To solve the problem, we need to analyze the relationship between the segments created by the altitude drawn from the right angle to the hypotenuse of the right triangle.
Let's denote:
- \( AB \) as the hypotenuse of the right triangle,
- \( C \) as the right angle (the vertex where the perpendicular from the right angle meets the hypotenuse),
- \( AD \) as the short segment of the hypotenuse which measures 2 units,
- \( DB \) as the long segment of the hypotenuse which measures 30 units.
The geometric mean relationship states that:
\[ AC^2 = AD \times AB \]
where \( AC \) is the short leg of the triangle, \( AD \) is the length of the short segment of the hypotenuse, and \( AB \) is the entire hypotenuse (which is \( AD + DB \)).
Now let's calculate \( AB \):
\[ AB = AD + DB = 2 + 30 = 32 \text{ units} \]
Next, we can use the geometric mean relationship for the short leg \( AC \):
\[ AC^2 = AD \times AB = 2 \times 32 \] \[ AC^2 = 64 \]
Now we take the square root to find \( AC \):
\[ AC = \sqrt{64} = 8 \text{ units} \]
Thus, the distance of the short leg of the original triangle is 8 units.
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