# What is the perimeter of a triangle with corners at (7 ,5 ), (1 ,6 ), and (4 ,8 )?

Perimeter of the triangle is $13.93 \left(2 \mathrm{dp}\right)$ unit
Side $A = \sqrt{{\left(7 - 1\right)}^{2} + {\left(5 - 6\right)}^{2}} = \sqrt{36 + 1} = \sqrt{37} = 6.08$
Side $B = \sqrt{{\left(1 - 4\right)}^{2} + {\left(6 - 8\right)}^{2}} = \sqrt{9 + 4} = \sqrt{13} = 3.61$
Side $C = \sqrt{{\left(4 - 7\right)}^{2} + {\left(8 - 5\right)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 4.24$
Perimeter of the triangle is $P = \left(6.08 + 3.61 + 4.24\right) = 13.93 \left(2 \mathrm{dp}\right)$unit [Ans]