A right triangle has a segment from the right angle that meets the hypotenuse at a right angle at point p. Point p splits the hypotenuse into two segments whose lengths are 16 units and 4 units. Find the distance from the perpendicular line to the right angle. (answer here) units

Answer 1

To solve the problem, we will use a few properties of right triangles and the relationships of the segments created when a perpendicular is drawn from the right angle to the hypotenuse.

Let's denote the right triangle as \( ABC \), with \( C \) as the right angle. The hypotenuse \( AB \) is split by point \( P \) at a right angle, which creates two segments: \( AP = 16 \) units and \( PB = 4 \) units.

Using the relationship involving the lengths of the segments on the hypotenuse and the perpendicular dropped from the right angle, we can apply the formula: \[ CP^2 = AP \cdot PB \]

Substituting the lengths of the segments: \[ CP^2 = 16 \cdot 4 \] \[ CP^2 = 64 \] \[ CP = \sqrt{64} = 8 \]

Thus, the distance from the right angle \( C \) to the hypotenuse \( AB \) (the length of the perpendicular segment \( CP \)) is \( \boxed{8} \) units.