Question

find tw.
w v
14 32
x u
16
t
tw =
submit

Explanation:

Step1: Identify Similar Triangles

Since \( XU \parallel WV \) (from the arrows indicating parallel lines), triangles \( TXU \) and \( TWV \) are similar by the Basic Proportionality Theorem (Thales' theorem).

Step2: Set Up Proportion

For similar triangles, the ratios of corresponding sides are equal. So, \(\frac{TU}{TV}=\frac{TX}{TW}\). We know \( TU = 16 \), \( TV=TU + UV=16 + 32 = 48 \), and \( TX=TW - XW \), but since \( XW = 14 \) and \( TX=TW - 14 \)? Wait, no, actually \( TW \) is the side we need, and \( TX \) is part of \( TW \). Wait, let's correct: \( TW \) is the side from \( T \) to \( W \), length 14? No, wait, the length from \( W \) to \( X \) to \( T \): \( WX \) is 14? Wait, no, the diagram: \( W \) to \( T \) has a segment \( WX \) of length 14? Wait, no, the label is 14 on \( WT \), with \( X \) between \( W \) and \( T \). So \( WT = WX + XT \), but \( XU \parallel WV \), so triangles \( TXU \sim TWV \). So the ratio of \( TU \) to \( TV \) is \( \frac{16}{16 + 32}=\frac{16}{48}=\frac{1}{3} \). Then the ratio of \( TX \) to \( TW \) should also be \( \frac{1}{3} \). Let \( TW = x \), then \( TX=x - 14 \)? Wait, no, wait: \( W \) to \( T \) is length \( x \), with \( X \) between \( W \) and \( T \), so \( TX=TW - WX=x - 14 \)? Wait, no, maybe I got the segments wrong. Wait, \( W \) to \( V \) is a horizontal line, \( T \) to \( V \) is length 32 + 16 = 48, \( T \) to \( U \) is 16, \( T \) to \( W \) is the side with length 14? No, the 14 is on \( WT \), so \( WT \) is the side from \( W \) to \( T \), with \( X \) on \( WT \), so \( WX = 14 \), \( XT \) is the other part. Wait, no, the correct proportion: since \( XU \parallel WV \), then \( \frac{TU}{TV}=\frac{TX}{TW} \). Let \( TW = x \), \( TX=x - 14 \)? No, that can't be. Wait, maybe \( WX \) is 14, so \( TW \) is the length from \( T \) to \( W \), so \( TX \) is from \( T \) to \( X \), so \( TX = TW - WX \)? No, \( W \) to \( T \): \( W \)---\( X \)---\( T \), so \( WT = WX + XT \), so \( XT=WT - WX \). Then in similar triangles \( TXU \) and \( TWV \), \( \frac{TU}{TV}=\frac{XT}{WT} \). So \( TU = 16 \), \( TV = 16 + 32 = 48 \), \( XT = WT - 14 \)? Wait, no, the 14 is the length of \( WX \), so \( WT \) is the side from \( W \) to \( T \), so \( WX = 14 \), so \( XT = WT - 14 \)? No, that would mean \( WT \) is longer than 14. Wait, maybe I mixed up the labels. Let's start over:

Triangles \( TXU \) and \( TWV \) are similar (AA similarity, since \( \angle T \) is common, and \( \angle TXU=\angle TWV \) because \( XU \parallel WV \), corresponding angles). So the ratio of sides: \( \frac{TU}{TV}=\frac{TX}{TW} \).

\( TU = 16 \), \( TV = TU + UV = 16 + 32 = 48 \), so \( \frac{16}{48}=\frac{TX}{TW} \), which simplifies to \( \frac{1}{3}=\frac{TX}{TW} \), so \( TW = 3 \times TX \).

But \( TW = TX + XW \), and \( XW = 14 \), so \( TX + 14 = 3 \times TX \).

Subtract \( TX \) from both sides: \( 14 = 2 \times TX \), so \( TX = 7 \). Then \( TW = TX + XW = 7 + 14 = 21 \)? Wait, no, that can't be. Wait, maybe the 14 is the length of \( XT \)? No, the diagram: \( W \) to \( T \) has a segment labeled 14, with \( X \) between \( W \) and \( T \). So \( WT = 14 \)? No, that doesn't make sense. Wait, maybe the 14 is the length of \( WX \), and \( TW \) is the side we need, with \( X \) on \( TW \), so \( TW = WX + XT \), and \( XT \) is related to \( TU \) and \( TV \).

Wait, let's use the correct proportion: in similar triangles, the ratio of the sides adjacent to the common angle are equal. So \( \frac{TU}{TV}=\frac{TX}{TW} \). Let \( TW = x \), \( TX = x - 14 \) (since \( WX = 14 \), so \( TX = TW - WX \)). Then:

\( \frac{16}{16 + 32}=\frac{x - 14}{x} \)

\( \frac{16}{48}=\frac{x - 14}{x} \)

Simplify \( \frac{1}{3}=\frac{x - 14}{x} \)

Cross - multiply: \( x = 3(x - 14) \)

\( x = 3x - 42 \)

Subtract \( x \) from both sides: \( 0 = 2x - 42 \)

Add 42 to both sides: \( 2x = 42 \)

Divide by 2: \( x = 21 \)

Wait, but that gives \( TW = 21 \), but let's check again. Wait, maybe the 14 is the length of \( XT \), not \( WX \). So \( XT = 14 \), then \( TX = 14 \), and \( TW = TX + XW \), but no, the diagram: \( W \)---\( X \)---\( T \), with \( WX = 14 \), \( XT \) is unknown, \( TU = 16 \), \( UV = 32 \). Then \( TV = 48 \), so ratio \( \frac{TU}{TV}=\frac{16}{48}=\frac{1}{3} \), so \( \frac{XT}{WT}=\frac{1}{3} \), \( XT = WT - WX = WT - 14 \), so \( \frac{WT - 14}{WT}=\frac{1}{3} \), solving gives \( WT = 21 \). Yes, that works.

Answer:

\( 21 \)