To find the length of the perpendicular drawn from the opposite vertex to the diagonal of a square, we start by understanding the geometry of the square and its diagonal.

Given:

• Length of the side of the square, \(s = 4\) units.

Step-by-step Solution:

1. The diagonal of the square can be found using the Pythagorean theorem. Since a square has equal sides, the diagonal \(d\) is given by:

d = $\sqrt{s^2 + s^2}$ = $\sqrt{2s^2}$ = $s\sqrt{2}$

• Substitute \(s = 4\):

1. The perpendicular from a vertex to the opposite diagonal of the square divides the square into two right triangles. This perpendicular acts as an altitude for the square when base is the diagonal.
2. The area of the square can be given as:

A = $s^2$ = $4^2$ = 16

1. The area can also be expressed using the diagonal as the base and the perpendicular as the height:

\text{Area} = $\frac{1}{2}$ $\times \text{diagonal} \times \text{perpendicular height}$

16 = $\frac{1}{2}$ $\times 4\sqrt{2} \times h$

• Solving for \(h\) (perpendicular height):

1. After Rationalizing the denominator:
   • So, length of the perpendicular from the opposite vertex to the diagonal is \(2\sqrt{2}\).

Thus, the correct answer is \(2\sqrt{2}\).