see the attached figure to better understand the problem
we know that the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
Let
[tex]A(-3,3)\ B(3,4))\ C(3,-3)[/tex]
Step 1
Find the distance AB
[tex]A(-3,3)\ B(3,4)[/tex]
substitute in the formula
[tex]d=\sqrt{(4-3)^{2}+(3+3)^{2}}[/tex]
[tex]d=\sqrt{(1)^{2}+(6)^{2}}[/tex]
[tex]dAB=\sqrt{37}\ units[/tex]
Step 2
Find the distance BC
[tex]B(3,4))\ C(3,-3)[/tex]
substitute in the formula
[tex]d=\sqrt{(-3-4)^{2}+(3-3)^{2}}[/tex]
[tex]d=\sqrt{(-7)^{2}+(0)^{2}}[/tex]
[tex]dBC=7\ units[/tex]
Step 3
Find the distance AC
[tex]A(-3,3)\ C(3,-3)[/tex]
substitute in the formula
[tex]d=\sqrt{(-3-3)^{2}+(3+3)^{2}}[/tex]
[tex]d=\sqrt{(-6)^{2}+(6)^{2}}[/tex]
[tex]dAC=\sqrt{72}\ units[/tex]
Step 4
Find the perimeter of the triangle
we know that
the perimeter of the triangle is the sum of the length sides of the triangle
[tex]P=AB+BC+AC[/tex]
substitute the values
[tex]P=\sqrt{37}\ units+7\ units+\sqrt{72}\ units=21.6\ units[/tex]
therefore
the answer is
the perimeter of the triangle is [tex]21.6\ units[/tex]
