The vertices of a triangle are located at P(0, 6), Q(8, 12), and R(3, -3). How do you find the perimeter of this triangle to the nearest tenth?

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Answer 1 · 2017-09-08T18:19:21

#color(blue)(35.3#

You need to find the lengths of #vec(RP) , vec(PQ) , vec(QR).#

Using distance formula: #sqrt((x_2 - x_1) + ( y_2 - y_1 ))#

Length of sides:

#vec(PR)# #sqrt((0 - 3)^2 + ( 6 - (-3))^2) = sqrt(90) #

#vec(PQ)# #sqrt((0 - 8)^2 + ( 6 - 12)^2) = sqrt(100) #

#vec(QR)# #sqrt((3 - 8)^2 + ( -3 - 12)^2) = sqrt(250) #

Perimeter is the sum of lengths of the sides.

#sqrt(90) + sqrt(100) + sqrt(250) = color(blue)(35.3)# to nearest tenth.