Question

geometric mean
question 12 of 12 (1 point) | question attempt: 4 of unlimited
in the figure below, find the exact value of z. (do not approximate your answer.)
z =

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 of the right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg. So, if we let the hypotenuse segment adjacent to leg \( z \) be \( x \) and the other segment be \( y \), and the hypotenuse be \( c = 5 \), and the leg be \( a = 3 \), then \( a^{2}=c\times x \)? Wait, no, actually, the leg \( z \) (wait, no, the leg with length 3 and the segment \( z \) and hypotenuse 5: the correct formula is that the leg (let's say length \( l \)) is the geometric mean of the hypotenuse (\( h \)) and the adjacent segment (\( s \)), so \( l^{2}=h\times s \). Wait, in the diagram, the right triangle has hypotenuse 5, one leg is 3, and the segment adjacent to the leg 3 is \( z \). So by the geometric mean theorem (altitude-on-hypotenuse theorem), the leg squared is equal to the hypotenuse times the adjacent segment. So \( 3^{2}=5\times z \)? Wait, no, wait. Wait, the altitude-on-hypotenuse theorem states that in a right triangle, 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. Wait, actually, let's label the triangle: let the right triangle be \( \triangle ABC \) with right angle at \( C \), and altitude \( CD \) to hypotenuse \( AB \), with \( AB = 5 \), \( BC = 3 \), and \( BD = z \). Then by the geometric mean theorem, \( BC^{2}=AB\times BD \). So \( 3^{2}=5\times z \)? Wait, no, that would be if \( CD \) is the altitude, but wait, no, the leg is \( BC = 3 \), hypotenuse \( AB = 5 \), and segment \( BD = z \). So the formula is \( BC^{2}=AB\times BD \), so \( 9 = 5z \)? Wait, that can't be, because then \( z = 9/5 \), but that seems off. Wait, maybe I mixed up the segments. Wait, actually, the correct formula is that in 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 segment of the hypotenuse adjacent to that leg. So if the hypotenuse is \( c \), and the two segments are \( m \) and \( n \) (so \( c = m + n \)), and the legs are \( a \) and \( b \), then \( a^{2}=c\times m \) and \( b^{2}=c\times n \). Wait, no, actually, the correct formula is \( a^{2}=m\times c \) where \( m \) is the segment adjacent to \( a \). Wait, let's check with Pythagoras. Suppose the hypotenuse is 5, one leg is 3, then the other leg (let's say \( AC \)) would be \( \sqrt{5^{2}-3^{2}}=\sqrt{25 - 9}=\sqrt{16}=4 \). Then the altitude \( CD \) can be found by area: area is \( (3\times4)/2 = 6 \), and also \( (5\times CD)/2 = 6 \), so \( CD = 12/5 \). Then the segments: \( AD = \sqrt{4^{2}-(12/5)^{2}}=\sqrt{16 - 144/25}=\sqrt{(400 - 144)/25}=\sqrt{256/25}=16/5 \), and \( BD = \sqrt{3^{2}-(12/5)^{2}}=\sqrt{9 - 144/25}=\sqrt{(225 - 144)/25}=\sqrt{81/25}=9/5 \). Wait, so then the leg \( BC = 3 \), the segment \( BD = 9/5 \), and hypotenuse \( AB = 5 \). So indeed, \( BC^{2}=9 \), and \( AB\times BD = 5\times(9/5)=9 \), so that works. Wait, so then \( z = BD = 9/5 \)? But that seems like a fraction. Wait, but maybe the diagram is different. Wait, the problem says "find the exact value of \( z \)". Wait, maybe I mislabeled the triangle. Wait, the diagram shows a right triangle with a right angle at the top, and an altitude to the hypotenuse, making two smaller right triangles. So the big triangle is right-angled at the top, so let's call the top vertex \( A \), bottom left \( B \), bottom right \( C \), and the foot of the altitude \( D \) on \( BC \) (wait, no, the hypotenuse is \( BC \) with length 5? Wait, no, the diagram shows a right triangle with a right angle at the top, and an altitude to the base (hypotenuse) of length 5, with one of the segments being \( z \), and the leg adjacent to \( z \) is 3. Wait, maybe the correct formula is that the leg (length 3) is the geometric mean of the hypotenuse segment \( z \) and the entire hypotenuse? No, that's not right. Wait, let's re-express: in a right triangle, if you have a right triangle with hypotenuse \( c \), and a segment \( s \) of the hypotenuse, and a leg \( l \) such that the leg is adjacent to the segment \( s \), then \( l^{2}=c\times s \). Wait, in the diagram, the hypotenuse is 5, the segment is \( z \), and the leg is 3. So \( 3^{2}=5\times z \), so \( z = 9/5 \). Wait, but that seems correct. Let's verify with Pythagoras. If \( z = 9/5 \), then the other segment of the hypotenuse is \( 5 - 9/5 = 16/5 \). Then the other leg (let's say \( x \)) would satisfy \( x^{2}=5\times(16/5)=16 \), so \( x = 4 \). Then the original right triangle has legs 3 and 4, hypotenuse 5, which is a Pythagorean triple (3-4-5 triangle). Then the altitude to the hypotenuse is \( (3\times4)/5 = 12/5 \), which matches the area calculation. Then the segments of the hypotenuse are \( (3^{2})/5 = 9/5 \) and \( (4^{2})/5 = 16/5 \), which are the lengths adjacent to each leg. So yes, the formula is that each leg squared is equal to the hypotenuse times the adjacent segment. So for the leg of length 3, the adjacent segment is \( z \), so \( 3^{2}=5\times z \), so \( z = 9/5 \).

Step1: Apply Geometric Mean Theorem

In a right triangle, a leg is the geometric mean of the hypotenuse and the adjacent hypotenuse segment. Let the hypotenuse be \( 5 \), the leg be \( 3 \), and the adjacent segment be \( z \). By the theorem:
\( 3^2 = 5 \times z \)

Step2: Solve for \( z \)

Simplify the equation:
\( 9 = 5z \)
Divide both sides by \( 5 \):
\( z = \frac{9}{5} \)

Answer:

\( \boxed{\dfrac{9}{5}} \)