Question

find uv.
uv =
submit

Explanation:

Step1: Identify Similar Triangles

Since \( VT \parallel WS \), triangles \( UVT \) and \( UWS \) are similar by the Basic Proportionality Theorem (Thales' theorem). So, the ratios of corresponding sides are equal.

Step2: Set Up Proportion

Let \( UV = x \). Then \( UW = UV + VW = x + 69 \)? Wait, no, wait. Wait, the sides: \( US \) is split into \( UT \) and \( TS \). Wait, \( US = UT + TS \)? Wait, no, looking at the diagram: \( UT \) and \( TS \)? Wait, no, the lengths: \( US \) has segments \( UT \) (wait, no, the right side: \( U \) to \( T \) to \( S \)? Wait, no, the diagram: \( U \) is the top vertex, \( S \) is the bottom right, \( W \) is bottom left. \( V \) is on \( UW \), \( T \) is on \( US \). So \( UW \) has length \( 69 \)? Wait, no, the left side: \( W \) to \( V \) to \( U \), with \( WV = 69 \)? Wait, no, the markings: the left side has length 69 (from \( W \) to the top segment), and the right side has length 54 (from \( S \) to the top segment), with \( TS = 36 \) and \( UT \) is the top part? Wait, no, let's re-examine.

Wait, the segments: \( UW \) (left side) is divided into \( UV \) and \( VW \), but the marking shows \( VW = 69 \)? No, the left side: from \( W \) to the top (the segment with tick marks) is 69, and the right side: from \( S \) to the top (tick marks) is 54, with \( TS = 36 \) and \( UT \) is the top part. Wait, actually, since \( VT \parallel WS \), triangles \( UVT \) and \( UWS \) are similar. So the ratio of \( UT \) to \( US \) should equal the ratio of \( UV \) to \( UW \). Wait, \( US = UT + TS \)? Wait, no, \( US \) is composed of \( UT \) and \( TS \)? Wait, the right side: the total length from \( U \) to \( S \) is \( UT + TS \), but the outer segment (from \( S \) to the top tick mark) is 54, and \( TS = 36 \). Wait, maybe the outer segments: \( UW \) (left) has length \( UV + VW = UV + 69 \)? No, that can't be. Wait, maybe the tick marks indicate that \( VW \) and the segment from \( S \) to the top are corresponding (same tick marks), so \( VW = 54 \)? No, the left tick mark is on \( W \) to \( V \) (length 69), right tick mark on \( S \) to \( T \)? No, the diagram: left side: \( W \) to \( V \) to \( U \), with \( WV = 69 \) (tick marks), right side: \( S \) to \( T \) to \( U \), with \( ST = 36 \) and \( TU \) (top) and \( TS = 36 \), and the outer segment (from \( S \) to the top tick mark) is 54. Wait, so \( US = UT + TS \), and the outer segment (from \( S \) to the top) is 54, so \( UT + TS = 54 \)? No, \( TS = 36 \), so \( UT = 54 - 36 = 18 \)? Wait, no, that doesn't make sense. Wait, maybe the ratio is \( \frac{UT}{US} = \frac{UV}{UW} \). Wait, \( US = 54 \)? No, the right side: the segment from \( S \) to the top (tick mark) is 54, and \( TS = 36 \), so \( UT = 54 - 36 = 18 \)? Wait, no, maybe the total length of \( US \) is 54, with \( TS = 36 \), so \( UT = 54 - 36 = 18 \)? Wait, no, that can't be. Wait, let's use the Basic Proportionality Theorem (Thales' theorem) which states that if a line is drawn parallel to one side of a triangle, intersecting the other two sides, then it divides those sides proportionally. So in triangle \( UWS \), \( VT \parallel WS \), so \( \frac{UV}{UW} = \frac{UT}{US} \).

Wait, \( UW \) is the left side: \( UV + VW = UV + 69 \). \( US \) is the right side: \( UT + TS \). But the outer segments (with tick marks) are \( VW = 69 \) and \( ST \) (wait, no, the right side's outer segment is 54, with \( TS = 36 \). Wait, maybe the tick marks indicate that \( VW \) (left outer) and \( ST \) (right outer) are corresponding, but no, the left outer is 69, right outer is 54, and \( TS = 36 \). Wait, perhaps \( UW = 69 + UV \) and \( US = 54 + UT \)? No, that's confusing. Wait, let's look at the lengths:

Wait, the key is that triangles \( UVT \) and \( UWS \) are similar. So the ratio of \( UT \) to \( US \) is equal to the ratio of \( UV \) to \( UW \). Wait, \( US = UT + TS \), but \( TS = 36 \), and the length from \( S \) to the top (tick mark) is 54, so \( UT = 54 - 36 = 18 \)? No, that's not right. Wait, maybe \( US \) is composed of \( UT \) (top part) and \( TS = 36 \), and the total length from \( U \) to \( S \) is \( UT + 36 \), but the outer segment (from \( S \) to the top tick mark) is 54, so \( UT = 54 - 36 = 18 \)? No, that's not. Wait, maybe the tick marks indicate that \( VW = 54 \) and \( ST = 69 \)? No, the diagram shows left side 69, right side 54, \( TS = 36 \).

Wait, let's start over. Since \( VT \parallel WS \), by the Basic Proportionality Theorem (Thales' theorem), \( \frac{UV}{UW} = \frac{UT}{US} \).

We know \( UW = UV + VW \), but \( VW = 69 \) (from the left side tick mark). \( US = UT + TS \), and \( TS = 36 \), and the right side tick mark segment is 54, so \( UT = 54 - 36 = 18 \)? No, that's not. Wait, maybe \( UW \) is the entire left side (69) and \( US \) is the entire right side (54), but \( TS = 36 \), so \( UT = US - TS = 54 - 36 = 18 \). Then, since \( VT \parallel WS \), triangles \( UVT \sim UWS \), so \( \frac{UV}{UW} = \frac{UT}{US} \). Wait, \( UW = 69 \)? No, \( UW \) is \( UV + VW \), but \( VW = 69 \)? No, the left side: from \( W \) to \( U \) is \( VW + UV \), with \( VW = 69 \). The right side: from \( S \) to \( U \) is \( ST + TU \), with \( ST = 36 \) and \( TU \) is the top part. The tick marks on the left ( \( W \) to \( V \)) and right ( \( S \) to \( T \))? No, the tick marks are on the outer segments: left outer ( \( W \) to the top) is 69, right outer ( \( S \) to the top) is 54. So \( VW = 69 \), \( ST = 54 \)? No, \( TS = 36 \), so \( ST = 36 \), and the outer segment ( \( S \) to top) is 54, so \( TU = 54 - 36 = 18 \). Then, \( UW = UV + VW = UV + 69 \), \( US = TU + ST = 18 + 36 = 54 \). Wait, no, \( US \) should be \( TU + ST = 18 + 36 = 54 \), which matches the outer segment (54). And \( UW = UV + VW = UV + 69 \). Then, by similarity, \( \frac{UV}{UW} = \frac{TU}{US} \). So \( \frac{UV}{UV + 69} = \frac{18}{54} \). Simplify \( \frac{18}{54} = \frac{1}{3} \). So:

\( \frac{UV}{UV + 69} = \frac{1}{3} \)

Cross-multiplying: \( 3 \cdot UV = UV + 69 \)

Subtract \( UV \) from both sides: \( 2 \cdot UV = 69 \)

Wait, that can't be, 69 is not divisible by 2. Wait, I must have messed up the segments.

Alternative approach: Maybe the tick marks indicate that \( UW = 69 \) and \( US = 54 \), and \( TS = 36 \), so \( UT = US - TS = 54 - 36 = 18 \). Then, since \( VT \parallel WS \), triangles \( UVT \) and \( UWS \) are similar, so \( \frac{UV}{UW} = \frac{UT}{US} \). So \( \frac{UV}{69} = \frac{18}{54} \). Simplify \( \frac{18}{54} = \frac{1}{3} \), so \( UV = 69 \times \frac{1}{3} = 23 \). Wait, that works. Let's check:

If \( UV = 23 \), then \( UW = UV + VW = 23 + 69 = 92 \)? No, wait, no, maybe \( UW \) is 69 (the entire left side), and \( US \) is 54 (entire right side), with \( TS = 36 \), so \( UT = 54 - 36 = 18 \). Then similarity ratio: \( \frac{UT}{US} = \frac{18}{54} = \frac{1}{3} \), so \( \frac{UV}{UW} = \frac{1}{3} \), so \( UV = \frac{1}{3} \times UW \). But \( UW \) is 69? Then \( UV = 23 \). Let's verify:

If \( UV = 23 \), \( UW = 23 + 69 = 92 \)? No, that's not. Wait, maybe \( UW \) is 69 (the segment from \( U \) to \( W \) is 69), and \( US \) is 54 (from \( U \) to \( S \) is 54), with \( TS = 36 \), so \( UT = US - TS = 54 - 36 = 18 \). Then, since \( VT \parallel WS \), triangles \( UVT \sim UWS \), so \( \frac{UV}{UW} = \frac{UT}{US} \). So \( \frac{UV}{69} = \frac{18}{54} \). \( 18/54 = 1/3 \), so \( UV = 69 \times (1/3) = 23 \). Yes, that makes sense. So the length of \( UV \) is 23.

Step3: Solve for \( UV \)

We set up the proportion using similar triangles:
\[ \frac{UV}{UW} = \frac{UT}{US} \]
We know \( UW = 69 \), \( US = 54 \), \( TS = 36 \), so \( UT = US - TS = 54 - 36 = 18 \). Substituting the values:
\[ \frac{UV}{69} = \frac{18}{54} \]
Simplify \( \frac{18}{54} = \frac{1}{3} \), so:
\[ UV = 69 \times \frac{1}{3} = 23 \]

Answer:

\( \boxed{23} \)