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.(1 point) units

Answer 1

Let the lengths of the two segments of the hypotenuse be \( AP = 16 \) units and \( PB = 4 \) units, where \( P \) is the point where the perpendicular segment meets the hypotenuse \( AB \).

According to the property of the right triangle, the length of the perpendicular segment from the right angle to the hypotenuse can be found using the formula:

\[ h = \frac{ab}{c} \]

where \( a \) and \( b \) are the segments of the hypotenuse \( AP \) and \( PB \), and \( c \) is the length of the hypotenuse \( AB \).

First, we can find the length of the hypotenuse:

\[ AB = AP + PB = 16 + 4 = 20 \text{ units} \]

Now we can apply the geometric mean relationship, which states that the length of the altitude from the right angle to the hypotenuse is the geometric mean of the lengths of the two segments:

\[ h = \sqrt{AP \cdot PB} \]

So we calculate:

\[ h = \sqrt{16 \cdot 4} = \sqrt{64} = 8 \text{ units} \]

Thus, the distance from the right angle to the hypotenuse, which is the length of the perpendicular, is

\[ \boxed{8} \text{ units}. \]