Explanation
Part A
Given the vertices of triangle ABC as A(2, 8), B(16, 2), and C6, 2). We can form the triangle
Therefore, the side can be found using the distance between two points.
[tex]\begin{gathered} AC=\sqrt{(6-2)^2+(2-8)^2}=\sqrt{16+36}=\sqrt{52} \\ CB=\sqrt{(2-6)^2+(16-2)^2}=\sqrt{100}=10 \\ AB=\sqrt{(16-2)^2+(2-8)^2}=\sqrt{232}= \end{gathered}[/tex]
Therefore, the perimeter of the triangle is
[tex]P=AC+CB+AB=\sqrt{52}+10+\sqrt{232}=32.44[/tex]
Answer: Perimeter=32.44 units
Part B
From the image below,
The perpendicular height of the triangle is 6 units. While the base is 10 units.
The area of the triangle becomes
[tex]Area=\frac{1}{2}\times perpendicular\text{ height}\times base=\frac{1}{2}\times6\times10=30\text{ square units}[/tex]
Area=30 square units
