Alfred · Accepted Answer

About #17.82#.

The perimeter is the sum of the lengths of the sides, so, if we call #A = (2,5)# #B = (9,2)# #C = (3,1)# we have #2p = AB+BC+AC#, where #2p# is the perimeter. The formula to find the length of a line, knowing its extreme points' coordinates, is #PQ = sqrt( (x_P-x_Q)^2 + (y_P-y_Q)^2)# Applying this formula to all points, we have #AB = sqrt((-7)^2+3^3)=sqrt(58)=7.62# #BC = sqrt(6^2+1^2)=sqrt(37)=6.08# #AC = sqrt( (-1)^2+4^2)=sqrt(17)=4.12# Thus, #AB+BC+AC=7.62+6.08+4.12=17.82# Note that all the values of the square roots are approximated!

Allison Allen · Answer

undefined To find the perimeter of a triangle with corners at the given coordinates, we need to calculate the distance between each pair of points and then sum these distances.

Using the distance formula: 
Distance = √((x2 - x1)^2 + (y2 - y1)^2)

Distance between (2, 5) and (9, 2):
√((9 - 2)^2 + (2 - 5)^2) = √(7^2 + (-3)^2) = √(49 + 9) = √58

Distance between (9, 2) and (3, 1):
√((3 - 9)^2 + (1 - 2)^2) = √((-6)^2 + (-1)^2) = √(36 + 1) = √37

Distance between (3, 1) and (2, 5):
√((2 - 3)^2 + (5 - 1)^2) = √((-1)^2 + 4^2) = √(1 + 16) = √17

Perimeter = √58 + √37 + √17 ≈ 7.62 + 6.08 + 4.12 ≈ 17.82