Question

use the geometric mean to find x round your answer to the nearest tenth

Explanation:

Step1: Recall geometric mean in right triangles

In a right triangle, when an altitude is drawn to the hypotenuse, the length of a leg is the geometric mean of the lengths of the hypotenuse segments adjacent to that leg. Here, the leg \( x \) is adjacent to segments of length 3 and \( 3 + 6=9 \)? Wait, no, wait. Wait, the hypotenuse is split into two segments: one is 3, the other is 6? Wait, no, looking at the diagram, the hypotenuse is divided into a segment of length 3 and the other part? Wait, no, the two segments of the hypotenuse are 3 and \( 3 + 6 \)? Wait, no, the triangle has a leg \( x \), and the hypotenuse is split into 3 and \( 3 + 6 \)? Wait, no, actually, the two segments of the hypotenuse are 3 and 6? Wait, no, the altitude divides the hypotenuse into two parts: let's say the hypotenuse is \( 3 + 6 = 9 \)? No, wait, the diagram shows one segment as 3, and the other part (the upper segment) as 6? Wait, no, the leg \( x \) is a leg of the smaller right triangle, and the two segments of the hypotenuse are 3 and \( 3 + 6 \)? Wait, no, I think the correct segments are 3 and \( 3 + 6 = 9 \)? No, wait, the geometric mean formula for a leg \( x \) is \( x=\sqrt{segment_1\times(segment_1 + segment_2)} \)? No, wait, no. The correct theorem is: In a right triangle, the length of a leg is the geometric mean of the length of the hypotenuse and the length of the adjacent segment. Wait, let's re - state: If we have a right triangle, and we draw an altitude from the right angle to the hypotenuse, then each leg is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg. So, if the hypotenuse is \( c\), and it is split into segments \( a\) and \( b\) (so \( c=a + b\)), then leg \( l_1=\sqrt{a\times c}\) and leg \( l_2=\sqrt{b\times c}\). Wait, no, actually, the correct formula is \( l_1=\sqrt{a\times(a + b)}\) where \( a\) is the adjacent segment and \( a + b\) is the hypotenuse? No, I think I made a mistake. Let's look at the diagram again. The large right triangle has a hypotenuse that is split into two segments: one is 3, and the other is 6? Wait, no, the two segments are 3 and \( 3+6 = 9\)? No, the diagram shows that one segment is 3, and the other part (the upper part) is 6? Wait, no, the leg \( x\) is adjacent to the segment of length 3 and the entire hypotenuse? Wait, no, the correct formula is: If the hypotenuse is divided into segments of length \( m\) and \( n\), then the length of a leg \( l\) is \( l=\sqrt{m\times(m + n)}\)? No, wait, the correct formula is \( l=\sqrt{m\times(m + n)}\) is wrong. The correct formula is that each leg is the geometric mean of the hypotenuse and the adjacent segment. So, if the hypotenuse is \( m + n\), and the adjacent segment to leg \( l\) is \( m\), then \( l=\sqrt{m\times(m + n)}\). Wait, in the diagram, the two segments of the hypotenuse are 3 and 6? Wait, no, the diagram has a segment of length 3, and the other segment (the one above the altitude) is 6? Wait, no, the total hypotenuse length is \( 3+6 = 9\)? Wait, no, the leg \( x\) is a leg of the right triangle, and the two segments of the hypotenuse are 3 and \( 3 + 6=9\)? No, I think the correct segments are 3 and 6. Wait, no, let's check the geometric mean formula for right triangles: In a right triangle, when an altitude is drawn to the hypotenuse, then \( leg^2=segment_1\times segment_2\)? No, that's for the altitude. Wait, no, the altitude \( h\) satisfies \( h = \sqrt{segment_1\times segment_2}\), and each leg \( l_1=\sqrt{segment_1\times hypotenuse}\), \( l_2=\sqrt{segment_2\times hypotenuse}\). Wait, I think I confused the formulas. Let's start over.

Given a right triangle with hypotenuse \( c\), and an altitude \( h\) to the hypotenuse, splitting \( c\) into segments \( a\) and \( b\) (so \( c=a + b\)). Then:

1. \( h=\sqrt{a\times b}\) (altitude is geometric mean of the two segments)
2. \( l_1=\sqrt{a\times c}\) (leg \( l_1\) is geometric mean of segment \( a\) and hypotenuse \( c\))
3. \( l_2=\sqrt{b\times c}\) (leg \( l_2\) is geometric mean of segment \( b\) and hypotenuse \( c\))

In the diagram, the leg \( x\) is adjacent to the segment of length 3, and the hypotenuse is \( 3 + 6=9\)? Wait, no, the two segments of the hypotenuse are 3 and 6? Wait, the diagram shows one segment as 3, and the other part (the upper segment) as 6. So the hypotenuse is \( 3+6 = 9\). Then, the leg \( x\) is adjacent to the segment of length 3, so \( x=\sqrt{3\times(3 + 6)}=\sqrt{3\times9}=\sqrt{27}\approx5.2\)? Wait, no, that can't be. Wait, maybe the two segments are 3 and 6, and the leg \( x\) is the geometric mean of 3 and \( 3 + 6\)? No, wait, maybe I got the segments wrong. Wait, the correct segments: the leg \( x\) is a leg of the right triangle, and the two segments of the hypotenuse are 3 and 6. Wait, no, the formula for the leg is \( x=\sqrt{3\times(3 + 6)}\)? No, let's check the theorem again. The geometric mean theorem (also known as the altitude - on - hypotenuse theorem) states that: In a right triangle, the altitude drawn to the hypotenuse is the geometric mean between the segments of the hypotenuse. Also, each leg of the right triangle is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg. So, if the hypotenuse is \( c\), and it is divided into segments \( m\) and \( n\) (so \( c=m + n\)), then for leg \( l_1\) (adjacent to segment \( m\)): \( l_1^2=m\times c=m\times(m + n)\), and for leg \( l_2\) (adjacent to segment \( n\)): \( l_2^2=n\times c=n\times(m + n)\).

In the given diagram, the two segments of the hypotenuse are \( m = 3\) and \( n=6\), so the hypotenuse \( c=3 + 6 = 9\). The leg \( x\) is adjacent to the segment \( m = 3\), so we use the formula \( x=\sqrt{m\times c}=\sqrt{3\times9}=\sqrt{27}\approx5.2\)? Wait, no, that's not right. Wait, maybe the segments are 3 and \( 3 + 6\) is wrong. Wait, maybe the two segments are 3 and 6, and the leg \( x\) is the geometric mean of 3 and \( 3+6\)? No, wait, let's look at the diagram again. The triangle has a right angle, and an altitude is drawn, creating two smaller right triangles. The leg \( x\) is a leg of the larger right triangle? No, the leg \( x\) is a leg of the smaller right triangle? Wait, no, the diagram shows that the leg \( x\) is a leg of the triangle with one segment of the hypotenuse as 3 and the other part (the upper segment) as 6. Wait, I think I made a mistake in the segments. Let's re - express: Let the hypotenuse of the large right triangle be \( AB\), with \( D\) as the foot of the altitude from \( C\) (the right angle) to \( AB\). Then \( AD = 3\), \( DB=6\), and \( AC=x\). Then by the geometric mean theorem, \( AC^{2}=AD\times AB\), where \( AB = AD+DB=3 + 6 = 9\). So \( x^{2}=3\times9 = 27\), so \( x=\sqrt{27}\approx5.2\) (rounded to the nearest tenth).

Step2: Calculate the square root

We have \( x=\sqrt{27}\). Since \( \sqrt{25}=5\) and \( \sqrt{36}=6\), \( \sqrt{27}\approx5.2\) (because \( 5.2^{2}=27.04\) which is close to 27).

Answer:

\( 5.2\)