# What is the perimeter of a triangle with corners at (1 ,2 ), (8 ,3 ), and (4 ,4 )?

Perimeter of triangle $= 14.8$ units
Side A = $\sqrt{{\left(8 - 1\right)}^{2} + {\left(3 - 2\right)}^{2}} = \sqrt{50} = 7.07$
Side B = $\sqrt{{\left(4 - 8\right)}^{2} + {\left(4 - 3\right)}^{2}} = \sqrt{17} = 4.12$
Side C = $\sqrt{{\left(1 - 4\right)}^{2} + {\left(2 - 4\right)}^{2}} = \sqrt{13} = 3.61$
Perimeter of triangle$= 7.07 + 4.12 + 3.61 = 14.8$ units [Ans]