Question

find uw.
w 16 x 8 y
v
16
u
uw =

Explanation:

Step1: Identify Midline Theorem

The segments with arrows are parallel, so by the Midline Theorem (or Basic Proportionality Theorem), \( X \) and \( V \) are midpoints? Wait, no—wait, the lengths \( WX = 16 \), \( XY = 8 \), and \( VU = 16 \). Wait, actually, the lines \( VX \) and \( UY \) are parallel, so the triangles or trapezoid? Wait, \( WX = 16 \), \( XY = 8 \), so \( WY = WX + XY = 24 \)? No, wait, maybe \( V \) divides \( UW \) such that \( VU = 16 \), and \( X \) divides \( WY \) into \( 16 \) and \( 8 \). Wait, the key is that the two parallel lines (with arrows) imply that \( \triangle UVY \sim \triangle WVX \) by AA similarity (since \( \angle U \) is common and the lines are parallel, so corresponding angles are equal). So the ratio of sides: \( XY / WX = VU / VW \)? Wait, no, \( WX = 16 \), \( XY = 8 \), so \( XY = \frac{1}{2} WX \)? Wait, \( 8 = \frac{1}{2} \times 16 \), so the ratio of \( XY \) to \( WX \) is \( 1:2 \). Then, since the lines are parallel, the ratio of \( VU \) to \( UW \) should be \( 1:2 \)? Wait, \( VU = 16 \), so if \( VU / UW = 1/2 \), then \( UW = 2 \times VU = 32 \)? Wait, no, wait: \( VU \) is part of \( UW \), so \( UW = UV + VW \)? No, \( V \) is on \( UW \), so \( UW = UV + VW \)? Wait, no, the diagram: \( U \) is at the bottom, \( V \) is on \( UW \), \( W \) is at the top. \( X \) is on \( WY \), \( Y \) is at the end. The two parallel lines are \( VX \) and \( UY \). So by the Basic Proportionality Theorem (Thales' theorem), \( \frac{XY}{WX} = \frac{VU}{VW} \). Wait, \( XY = 8 \), \( WX = 16 \), so \( \frac{8}{16} = \frac{1}{2} \). So \( \frac{VU}{VW} = \frac{1}{2} \), so \( VW = 2 \times VU \). But \( VU = 16 \), so \( VW = 32 \)? No, wait, \( UW = UV + VW \)? No, \( V \) is between \( U \) and \( W \), so \( UW = UV + VW \)? Wait, no, \( U \) to \( V \) is \( 16 \), \( V \) to \( W \) is \( x \), so \( UW = 16 + x \). But from the ratio, \( \frac{XY}{WX} = \frac{VU}{VW} \), so \( \frac{8}{16} = \frac{16}{VW} \)? Wait, that would be \( \frac{1}{2} = \frac{16}{VW} \), so \( VW = 32 \). Then \( UW = UV + VW = 16 + 32 = 48 \)? No, that can't be. Wait, maybe I got the ratio reversed. Wait, \( WX = 16 \), \( XY = 8 \), so \( WX = 2 \times XY \). Therefore, the ratio of similarity is \( 2:1 \) for \( \triangle WVX \) and \( \triangle WUY \)? No, wait, \( X \) is closer to \( W \), \( Y \) is further. Wait, maybe \( V \) is the midpoint? No, \( VU = 16 \), so if \( UW = UV + VW \), and \( VW = UV \), then \( UW = 16 + 16 = 32 \). Wait, that makes sense: because \( XY = 8 \), \( WX = 16 \), so \( WX = 2 \times XY \), so the segment \( VU \) (which is 16) should be half of \( UW \)? No, wait, if \( XY \) is half of \( WX \), then \( VU \) is half of \( UW \), so \( UW = 2 \times VU = 32 \). Yes, that makes sense. So \( UW = 32 \).

Step2: Calculate UW

Given \( VU = 16 \) and the ratio from the parallel lines (since \( XY = \frac{1}{2} WX \), the corresponding segment \( VU = \frac{1}{2} UW \)), so \( UW = 2 \times VU = 2 \times 16 = 32 \).

Answer:

\( 32 \)