Answer:
The perimeter of the triangle is 21.8 unit.
Step-by-step explanation:
Given : Triangle ABC has vertices at (-4,0) (-1,6) and (3,-1).
To find : What is the perimeter of triangle ABC?
Solution :
Vertices of the triangle : A(-4, 0) ,B(-1, 6), C(3, -1)
First we find the distance between the points or length of the sides of triangle.
Length of AB - A(-4, 0) ,B(-1, 6)
\(AB=\sqrt{(6 - 0)^2 + (-1 + 4)^2}\)
\(AB=\sqrt{(6)^2 + (3)^2}\)
\(AB=\sqrt{36+9}\)
\(AB=\sqrt{45}\)
Length of BC - B(-1, 6), C(3, -1)
\(BC=\sqrt{(-1-6)^2 + (3+1)^2}\)
\(BC=\sqrt{(-7)^2 + (4)^2}\)
\(BC=\sqrt{49+16}\)
\(BC=\sqrt{65}\)
Length of AC - A(-4, 0), C(3, -1)
\(AC=\sqrt{(0+1)^2 + (-4-3)^2}\)
\(AC=\sqrt{(1)^2 + (-7)^2}\)
\(AC=\sqrt{1+49}\)
\(AC=\sqrt{50}\)
Perimeter of the triangle is sum of all sides of the triangle,
P=Length of (AB+BC+AC)
\(P = \sqrt{45} + \sqrt{65} + \sqrt{50}\)
\(P =6.708 + 8.062 +7.071\)
\(P =21.841\)
Nearest tenth, The perimeter of the triangle is 21.8 unit.
