Question

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

Explanation:

Step1: Recall Geometric Mean Theorem

In a right triangle, the altitude to the hypotenuse relates the segments. Also, for a right triangle, if we have a right triangle with an altitude, the length of a leg is the geometric mean of the hypotenuse segment adjacent to it and the hypotenuse. Wait, more accurately, in a right triangle, when you draw an altitude from the right angle to the hypotenuse, the length of each leg is the geometric mean of the length of the hypotenuse segment adjacent to that leg and the length of the entire hypotenuse. Also, another form: In triangle \( QTS \), right - angled at \( T \), and \( TR \perp QS \), then \( TS^{2}=SR\times SQ \)? Wait, no, let's correct. The Geometric Mean (Leg) Theorem states that in a right triangle, the square of a leg is equal to the product of the hypotenuse and the segment of the hypotenuse adjacent to that leg. Wait, actually, if we have right triangle \( QTS \) with right angle at \( T \), and \( TR \) is the altitude to hypotenuse \( QS \), then \( TS^{2}=SR\times SQ \)? Wait, \( QS=QR + RS=6 + 12 = 18 \)? No, wait, no. Wait, the correct formula is: If in right triangle \( ABC \), right - angled at \( C \), and \( CD\perp AB \), then \( AC^{2}=AD\times AB \) and \( BC^{2}=BD\times AB \). So in our case, triangle \( QTS \) is right - angled at \( T \), and \( TR\perp QS \). So \( TS^{2}=SR\times SQ \)? Wait, \( SQ=QR + RS=6 + 12 = 18 \)? No, wait, \( QS \) is the hypotenuse, \( QR = 6 \), \( RS = 12 \), so \( SQ=QR + RS=18 \)? Wait, no, \( Q \)---\( R \)---\( S \), so \( QS=QR + RS = 6+12 = 18 \), and \( TS \) is a leg, \( SR = 12 \), \( SQ=18 \)? Wait, no, maybe I mixed up. Wait, the correct formula is \( TS^{2}=SR\times SQ \)? Wait, no, let's look at the triangle. Let's denote: In right triangle \( QTS \), right angle at \( T \), \( TR\perp QS \). Then, by the geometric mean theorem (leg - leg), \( TS^{2}=SR\times SQ \)? Wait, no, \( SQ \) is \( QR + RS=6 + 12 = 18 \), \( SR = 12 \), so \( TS^{2}=12\times18 \)? Wait, no, that can't be. Wait, maybe the triangle is such that \( TS \) is related to \( 3x \) and the segments. Wait, another approach: Let's assume that triangle \( TRS \) and triangle \( QTS \) are similar. Because \( \angle TSR=\angle QST \) (common angle) and \( \angle TRS=\angle QTS = 90^{\circ} \), so by AA similarity, \( \triangle TRS\sim\triangle QTS \). Therefore, the ratios of corresponding sides are equal. So \( \frac{TS}{QS}=\frac{SR}{TS} \), which gives \( TS^{2}=SR\times QS \). \( QS=QR + RS=6 + 12 = 18 \), \( SR = 12 \), so \( TS^{2}=12\times18=216 \), then \( TS=\sqrt{216}=\sqrt{36\times6}=6\sqrt{6} \). Wait, but also, we see that \( TS = 3x \). Let's check. If \( TS = 6\sqrt{6} \), then \( 3x=6\sqrt{6}\), so \( x = 2\sqrt{6} \), but we need \( TS \). Wait, let's verify the similarity again. \( \triangle TRS \) and \( \triangle QTS \): \( \angle TRS=\angle QTS = 90^{\circ} \), \( \angle S=\angle S \), so yes, similar. So \( \frac{TS}{QS}=\frac{SR}{TS} \), so \( TS^{2}=SR\times QS \). \( QS=6 + 12 = 18 \), \( SR = 12 \), so \( TS^{2}=12\times18 = 216 \), \( TS=\sqrt{216}=6\sqrt{6} \).

Step2: Calculate the length of \( TS \)

We know from the geometric mean theorem (leg - hypotenuse segment relationship) that for right triangle \( QTS \) with altitude \( TR \) to hypotenuse \( QS \), \( TS^{2}=SR\times QS \).
First, find \( QS \): \( QS=QR + RS=6 + 12 = 18 \).
Then, substitute \( SR = 12 \) and \( QS = 18 \) into the formula: \( TS^{2}=12\times18=216 \).
Take the square root of both sides: \( TS=\sqrt{216}=\sqrt{36\times6}=6\sqrt{6} \).

Answer:

\( 6\sqrt{6} \) units