To solve for the distance of the short leg of the right triangle formed by a hypotenuse that consists of a short segment of 2 units and a long segment of 30 units, we can use the geometric mean relationships that arise when a perpendicular line is drawn from the right angle to the hypotenuse.
In a right triangle, if we let \( a \) and \( b \) be the lengths of the legs and \( c \) be the length of the hypotenuse, the relationship involving the segments of the hypotenuse divided by the altitude (the perpendicular from the right angle) is given by:
- If the hypotenuse is divided into segments of length \( m \) and \( n \) by the altitude, the legs \( a \) and \( b \) are related by the following relationships: \[ a = \sqrt{m \cdot c} \quad \text{and} \quad b = \sqrt{n \cdot c} \]
Here, let’s define:
- The total length of the hypotenuse \( c = 32 \) (since \( 2 + 30 = 32 \))
- The short segment \( m = 2 \)
- The long segment \( n = 30 \)
Now we can find the lengths of the legs:
For the short leg \( a \): \[ a = \sqrt{m \cdot c} = \sqrt{2 \cdot 32} = \sqrt{64} = 8 \]
Thus, the distance of the short leg of the original triangle is 8 units.
Final response: 8 units
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