Question

applying the geometric mean (leg) theorem
what is the value of q?
○ $4\sqrt{5}$
○ $2\sqrt{14}$
○ $20\sqrt{5}$
○ $64\sqrt{5}$
(diagram: triangle qsr with right angles at s and t; segment qt = 10, tr = 4; sides labeled r, s, q)

Explanation:

Step1: Recall Geometric Mean (Leg) Theorem

The Geometric Mean (Leg) Theorem states that in a right triangle, the length of a leg is the geometric mean of the length of the hypotenuse segment adjacent to that leg and the length of the entire hypotenuse. For leg \( q \) (which is a leg of the right triangle \( \triangle SQR \) with altitude \( ST \)), the adjacent segment to \( q \) is \( 4 \) and the entire hypotenuse segment from \( Q \) to \( R \) is \( 10 + 4=14 \)? Wait, no, wait. Wait, actually, the hypotenuse of the large right triangle \( \triangle SQR \) is \( QR \), which is \( QT + TR = 10 + 4 = 14 \)? Wait, no, looking at the diagram: \( QT = 10 \), \( TR = 4 \), so \( QR = 10 + 4 = 14 \). The leg \( q \) (which is \( SR \)) is a leg of the right triangle, and the two segments of the hypotenuse are \( TR = 4 \) and \( QR = 14 \)? Wait, no, the Geometric Mean (Leg) Theorem formula is: if in a right triangle, an altitude is drawn to the hypotenuse, then each leg is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg. So for leg \( SR = q \), the adjacent segment is \( TR = 4 \), and the hypotenuse is \( QR = QT + TR = 10 + 4 = 14 \)? Wait, no, wait, maybe I mixed up. Wait, the large triangle is \( \triangle SQR \), right-angled at \( S \), and \( ST \) is the altitude to hypotenuse \( QR \). Then, by the Geometric Mean (Leg) Theorem, \( SR^2 = TR \times QR \). Wait, \( QR = QT + TR = 10 + 4 = 14 \), \( TR = 4 \). So \( q^2 = 4 \times 14 \)? Wait, no, that can't be. Wait, no, maybe \( QT = 10 \), \( TR = 4 \), so the two segments of the hypotenuse are \( QT = 10 \) and \( TR = 4 \). Then, the leg \( SR = q \) is adjacent to \( TR = 4 \), so the formula is \( SR^2 = TR \times QR \), where \( QR = QT + TR = 14 \). Wait, but let's check the theorem again. The Geometric Mean (Leg) Theorem: 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. So, if we have right triangle \( ABC \), right-angled at \( C \), and altitude \( CD \) to hypotenuse \( AB \), then \( AC^2 = AD \times AB \) and \( BC^2 = BD \times AB \). So in our case, right triangle \( SQR \), right-angled at \( S \), altitude \( ST \) to hypotenuse \( QR \). So \( SR^2 = TR \times QR \). \( QR = QT + TR = 10 + 4 = 14 \), \( TR = 4 \). So \( q^2 = 4 \times 14 \)? Wait, no, that would be \( q^2 = 56 \), so \( q = \sqrt{56} = 2\sqrt{14} \). Wait, but let's confirm. Wait, maybe \( QT = 10 \), \( TR = 4 \), so the two segments are \( 10 \) and \( 4 \). Then, the leg \( SR = q \) is adjacent to \( TR = 4 \), so the formula is \( SR^2 = TR \times (TR + QT) \), which is \( 4 \times (4 + 10) = 4 \times 14 = 56 \). So \( q = \sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14} \).

Step2: Calculate \( q \)

We have \( q^2 = 4 \times 14 = 56 \)
Then \( q = \sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14} \)

Answer:

\( 2\sqrt{14} \) (corresponding to the option \( 2\sqrt{14} \))