Question

there is a right triangle - related diagram (a triangle with a perpendicular segment to the base, base segments labeled 3 and 9, and a side labeled x, with a right angle at the top and the perpendicular to the base).

Explanation:

Step1: Identify the geometric theorem

This is a right triangle with an altitude to the hypotenuse, so we use the geometric mean theorem (or altitude-on-hypotenuse theorem), which 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 hypotenuse. Let the hypotenuse segments be \( 3 \) and \( 9 \), and the leg be \( x \). The adjacent segment to \( x \) is \( 9 \), and the hypotenuse (if we consider the whole, but actually the theorem is \( \text{leg}^2=\text{segment}_1\times\text{segment}_2 \) where segment\(_1\) and segment\(_2\) are the segments of the hypotenuse divided by the altitude. Wait, actually, the correct formula is that for a right triangle, when you draw an altitude from the right angle to the hypotenuse, then each leg is the geometric mean of the hypotenuse and the adjacent segment. So if the two segments of the hypotenuse are \( a = 3 \) and \( b = 9 \), and the leg opposite to the segment \( b \) is \( x \), then \( x^2= b\times(a + b) \)? Wait no, wait the hypotenuse is \( 3 + 9=12 \)? Wait no, the altitude divides the hypotenuse into two parts: let's say the hypotenuse is \( c = 3 + 9 = 12 \), and the altitude is \( h \), and the legs are \( l_1 \) and \( l_2 \). Then the correct theorem is \( l_1^2=3\times12 \)? No, no, I messed up. Wait, the geometric mean theorem: In a right triangle, the altitude drawn to the hypotenuse is the geometric mean between the segments of the hypotenuse. Also, each leg is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg. So if the hypotenuse is divided into segments of length \( m \) and \( n \), and the legs are \( p \) and \( q \), then \( p^2=m\times(m + n) \)? No, no, \( m \) and \( n \) are the two segments, so \( p^2=m\times(m + n) \) is wrong. Wait, let's denote: Let the right triangle have hypotenuse \( AB \), and altitude \( CD \) from right angle \( C \) to \( AB \), so \( AD = m \), \( DB = n \), then \( AC^2=AD\times AB \), \( BC^2=DB\times AB \), and \( CD^2=AD\times DB \). Ah, right! So \( AB = AD + DB = m + n \). So in our case, the two segments of the hypotenuse are \( 3 \) and \( 9 \), so \( AB = 3 + 9 = 12 \). The leg \( x \) is adjacent to the segment \( 9 \), so \( x^2=9\times12 \)? Wait no, wait the leg adjacent to segment \( 9 \) is the one whose square is equal to the product of the hypotenuse and the adjacent segment. Wait, no, let's look at the diagram: the right angle is at the top, and the altitude is drawn to the hypotenuse, dividing it into \( 3 \) and \( 9 \). So the leg \( x \) is connected to the segment of length \( 9 \). So according to the theorem, \( x^2=9\times(3 + 9) \)? Wait, no, \( AB = 3 + 9 = 12 \), and the leg \( x \) is \( BC \), where \( DB = 9 \), so \( BC^2=DB\times AB = 9\times12 = 108 \)? Wait, but that would make \( x=\sqrt{108}=6\sqrt{3}\approx10.392 \), but maybe I got the segments wrong. Wait, maybe the two segments are \( 3 \) and \( 9 \), and the leg is adjacent to \( 9 \), but actually, the correct formula is that if the altitude divides the hypotenuse into segments \( a \) and \( b \), then the leg corresponding to segment \( b \) is \( \sqrt{b\times(a + b)} \)? No, no, the hypotenuse is \( a + b \), so \( a = 3 \), \( b = 9 \), hypotenuse \( c = a + b = 12 \). Then the leg adjacent to \( b \) (which is \( 9 \)) is \( \sqrt{b\times c} \)? Wait, no, let's check with a simple case: if \( a = 1 \), \( b = 1 \), hypotenuse \( 2 \), then the leg should be \( \sqrt{2} \), and \( \sqrt{1\times2}=\sqrt{2} \), which is correct. So yes, the formula is \( \text{leg}^2=\text{segment}\times\text{hypotenuse} \). So in our problem, the segment adjacent to \( x \) is \( 9 \), and the hypotenuse is \( 3 + 9 = 12 \). So \( x^2=9\times12 \).

Step2: Calculate \( x \)

First, compute \( 9\times12 = 108 \). Then take the square root of \( 108 \). Simplify \( \sqrt{108} \): \( 108 = 36\times3 \), so \( \sqrt{108}=\sqrt{36\times3}=6\sqrt{3}\approx10.392 \). Wait, but maybe I made a mistake in the segments. Wait, maybe the two segments are \( 3 \) and \( 9 \), and the leg is the geometric mean of \( 9 \) and \( (3 + 9) \)? Wait, no, the hypotenuse is \( 3 + 9 = 12 \), so \( x^2=9\times12 \), so \( x=\sqrt{108}=6\sqrt{3} \). Alternatively, maybe the problem is that the two segments are \( 3 \) and \( 9 \), and the leg is \( x \), so \( x^2=3\times(3 + 9) \)? No, that would be if the segment adjacent is \( 3 \). Wait, looking at the diagram: the left segment is \( 3 \), the right segment is \( 9 \), and the leg \( x \) is on the right triangle (the one with segment \( 9 \)). So the leg \( x \) is opposite to the segment \( 3 \)? No, the right angle at the top, so the two legs are the left and right sides, and the altitude is in the middle. So the left leg: square is \( 3\times(3 + 9) \), the right leg (x) is \( 9\times(3 + 9) \). So yes, \( x^2=9\times12 = 108 \), so \( x=\sqrt{108}=6\sqrt{3} \). Wait, but maybe the problem is that the hypotenuse is not \( 3 + 9 \), but the two segments are \( 3 \) and \( 9 \), and the leg is \( x \), so the formula is \( x^2=9\times(3 + 9) \). So calculating that: \( 9\times12 = 108 \), so \( x=\sqrt{108}=6\sqrt{3}\approx10.39 \).

Answer:

\( 6\sqrt{3} \) (or approximately \( 10.39 \))