Question

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

Answer 1

$XY= \sqrt{(21-(-3))^2+(-6-(-6))^2} = \sqrt{(21+3)^2+(12-12)^2} = \\ = \sqrt{24^2+0} = \sqrt{24^2} =24 \\ YZ= \sqrt{(21-21)^2+(4-(-6))^2} = \sqrt{0^2+(4+6)^2} = \sqrt{10^2} =10 \\ XZ= \sqrt{(21-(-3))^2+(4-(-6))^2} = \sqrt{(21+3)^2+(4+6)^2} = \\ =\sqrt{24^2+10^2} = \sqrt{576+100} = \sqrt{676} =26 \\ \\ P=XY+YZ+XZ=24+26+10=60$