Question

The vertices of AABC are A(2,8), B(16, 2), and C(6, 2). The perimeter of triangleABC Is blank units

Answer 1

Explanation

We are given the following vertices of a triangle:

`\begin{gathered} A(2,8) \\ B(16,2) \\ C(6,2) \end{gathered}`

We are required to determine the perimeter of the triangle.

This is achieved thus:

We know that the distance between two points is given as:

`D=√((x_2-x_1)^2+(y_2-y_1)^2)`

To determine the perimeter of the triangle ABC, we need to determine the distances of AB, AC, and BC.

Therefore, we have:

`\begin{gathered} A(2,8)\to(x_1,y_1) \\ B(16,2)\to(x_2,y_2) \\ \\ \therefore AB=√((16-2)^2+(2-8)^2) \\ AB=√((14)^2+(-6)^2) \\ AB=√(196+36)=√(232) \\ AB=2√(58)\approx15.23units \end{gathered}`
`\begin{gathered} A(2,8) \\ C(6,2) \\ \\ \therefore AC=√((6-2)^2+(2-8)^2) \\ AC=√((4)^2+(-6)^2) \\ AC=√(16+36)=√(52) \\ AC=2√(13)\approx7.21units \end{gathered}`
`\begin{gathered} B(16,2)\to(x_1,y_1) \\ C(6,2)\to(x_2,y_2) \\ \\ \therefore BC=√((6-16)^2+(2-2)^2) \\ BC=√((-10)^2+0^2) \\ BC=√(100) \\ BC=10units \end{gathered}`

Therefore, the perimeter becomes:

`\begin{gathered} P=AB+AC+BC \\ P=15.23+7.21+10 \\ P=32.44units \end{gathered}`

Hence, the perimeter is:

`P=32.44un\imaginaryI ts`