The vertices of ∆XYZ are X(−3, −6), Y(21, −6), and Z(21, 4). What is the perimeter of the triangle?

Answer 1

X(−3, −6), Y(21, −6), and Z(21, 4)

Use the distance formula to find the length of each side:

distance = √[(x₂ - x₁)² + (y₂ - y₁)²]

XY = √[(21 - (-3))² + ((-6) - (-6))²] = √(24² + 0²) = √576 = 24

YZ = √[(21 - 21)² + ((4) - (-6))²] = √(0² + 10²) = √100 = 10

XZ = √[(21 - (-3))² + ((4) - (-6))²] = √(24² + 10²) = √676 = 26

perimeter = XY + YZ + XZ = 24 + 10 + 26 = 60

Answer 2

sides are 10, 24 and 26 which add to 60 for the perimeter

It's a right triangle in the ratio of 5, 12 and 13. Just twice those numbers.

25+144=169 = the square of the hypotenuse

the height of the right triangle is 4--6 = 10

the base = 21--3 = 24