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What is the perimeter of a triangle with corners at #(1 ,4 )#, #(6 ,3 )#, and #(4 ,2 )#?

Answer 1

Caroline Aucoin

Perimeter = 10.995

Let a, b, c be the sides. Sum of the three sides will give the perimeter.

#a=sqrt((4-6)^2+(2-3)^2)=sqrt5=2.236# #b=sqrt((4-1)^2+(2-4)^2)=sqrt13=3.606# #c=sqrt((1-6)^2+(4-3)^2)=sqrt26=5.099#

Perimeter #=a+b+c=2.236+3.606+5.099= #10.995

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Answer 2

Sadie Allen

To find the perimeter of the triangle, we need to calculate the distance between each pair of points and then sum these distances. Using the distance formula (d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}), we can compute the distances between the given points:

Between (1, 4) and (6, 3): (d_1 = \sqrt{(6 - 1)^2 + (3 - 4)^2} = \sqrt{5^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26})
Between (1, 4) and (4, 2): (d_2 = \sqrt{(4 - 1)^2 + (2 - 4)^2} = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13})
Between (6, 3) and (4, 2): (d_3 = \sqrt{(4 - 6)^2 + (2 - 3)^2} = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5})

Now, we sum these distances to find the perimeter:

Perimeter = (d_1 + d_2 + d_3 = \sqrt{26} + \sqrt{13} + \sqrt{5})

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