What is the perimeter of a triangle with corners at #(7 ,2 )#, #(8 ,3 )#, and #(4 ,4 )#?

Question

What is the perimeter of a triangle with corners at #(7 ,2  )#, #(8 ,3  )#, and #(4  ,4   )#?

Wyatt Anderson · Accepted Answer

#9.14# #"units"# To find the perimeter of a triangle, add up the lengths of all of its sides. Use the distance formula, #d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#. Find the distance between #(7,2)# and #(8,3)#: #d=sqrt((8-7)^2+(3-2)^2) = sqrt(1^2+1^2)=sqrt2# Find the distance between #(8,3)# and #(4,4)#: #d=sqrt((4-8)^2+(4-3)^2) = sqrt((-4)^2+1^2)=sqrt17# Find the distance between #(7,2)# and #(4,4)#: #d=sqrt((4-7)^2+(4-2)^2) = sqrt((-3)^2+2^2)=sqrt13# The perimeter is #sqrt2+sqrt17+sqrt13#, or about #9.14# #"units"#.

Allison Allen · Answer

undefined To find the perimeter of a triangle with corners at (7, 2), (8, 3), and (4, 4), you need to calculate the lengths of the three sides of the triangle using the distance formula, and then sum them up.

The distance formula between two points (x1, y1) and (x2, y2) is given by:

$$ d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2} $$

Using this formula for each pair of points, we get:

1. Between (7, 2) and (8, 3):
$$ d_1 = \sqrt{(8 - 7)^2 + (3 - 2)^2} = \sqrt{1^2 + 1^2} = \sqrt{2} $$

2. Between (8, 3) and (4, 4):
$$ d_2 = \sqrt{(4 - 8)^2 + (4 - 3)^2} = \sqrt{(-4)^2 + 1^2} = \sqrt{17} $$

3. Between (4, 4) and (7, 2):
$$ d_3 = \sqrt{(7 - 4)^2 + (2 - 4)^2} = \sqrt{3^2 + (-2)^2} = \sqrt{13} $$

Now, summing up the lengths of the three sides:

$$ Perimeter = d_1 + d_2 + d_3 = \sqrt{2} + \sqrt{17} + \sqrt{13} $$

So, the perimeter of the triangle is $ \sqrt{2} + \sqrt{17} + \sqrt{13} $ units.