Question

finding a leg length
what is the length of side ts?
6√6 units
24 units
8 units
2√6 units

Explanation:

Step1: Recall Geometric Mean Theorem

In a right triangle, the altitude to the hypotenuse relates the segments. For right triangle \( \triangle QTS \) with altitude \( TR \), we have \( TS^2 = SR \times SQ \). First, find \( SQ \): \( SQ = QR + RS = 6 + 12 = 18 \)? Wait, no, wait—wait, actually, the Geometric Mean (Leg) Theorem: in a right triangle, each leg is the geometric mean of the hypotenuse and the adjacent segment. Wait, correct theorem: If in right \( \triangle QTS \), right-angled at \( T \), and \( TR \perp QS \), then \( TS^2 = SR \times SQ \). Wait, \( SQ \) is \( QR + RS = 6 + 12 = 18 \)? No, wait, \( QS = QR + RS = 6 + 12 = 18 \)? Wait, no, the segments are \( QR = 6 \), \( RS = 12 \), so \( SQ = QR + RS = 18 \)? Wait, no, \( QS \) is the hypotenuse? Wait, no, \( \triangle QTS \) is right-angled at \( T \), so \( QT \) and \( TS \) are legs, \( QS \) is hypotenuse. Then \( TR \) is altitude to hypotenuse \( QS \). So by Geometric Mean Theorem, \( TS^2 = RS \times QS \). Wait, \( QS = QR + RS = 6 + 12 = 18 \)? Wait, no, \( QR = 6 \), \( RS = 12 \), so \( QS = 6 + 12 = 18 \). Then \( TS^2 = 12 \times 18 \)? Wait, no, wait: the Geometric Mean Theorem (Leg) states that \( \text{leg}^2 = \text{adjacent segment} \times \text{hypotenuse} \). Wait, adjacent segment to \( TS \) is \( RS = 12 \), and hypotenuse is \( QS = 6 + 12 = 18 \)? Wait, no, that can't be. Wait, maybe I mixed up. Let's re-express: In right \( \triangle QTS \), right angle at \( T \), \( TR \perp QS \) (so \( R \) is on \( QS \)). Then:

• \( QT^2 = QR \times QS \)
• \( TS^2 = RS \times QS \)

Yes, that's the Leg Theorem. So \( QS = QR + RS = 6 + 12 = 18 \). Then \( TS^2 = RS \times QS = 12 \times 18 = 216 \). Then \( TS = \sqrt{216} = \sqrt{36 \times 6} = 6\sqrt{6} \)? Wait, but wait, maybe I made a mistake. Wait, no, wait: \( QS \) is \( QR + RS = 6 + 12 = 18 \)? Wait, but the options include \( 6\sqrt{6} \), which is \( \sqrt{216} \). Wait, let's check again.

Wait, another way: Let \( TR = h \), \( TS = 3x \) (from the diagram, \( TS \) is labeled \( 3x \)). Wait, maybe using similar triangles. \( \triangle TRS \sim \triangle QTS \) (AA similarity: right angle, common angle at \( S \)). So \( \frac{TS}{QS} = \frac{RS}{TS} \), so \( TS^2 = RS \times QS \). \( QS = 6 + 12 = 18 \), \( RS = 12 \), so \( TS^2 = 12 \times 18 = 216 \), so \( TS = \sqrt{216} = 6\sqrt{6} \). Wait, but let's check the options. One of the options is \( 6\sqrt{6} \) units.

Wait, but wait, maybe I messed up \( QS \). Wait, \( QR = 6 \), \( RS = 12 \), so \( QS = 6 + 12 = 18 \). Then \( TS^2 = 12 \times 18 = 216 \), so \( TS = \sqrt{216} = 6\sqrt{6} \). So that's the first option.

Step2: Calculate \( TS \)

Using the Geometric Mean (Leg) Theorem: \( TS^2 = RS \times QS \).
\( QS = QR + RS = 6 + 12 = 18 \), \( RS = 12 \).
Thus, \( TS^2 = 12 \times 18 = 216 \).
Take square root: \( TS = \sqrt{216} = \sqrt{36 \times 6} = 6\sqrt{6} \).

Answer:

\( 6\sqrt{6} \) units (the first option)