Given a right triangle \( \triangle ABC \) where \( \angle C \) is the right angle, we denote \( P \) as the point where the altitude from \( C \) meets the hypotenuse \( AB \) at a right angle. Let \( AP = 16 \) units and \( PB = 4 \) units. The total length of the hypotenuse \( AB \) is:
\[ AB = AP + PB = 16 + 4 = 20 \text{ units} \]
In a right triangle, the length of the altitude \( CP \) from the right angle to the hypotenuse can be calculated using the formula for the area of the triangle. The area \( A \) of the triangle can be expressed in two ways:
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Using the base \( AB \) and height \( CP \): \[ A = \frac{1}{2} \cdot AB \cdot CP = \frac{1}{2} \cdot 20 \cdot CP = 10 \cdot CP \]
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Using the segments \( AP \) and \( PB \): \[ A = \frac{1}{2} \cdot AP \cdot PB = \frac{1}{2} \cdot 16 \cdot 4 = 32 \]
Setting these two expressions for the area equal to each other, we have:
\[ 10 \cdot CP = 32 \]
Now, solve for \( CP \):
\[ CP = \frac{32}{10} = 3.2 \text{ units} \]
Thus, the distance from the perpendicular line to the right angle \( C \) is
\[ \boxed{3.2} \text{ units} \]