Type the correct answer in each box. If necessary, round your answer(s) to the nearest hundredth. The vertices of ∆ABC are A(-2, 2), B(6, 2), and C(0, 8). The perimeter of ∆ABC is units, and its area is square units.

Answer:[tex]P=8+2\sqrt{10}+6\sqrt{2}\ units[/tex][tex]A=24\ un^2.[/tex]Step-by-step explanation:Plot the vertices of the triangle ABC on the coordinate plane and find the sides AB, BC and AC lengths:[tex]AB=\sqrt{(-2-6)^2+(2-2)^2}=\sqrt{(-8)^2+0^2}=\sqrt{64+0}=8\\ \\AC=\sqrt{(-2-0)^2+(2-8)^2}=\sqrt{(-2)^2+(-6)^2}=\sqrt{4+36}=2\sqrt{10}\\ \\BC=\sqrt{(6-0)^2+(2-8)^2}=\sqrt{6^2+(-6)^2}=\sqrt{36+36}=6\sqrt{2}[/tex]So, the perimeter of the triangle ABC is [tex]P=8+2\sqrt{10}+6\sqrt{2}\ units[/tex]To find the area of the triangle, use the formula[tex]A=\dfrac{1}{2}\cdot \text{Base}\cdot \text{Height}[/tex]In your case, AB is the base and the height is 6 units long (see attached diagram). Therefore,[tex]A=\dfrac{1}{2}\cdot 8\cdot 6=24\ un^2.[/tex]