The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is :

1. 12 units

2. 6 units

3. 5 units

4. 10 units

Let (0, 4) = (x₁, y₁), (0, 0) = (x₂, y₂) and (3, 0) = (x₃, y₃)

Distance between the given points = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

The perimeter of a triangle

$= \sqrt{(x$2 - x1)^2 + (y2 - y1)^2} + \sqrt{(x3 - x2)^2 + (y3 - y2)^2} + \sqrt{(x3 - x1)^2 + (y3 - y1)^2}\\[1em] = \sqrt{(0 - 0)^2 + (0 - 4)^2} + \sqrt{(3 - 0)^2 + (0 - 0)^2} + \sqrt{(3 - 0)^2 + (0 - 4)^2}\\[1em] = \sqrt{(- 4)^2} + \sqrt{(3)^2} + \sqrt{(3)^2 + (- 4)^2}\\[1em] = \sqrt{16} + \sqrt{9} + \sqrt{9 + 16}\\[1em] = 4 + 3 + \sqrt{25}\\[1em] = 4 + 3 + 5\\[1em] = 12 \text{units}

Hence, option 1 is the correct option.