Question

A right triangle is drawn on a coordinate plane with a line drawn from the right angle that is perpendicular to the hypotenuse. The hypotenuse consists of a short line segment measuring 2 units and a long line segment measuring 30 units. Using the geometric mean, what is the distance of the short leg of the original triangle? (1 point)
5.3 units
8 units
60 units
28 units

Answer 1

To find the distance of the short leg of the original triangle, we can use the concept of the geometric mean in the context of a right triangle with a right angle at \(C\) and hypotenuse \(AB\).

Let’s denote:
- The lengths of the segments on hypotenuse \(AB\) as \(AD\) (short segment, 2 units) and \(DB\) (long segment, 30 units).
- The length of the altitude from point \(C\) to hypotenuse \(AB\) as \(h\).

By the geometric mean theorem (or the altitude-on-hypotenuse theorem), we have:

\[
h^2 = AD \cdot DB
\]

Substituting the lengths of the segments:

\[
h^2 = 2 \cdot 30 = 60
\]

Now take the square root to find \(h\):

\[
h = \sqrt{60} = \sqrt{4 \cdot 15} = 2\sqrt{15}
\]

Next, we need to determine the lengths of the legs of the triangle. The relationship between the hypotenuse \(AB\) and the legs \(AC\) and \(BC\) can be expressed with respect to the segments on the hypotenuse:

Let \(AC\) be the length of one leg and \(BC\) be the length of the other leg. From the properties of the right triangle:

\[
AC^2 = AD \cdot AB
\]
\[
BC^2 = DB \cdot AB
\]

Since \(AD = 2\) and \(DB = 30\), we need \(AB\):

\[
AB = AD + DB = 2 + 30 = 32 \text{ units}
\]

Now calculate \(AC^2\) and \(BC^2\):

For \(AC\):

\[
AC^2 = AD \cdot AB = 2 \cdot 32 = 64
\]
\[
AC = \sqrt{64} = 8 \text{ units}
\]

For \(BC\):

\[
BC^2 = DB \cdot AB = 30 \cdot 32 = 960
\]
\[
BC = \sqrt{960} = 4\sqrt{60} = 4 \cdot 2 \sqrt{15} = 8\sqrt{15} \text{ units}
\]

Thus, the length of the short leg \(AC\) is \(8\) units.

The distance of the short leg of the original triangle is therefore:

**8 units**.