Question

question 11 (1 point)
find the value of x.
(there is a right triangle with a blue line segment of length 25 (the hypotenuse), a segment of length 9 on the hypotenuse, and a red perpendicular segment inside the triangle, and the leg of length x is to be found.)
blank 1: input box

Explanation:

Step1: Identify the geometric theorem

This is a right triangle with an altitude to the hypotenuse, so we use the geometric mean (altitude-on-hypotenuse) theorem. The theorem states that in a right triangle, the length of a leg (x) is the geometric mean of the hypotenuse and the adjacent segment of the hypotenuse. First, find the length of the segment adjacent to \( x \). The hypotenuse is 25, and the other segment is 9, so the adjacent segment to \( x \) is \( 25 - 9 = 16 \)? Wait, no, wait. Wait, actually, the leg \( x \) is adjacent to the segment of length \( 25 - 9 \)? Wait, no, let's re-examine. Wait, the hypotenuse is 25, and the segment adjacent to \( x \) (the leg) is actually the segment from the right angle to the foot of the altitude. Wait, no, the correct 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, let's denote the hypotenuse as \( c = 25 \), the segment adjacent to \( x \) as \( a \), and the other segment as \( b = 9 \). Wait, actually, the two segments of the hypotenuse are \( b = 9 \) and \( a = 25 - 9 = 16 \)? Wait, no, maybe I got the segments wrong. Wait, the altitude divides the hypotenuse into two segments: one of length 9, and the other of length \( 25 - 9 = 16 \)? Wait, no, the hypotenuse is 25, so the two segments are 9 and \( 25 - 9 = 16 \)? Wait, no, actually, the leg \( x \) is adjacent to the segment of length \( 25 - 9 \)? Wait, no, let's recall the theorem correctly. The formula is: \( x^2 = (segment\ adjacent\ to\ x) \times hypotenuse \)? Wait, no, the correct formula is: 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 the hypotenuse is \( c \), and the segment adjacent to leg \( x \) is \( m \), then \( x^2 = c \times m \)? Wait, no, that's not right. Wait, the correct theorem is: If an altitude is drawn to the hypotenuse of a right triangle, then (1) the length of the altitude is the geometric mean of the lengths of the two segments of the hypotenuse, and (2) the length of each leg is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg. So, let's denote: hypotenuse \( c = 25 \), segment adjacent to \( x \) is \( m \), and the other segment is \( n = 9 \). Then, \( x^2 = c \times m \), but wait, \( c = m + n \), so \( m = c - n = 25 - 9 = 16 \). Then, \( x^2 = 25 \times 16 \)? Wait, no, that can't be. Wait, no, I think I mixed up the segments. Wait, actually, the two segments of the hypotenuse are \( n = 9 \) and \( m = 25 - 9 = 16 \), and the leg \( x \) is adjacent to \( m = 16 \)? Wait, no, let's take an example. Suppose the hypotenuse is \( c \), divided into segments \( p \) and \( q \), with \( p + q = c \). Then, the leg adjacent to \( p \) is \( \sqrt{c \times p} \), and the leg adjacent to \( q \) is \( \sqrt{c \times q} \). Wait, in this problem, the segment adjacent to \( x \) is \( 25 - 9 = 16 \)? Wait, no, the segment marked 9 is adjacent to the other leg. Wait, the leg \( x \) is adjacent to the segment of length \( 25 - 9 = 16 \)? Wait, maybe the correct segments are: the hypotenuse is 25, one segment is 9, so the other segment is \( 25 - 9 = 16 \). Then, the leg \( x \) is adjacent to the segment of length 16? Wait, no, let's check the formula again. The correct formula is: \( x^2 = (segment\ adjacent\ to\ x) \times hypotenuse \)? No, that's not. Wait, the correct formula is \( x^2 = (segment\ adjacent\ to\ x) \times (hypotenuse) \)? Wait, no, let's derive it. Let’s call the right triangle \( ABC \), right-angled at \( C \), with altitude \( CD \) to hypotenuse \( AB \), where \( AB = 25 \), \( AD = 9 \), \( DB = x \)? Wait, no, in the diagram, the right angle is at the bottom, and the altitude is drawn to the hypotenuse, creating two smaller right triangles. Wait, the leg \( x \) is one of the legs of the original right triangle, and the hypotenuse is 25. The altitude divides the hypotenuse into two segments: one of length 9, and the other of length \( 25 - 9 = 16 \). Then, the leg \( x \) is adjacent to the segment of length 16? Wait, no, the leg \( x \) is opposite to the segment of length 9? Wait, I think I made a mistake. Let's look at the diagram again. The hypotenuse is 25, the segment from the top vertex to the foot of the altitude is 9, so the segment from the foot of the altitude to the right angle vertex (the other end of the hypotenuse) is \( 25 - 9 = 16 \). Then, the leg \( x \) (the vertical leg) is adjacent to the segment of length 16? Wait, no, the correct formula is that the leg \( x \) (vertical leg) is the geometric mean of the hypotenuse (25) and the segment of the hypotenuse adjacent to \( x \), which is \( 25 - 9 = 16 \)? Wait, no, that would be \( x^2 = 25 \times 16 \), which is \( x^2 = 400 \), so \( x = 20 \). Wait, that makes sense. Let's verify. If \( x = 20 \), then the other leg (horizontal) can be found by \( \sqrt{25^2 - 20^2} = \sqrt{625 - 400} = \sqrt{225} = 15 \). Then, the altitude to the hypotenuse would be \( \frac{15 \times 20}{25} = 12 \). Then, the two segments of the hypotenuse would be \( \frac{15^2}{25} = 9 \) and \( \frac{20^2}{25} = 16 \), which matches the diagram (one segment is 9, the other is 16, since 25 - 9 = 16). Yes, that works. So the formula is: \( x^2 = (segment\ adjacent\ to\ x) \times hypotenuse \)? Wait, no, the segment adjacent to \( x \) is 16, and hypotenuse is 25, so \( x^2 = 25 \times 16 \)? Wait, no, \( x^2 = 16 \times 25 \)? Wait, no, the correct formula is that each leg is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg. So, if the leg is \( x \), and the adjacent segment is \( m \), then \( x^2 = c \times m \), where \( c \) is the hypotenuse. In this case, \( m = 25 - 9 = 16 \), \( c = 25 \), so \( x^2 = 25 \times 16 \)? Wait, no, that would be \( x^2 = 400 \), so \( x = 20 \). Yes, that's correct. So let's do the steps:

Step1: Determine the segment adjacent to \( x \)

The hypotenuse is 25, and one segment is 9, so the other segment (adjacent to \( x \)) is \( 25 - 9 = 16 \).

Step2: Apply the geometric mean theorem

By the geometric mean theorem (leg - hypotenuse segment), \( x^2 = (hypotenuse) \times (segment\ adjacent\ to\ x) \). So \( x^2 = 25 \times 16 \).

Step3: Calculate \( x \)

\( x^2 = 400 \), so \( x = \sqrt{400} = 20 \).

Answer:

20