To solve for the length of the short leg of the right triangle using the given lengths of the segments on the hypotenuse, we can utilize the geometric mean in the context of a right triangle.
Let \( a \) be the length of the short leg, \( b \) be the length of the long leg, and \( c \) be the length of the hypotenuse. According to the information provided, we have:
-
The hypotenuse consists of two segments: a short segment measuring 2 units and a long segment measuring 30 units. Therefore, the total length of the hypotenuse \( c \) is:
\[ c = 2 + 30 = 32 \text{ units} \]
-
The geometric mean theorem states that the altitude (let's call it \( h \)) drawn from the right angle to the hypotenuse is the geometric mean of the two segments it creates on the hypotenuse. If we let \( x = 2 \) (short segment) and \( y = 30 \) (long segment), then:
\[ h^2 = x \cdot y \]
Therefore, we compute:
\[ h^2 = 2 \cdot 30 = 60 \] \[ h = \sqrt{60} = \sqrt{4 \cdot 15} = 2\sqrt{15} \]
-
The geometric mean can also relate the lengths of the legs of the triangle and the segments of the hypotenuse:
Using the formula:
\[ a^2 = x \cdot c \quad \text{and} \quad b^2 = y \cdot c \]
We find the length of the short leg \( a \):
\[ a^2 = x \cdot c = 2 \cdot 32 = 64 \implies a = \sqrt{64} = 8 \text{ units} \]
Thus, the distance of the short leg of the original triangle is \( \boxed{8} \) units.
3 answers