# What is the perimeter of a triangle with corners at (8 ,5 ), (9 ,9 ), and (3 ,4 )?

Perimeter of the triangle is $A B + B C + C A = 17.03 \left(2 \mathrm{dp}\right)$
Let $A \left(8 , 5\right) B \left(9 , 9\right) C \left(3 , 4\right)$are the vertices of the triangle. Sides AB=sqrt((9-5)^2+(9-8)^2) =sqrt 17=4.12 ; BC= sqrt((4-9)^2+(3-9)^2) =sqrt 61 =7.81 ; CA= sqrt((5-4)^2+(8-3)^2) =sqrt 26=5.1 :.Perimeter of the triangle is $A B + B C + C A = 17.03 \left(2 \mathrm{dp}\right)$[Ans]