Answer:
\(9+\sqrt{41}\)
Step-by-step explanation:
To find the perimeter of the triangle with vertices (-3, 5), (2, 1) and (-3, 1), calculate the distances between these points using the distance formula, and then sum them up.
\(\boxed{\begin{array}{l}\underline{\sf Distance \;Formula}\\\\d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\\textsf{where:}\\ \phantom{ww}\bullet\;\;d\;\textsf{is the distance between two points.} \\\phantom{ww}\bullet\;\;\textsf{$(x_1,y_1)$ and $(x_2,y_2)$ are the two points.}\end{array}}\)
Let A = (-3, 5), B = (2, 1), and C = (-3, 1). Therefore:
\(AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}\\\\AB=\sqrt{(2-(-3))^2+(1-5)^2}\\\\AB=\sqrt{(5)^2+(-4)^2}\\\\AB=\sqrt{25+16}\\\\AB=\sqrt{41}\)
\(BC=\sqrt{(x_C-x_B)^2+(y_C-y_B)^2}\\\\BC=\sqrt{(-3-2)^2+(1-1)^2}\\\\BC=\sqrt{(-5)^2+(0)^2}\\\\BC=\sqrt{25+0}\\\\BC=\sqrt{25}\\\\BC=5\)
\(AC=\sqrt{(x_C-x_A)^2+(y_C-y_A)^2}\\\\AC=\sqrt{(-3-(-3))^2+(1-5)^2}\\\\AC=\sqrt{(0)^2+(-4)^2}\\\\AC=\sqrt{0+16}\\\\AC=\sqrt{16}\\\\AC=4\)
Now, we can find the perimeter by summing up these distances:
\(\textsf{Perimeter}=AB+BC+AC\\\\\textsf{Perimeter}=\sqrt{41}+5+4\\\\\textsf{Perimeter}=9+\sqrt{41}\)
So, the exact perimeter of the triangle with vertices (-3, 5), (2, 1) and (-3, 1) is:
\(\Large\boxed{\boxed{9+\sqrt{41}}}\)
