Question

Find the perimeter of triangle ABC. Round your answer to two decimal places.

Answer 1

Check the picture below.

`~\hfill \stackrel{\textit{\large distance between 2 points}}{d = √(( x_2- x_1)^2 + ( y_2- y_1)^2)}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{-7}~,~\stackrel{y_1}{2})\qquad B(\stackrel{x_2}{3}~,~\stackrel{y_2}{-8}) ~\hfill AB=√((~~ 3- (-7)~~)^2 + (~~ -8- 2~~)^2) \\\\\\ ~\hfill AB=√(( 10)^2 + ( -10)^2) \implies \boxed{AB=√( 200 )}`

`B(\stackrel{x_1}{3}~,~\stackrel{y_1}{-8})\qquad C(\stackrel{x_2}{-4}~,~\stackrel{y_2}{-9}) ~\hfill BC=√((~~ -4- 3~~)^2 + (~~ -9- (-8)~~)^2) \\\\\\ ~\hfill BC=√(( -7)^2 + ( -1)^2) \implies \boxed{BC=√( 50 )} \\\\\\ C(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-9})\qquad A(\stackrel{x_2}{-7}~,~\stackrel{y_2}{2}) ~\hfill CA=√((~~ -7- (-4)~~)^2 + (~~ 2- (-9)~~)^2) \\\\\\ ~\hfill CA=√(( -3)^2 + (11)^2) \implies \boxed{CA=√( 130 )} \\\\[-0.35em] ~\dotfill`

`\stackrel{\textit{\LARGE perimeter}}{√(200)~~ + ~~√(50)~~ + ~~√(130)} ~~ \approx ~~ \text{\LARGE 32.61}`