Answer:
AB = √ ((0 - -4)2 + (0 - 2)2 ) = √(16 + 4) = √ 20 = 2√5
BC = √ ((0 - 5)2 + (0 - -5)2 ) = √(25 + 25) = √ 50 = 5√2
AC = √ ((-4 - 5)2 + (2 - -5)2 ) = √(81 + 49) = √ 130
2√5 + 5√2 + √ 130
Teacher Notes:
Students should be encouraged to solve this problem through the use of the Pythagorean Theorem.