348,170 views
21 votes
21 votes
Find the perimeter of the triangle whose vertices are the following specified points in the plane.

(0,−4),(−1,−2),(5,-3)

User Dana
by
2.5k points

1 Answer

26 votes
26 votes

Answer:

Perimeter = 5.67

Explanation:

AB = sqrt{-1 - (0)} + (-2 - (-4)}

Multiply -1 by 0

sqrt(-1 + 0 - 2 - (-4)

-1 by -4

sqrt(-1 + 0 - 2 + 4

Add -1 and 0

sqrt(-1 - 2 + 4)

Subtract 2 from -1

sqrt(-3 + 4)

Add

Sqrt(1) > 1

BC = sqrt{5 - (-1)} + {-3 - (-2)}

Multiply -1 by -1

sqrt(5 + 1 - 3 - (-2)

-1 by -2

sqrt(5 + 1 - 3 + 2

Add 5 and 1

sqrt(6 - 3 + 2)

Add 3 and 2

Sqrt(5) > 2.23

AC = sqrt{5 - 0} + {-3 - (-4)}

Multiply -1 by 0

sqrt(5 + 0 - 3 - (-4)

-1 by -4

sqrt(5 + 0 - 3 + 4)

Add 5 and 0

sqrt(5 - 3 + 4)

Subtract 3 from 5

sqrt(2 + 4)

Add

Sqrt(6) > 2.44

Perimeter = AB + BC + AC


Sqrt(1) + Sqrt(5) + Sqrt(6) = 1 + 2.23 + 2.44 = 5.67

User Kanishka Dilshan
by
2.7k points