Question

Triangle ABC has vertices at (5, 0), (- 1, - 4) , and (3, - 6) . What is the perimeter of ABC , rounded to two decimal places ?

Answer 1

18.01 units to 2 dec. places.

Explanation:

Calculate the length of the 3 lines of the triangle

AB = √ [(5 --1)^2 + ( 0 - -4)^2]

= √(36 + 16)

= √52

= 7.211.

BC = √ [(-1-3)^2 + (-4- -6)^2]

= √(16 + 4)

= √20

= 4.472.

CA = √ [(3-5)^2 + (-6-0)^2]

= √(4 + 36)

= √40

= 6.325.

So the perimeter = 7.211 + 4.472 + 6.325

= 18.008