What is the perimeter of a triangle with corners at #(1 ,5 )#, #(6 , 2 )#, and #(2 ,7 )#?

Question

What is the perimeter of a triangle with corners at #(1  ,5 )#, #(6  , 2  )#, and #(2  ,7 )#?

Noah Atkinson · Accepted Answer

Perimeter #P=sqrt(41)+sqrt(5)+sqrt(34)#
Perimeter #P=14.47014411" "#units

let the points be #A(1, 5)#, #B(6, 2)#, #C(2, 7)# compute lengths of sides a, b , c #a=sqrt((x_B-x_C)^2+(y_B-y_C)^2)# #a=sqrt((6-2)^2+(2-7)^2)# #a=sqrt((4)^2+(-5)^2)# #a=sqrt(16+25)# #a=sqrt(41)# #b=sqrt((x_A-x_C)^2+(y_A-y_C)^2)# #b=sqrt((1-2)^2+(5-7)^2)# #b=sqrt((-1)^2+(-2)^2)# #b=sqrt((1+4)# #b=sqrt(5)# #c=sqrt((x_A-x_B)^2+(y_A-y_B)^2)# #c=sqrt((1-6)^2+(5-2)^2)# #c=sqrt((-5)^2+(3)^2)# #c=sqrt(25+9)# #c=sqrt(34)# Perimeter #P=sqrt(41)+sqrt(5)+sqrt(34)# Perimeter #P=14.47014411" "#units

Sadie Allen · Answer

undefined To find the perimeter of the triangle with corners at (1, 5), (6, 2), and (2, 7), you need to calculate the sum of the lengths of its three sides.

Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, where $(x_1, y_1)$ and $(x_2, y_2)$ are the coordinates of two points:

For side 1: $(1, 5)$ to $(6, 2)$
$$d_1 = \sqrt{(6 - 1)^2 + (2 - 5)^2} = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}$$

For side 2: $(6, 2)$ to $(2, 7)$
$$d_2 = \sqrt{(2 - 6)^2 + (7 - 2)^2} = \sqrt{(-4)^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41}$$

For side 3: $(2, 7)$ to $(1, 5)$
$$d_3 = \sqrt{(1 - 2)^2 + (5 - 7)^2} = \sqrt{(-1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$$

The perimeter, $P$, is the sum of these distances:
$$P = d_1 + d_2 + d_3 = \sqrt{34} + \sqrt{41} + \sqrt{5}$$

Therefore, the perimeter of the triangle is $P = \sqrt{34} + \sqrt{41} + \sqrt{5}$.