What is the perimeter of a triangle with vertices located at (1, 3), (2, 6), and (0, 4), rounded to the nearest hundredth?A.7.40 unitsB.9.07 unitsC.8.49 unitsD.7.07 units

To find the perimeter of a triangle with given vertices, we first need to find the distance between each pair of points. The distance between two points (x1, y1) and (x2, y2) can be found using the distance formula:

d = sqrt[(x2-x1)² + (y2-y1)²]

Let's calculate the distance between each pair of points:

1. Distance between (1, 3) and (2, 6): d1 = sqrt[(2-1)² + (6-3)²] = sqrt[1 + 9] = sqrt[10] ≈ 3.16 units

2. Distance between (2, 6) and (0, 4): d2 = sqrt[(0-2)² + (4-6)²] = sqrt[4 + 4] = sqrt[8] ≈ 2.83 units

3. Distance between (0, 4) and (1, 3): d3 = sqrt[(1-0)² + (3-4)²] = sqrt[1 + 1] = sqrt[2] ≈ 1.41 units

Now, add all these distances to get the perimeter of the triangle:

Perimeter = d1 + d2 + d3 = 3.16 units + 2.83 units + 1.41 units = 7.40 units

So, the perimeter of the triangle is approximately 7.40 units. The correct answer is A.7.40 units.