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)

Question

units

Answer 1

To solve this problem, we will use the properties of a right triangle and the geometric mean.

Let \( A \) and \( B \) be the points where the segments on the hypotenuse intersect point \( P \), so:

• \( AP = 16 \)
• \( PB = 4 \)

The length of the hypotenuse \( AB \) is given by:

\[ AB = AP + PB = 16 + 4 = 20 \]

Since \( P \) is the foot of the altitude from the right angle (let's call the right angle point \( C \)), we need to find the length of the segments from point \( C \) to point \( P \).

In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of the altitude \( CP \) can be found using the formula:

\[ CP = \frac{AB}{2} \cdot \sin C \]

However, we will also use the geometric mean. The relationship between the sides can be expressed as:

\[ CP^2 = AP \cdot PB \]

Substituting the values of \( AP \) and \( PB \):

\[ CP^2 = 16 \cdot 4 = 64 \]

Taking the square root, we find:

\[ CP = \sqrt{64} = 8 \]

Thus, the distance from the perpendicular line to the right angle, which is the length of the altitude \( CP \), is

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