Part A.
The figure has three vertices and therefore it must be a triangle.
Part B.
Let us first find the lengths of the sides of this triangle in order to find its perimeter.
The length from G(-4, 1) and H(0, -2) is
[tex]\sqrt[]{(-4-0)^2+(1-(-2))^2}=\sqrt[]{4^2+3^2=5}[/tex]
The length from G(-4, 1) and I(4, 1) is
[tex]\sqrt[]{(-4-(-4))^2+(1-1))^2}=\sqrt[]{8^2}=8[/tex]
And finally, the length between the point H(0,-2) and I(4, 1) is
[tex]\sqrt[]{(0-4)^2+(-2-1)^2}=\sqrt[]{4^2+(-3)^2}=5[/tex]
Hence, the side lengths are 5, 8, and 5; therefore, the perimeter is
[tex]5+8+5=\textcolor{#FF7968}{18.}[/tex]
The perimeter is 18 units.
And the area is
[tex]A=(\frac{8}{2})\times3=12[/tex]
