Question

applying the geometric mean (altitude) theorem
what is the value of ( k )?
( k = square )
diagram: triangle ( lnm ) with right angles at ( n ) and ( o ); segment ( lo = 8 ), ( on = 4 ), ( om = k )

Explanation:

Step1: Recall Geometric Mean Theorem

The Geometric Mean (Altitude) Theorem states that in a right triangle, the length of the altitude drawn to the hypotenuse is the geometric mean of the lengths of the two 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. Here, we can use the theorem for the segments of the hypotenuse. If we consider the hypotenuse \( LM \) with segments \( LO = 8 \) and \( OM = k \), and the altitude \( NO = 4 \), then by the geometric mean theorem, \( NO^2=LO\times OM \).

Step2: Substitute the known values

We know \( NO = 4 \) and \( LO = 8 \), so substitute into the formula \( 4^2=8\times k \).

Step3: Solve for \( k \)

First, calculate \( 4^2 = 16 \), so the equation becomes \( 16 = 8k \). Then, divide both sides by 8: \( k=\frac{16}{8}=2 \)? Wait, no, wait. Wait, maybe I mixed up the segments. Wait, actually, the correct application: In a right triangle, when you draw an altitude from the right angle to the hypotenuse, the square of the altitude is equal to the product of the lengths of the two segments of the hypotenuse. Wait, but also, the leg squared is equal to the product of the hypotenuse segment and the whole hypotenuse. Wait, maybe I made a mistake. Wait, let's re - examine the triangle. Let's assume triangle \( LNM \) is right - angled at \( N \), and \( NO \) is the altitude to hypotenuse \( LM \). Then, the segments of the hypotenuse are \( LO \) and \( OM \), with \( LO = 8 \), \( OM = k \), and altitude \( NO = 4 \). Then by the geometric mean (altitude) theorem: \( NO^{2}=LO\times OM \). So \( 4^{2}=8\times k \), \( 16 = 8k \), \( k=\frac{16}{8}=2 \)? Wait, that seems too small. Wait, maybe the segments are \( LO \) and \( OM \), but maybe \( LO \) is one segment, and the other segment is \( k \), and the altitude is 4. Wait, no, maybe the formula is that the length of the segment adjacent to the leg? Wait, no, let's recall the exact theorem: In a right triangle, if an altitude is drawn from the right angle to the hypotenuse, then:

1. The length of the altitude is the geometric mean of the lengths of the two segments of the hypotenuse. That is, \( h^{2}=p\times q \), where \( h \) is the altitude, \( p \) and \( q \) are the segments of the hypotenuse.

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. That is, \( a^{2}=c\times p \) and \( b^{2}=c\times q \), where \( a \) and \( b \) are the legs, \( c \) is the hypotenuse, \( p \) and \( q \) are the segments.

Wait, in the diagram, \( LO = 8 \), \( OM = k \), \( NO = 4 \). So according to the first part of the theorem, \( NO^{2}=LO\times OM \), so \( 4^{2}=8\times k \), so \( 16 = 8k \), so \( k = 2 \)? Wait, but maybe I got the segments wrong. Wait, maybe \( LO \) is not 8, but the whole hypotenuse? No, the diagram shows \( LO = 8 \), \( OM = k \), so the hypotenuse \( LM=LO + OM=8 + k \). And \( NO \) is the altitude. Wait, maybe the leg \( LN \) has length \( m \), and \( LN^{2}=LO\times LM \), and \( MN^{2}=OM\times LM \), and \( NO^{2}=LO\times OM \). So using \( NO^{2}=LO\times OM \), \( 4^{2}=8\times k \), so \( k = \frac{16}{8}=2 \). Wait, but that seems odd. Wait, maybe the problem is that \( LO = 8 \), and \( NO = 4 \), and we need to find \( OM = k \). So by the geometric mean theorem, \( NO^2=LO\times OM \), so \( 4^2 = 8\times k \), so \( 16=8k \), so \( k = 2 \). Wait, but maybe I made a mistake in the theorem. Let me check again. The geometric mean theorem (also called the altitude - on - hypotenuse theorem) states that in a right - angled triangle, the altitude drawn to the hypotenuse is the geometric mean between the segments into which it divides the hypotenuse. So if the hypotenuse is divided into segments of length \( x \) and \( y \), and the altitude is \( h \), then \( h^{2}=x\times y \). In this case, \( x = 8 \), \( h = 4 \), so \( 4^{2}=8\times y \) (where \( y = k \)), so \( 16 = 8y \), so \( y=k = 2 \). Wait, but that seems too small. Wait, maybe the segment \( LO \) is not 8, but the other segment? Wait, maybe the diagram is such that \( LO = 4 \) and \( OM = k \), but no, the diagram shows \( LO = 8 \), \( NO = 4 \). So according to the theorem, \( k = 2 \). Wait, but maybe I have the formula reversed. Let's think of similar triangles. Triangle \( LNO \) is similar to triangle \( NMO \) (by AA similarity, since both are right - angled and share angle \( M \)). So the ratio of corresponding sides is equal. So \( \frac{LO}{NO}=\frac{NO}{OM} \), which is the same as \( NO^{2}=LO\times OM \). So \( 4^{2}=8\times k \), so \( k=\frac{16}{8}=2 \).

Answer:

\( k = 2 \)