Question

tu || sv. find uv. 60 u v w s 36 12 t uv =

Explanation:

Step1: Identify Similar Triangles

Since \( TU \parallel SV \), by the Basic Proportionality Theorem (Thales' theorem) or AA similarity criterion, \(\triangle TUV \sim \triangle SWV\) (or we can consider the segments on the sides). The ratio of corresponding sides should be equal. The sides \( TS = 12 \) and \( TW = TS + SW = 12 + 36 = 48 \)? Wait, no, looking at the diagram, the sides with lengths 12 and 36 (with the same tick marks) – actually, the segments on the side \( TW \) (or \( TT' \)? Wait, the diagram has \( T \), \( S \), and the side with length 36 (from \( S \) to the other end with tick mark) and \( TS = 12 \). So the ratio of \( TS \) to the total length (let's say \( TW \) related to \( TU \))? Wait, actually, since \( TU \parallel SV \), the triangles \( \triangle TUV \) and \( \triangle SWV \) are similar? Wait, no, maybe \( \triangle UTV \) and \( \triangle VSW \)? Wait, better to use the segment ratios. The length from \( T \) to \( S \) is 12, and from \( S \) to the other end (let's say \( W \)'s side) is 36? Wait, the two sides with tick marks: the top side \( UW \) (wait, no, the top side is \( U \) to \( W \) with \( V \) in between, and the side from \( T \) to the right has length 36 (with tick mark) and \( TS = 12 \). So the ratio of \( TS \) to \( TW \) (if \( TW = TS + SW \), but \( SW \) is 36? Wait, no, the two segments on the side (the one with \( T \), \( S \)): \( TS = 12 \), and the other segment (from \( S \) to the end with tick mark) is 36. So the ratio of \( TS \) to the total length (12 + 36 = 48)? Wait, no, the top side \( UW \) (or \( U \) to \( W \)) has length 60? Wait, the top horizontal segment (with tick mark) is 60, from \( U \) to \( W \)? Wait, the diagram: \( U \)---\( V \)---\( W \) (top side), length 60? Wait, no, the top segment is \( U \) to \( W \) with \( V \) in between, and the length from \( U \) to \( W \) is 60? Wait, the problem is to find \( UV \). Let's denote \( UV = x \), then \( VW = 60 - x \) (since \( UW = 60 \)). Now, since \( TU \parallel SV \), by the Basic Proportionality Theorem (Thales' theorem), the line \( SV \) divides the sides \( TU \) and \( TW \) proportionally? Wait, no, \( SV \) is parallel to \( TU \), so in triangle \( TUV \), \( SV \) is parallel to \( TU \), so the ratio of \( TS \) to \( TV \)? Wait, maybe the sides: \( TS = 12 \), \( TW = 12 + 36 = 48 \)? Wait, no, the side with length 36 is from \( S \) to the end (let's say \( W \)) with the same tick mark as the top side (60). So the ratio of the segments on the side (12 and 36) is \( 12:36 = 1:3 \). Wait, no, \( 12 + 36 = 48 \), so \( TS:TW = 12:48 = 1:4 \)? Wait, no, the top side is 60, and we need to find \( UV \). Let's use similar triangles. Since \( TU \parallel SV \), \( \triangle TUV \sim \triangle VSW \) (angle-angle similarity: \( \angle TUV = \angle VSW \) (alternate interior angles), \( \angle UTV = \angle SVW \) (alternate interior angles)). So the ratio of corresponding sides is equal. So \( \frac{TS}{TW} = \frac{UV}{VW} \)? Wait, no, \( TS = 12 \), \( SW = 36 \), so \( TS:SW = 12:36 = 1:3 \). Wait, \( UV \) and \( VW \) are segments on \( UW \) (length 60). Let \( UV = x \), then \( VW = 60 - x \). Then, by similar triangles, \( \frac{TS}{SW} = \frac{UV}{VW} \)? Wait, no, \( TS \) is 12, \( SW \) is 36, so \( \frac{TS}{SW} = \frac{12}{36} = \frac{1}{3} \). So \( \frac{UV}{VW} = \frac{1}{3} \)? Wait, no, maybe the other way. Wait, \( TS \) is part of \( TW \), and \( UV \) is part of \( UW \). Wait, let's re-express: the two triangles are similar, so the ratio of \( TS \) to \( TW \) (if \( TW = TS + SW = 12 + 36 = 48 \)) and the ratio of \( UV \) to \( UW \) (which is 60). Wait, no, \( UW \) is 60? Wait, the top segment (with tick mark) is 60, from \( U \) to \( W \). So \( U \)---\( V \)---\( W \), length \( UV + VW = 60 \). The side from \( T \) to \( S \) is 12, and from \( S \) to the end (let's say \( W \)'s side) is 36. So the ratio of \( TS \) to \( SW \) is \( 12:36 = 1:3 \). Since \( TU \parallel SV \), the triangles \( \triangle UTV \) and \( \triangle VSW \) are similar, so \( \frac{UV}{VW} = \frac{TS}{SW} \)? Wait, no, \( \frac{UV}{VW} = \frac{TS}{SW} \) would be \( \frac{x}{60 - x} = \frac{12}{36} = \frac{1}{3} \). Then solving \( 3x = 60 - x \), \( 4x = 60 \), \( x = 15 \)? Wait, that can't be. Wait, maybe the ratio is reversed. Let's think again. The length from \( T \) to \( S \) is 12, and from \( S \) to the other end (let's say \( W \)) is 36, so the total length from \( T \) to \( W \) (along that side) is \( 12 + 36 = 48 \). The top side is \( U \) to \( W \), length 60. Since \( TU \parallel SV \), the triangles \( \triangle TUW \) and \( \triangle SVW \) are similar? Wait, \( TU \parallel SV \), so \( \angle TUW = \angle SVW \) (corresponding angles), and \( \angle TWU = \angle SWV \) (common angle). So \( \triangle TUW \sim \triangle SVW \). Then the ratio of sides is \( \frac{TW}{SW} = \frac{TU}{SV} = \frac{UW}{VW} \). Wait, \( TW = TS + SW = 12 + 36 = 48 \), \( SW = 36 \). So \( \frac{TW}{SW} = \frac{48}{36} = \frac{4}{3} \). Then \( \frac{UW}{VW} = \frac{4}{3} \). But \( UW = UV + VW = 60 \), so \( \frac{60}{VW} = \frac{4}{3} \), so \( VW = \frac{60 \times 3}{4} = 45 \), then \( UV = 60 - 45 = 15 \)? No, that's the same. Wait, but that seems low. Wait, maybe the top length is not \( UW \) but \( U \) to the other end. Wait, the diagram: the top horizontal segment (with tick mark) is 60, from \( U \) to \( W \). The side from \( T \) to the right has length 36 (with tick mark) and \( TS = 12 \). So the two segments on the side (12 and 36) have a ratio of \( 12:36 = 1:3 \). Since \( TU \parallel SV \), the segments \( UV \) and \( VW \) should have the same ratio? Wait, no, the ratio of the segments on the transversal (the side with \( T \), \( S \)) is \( 12:36 = 1:3 \), so the ratio of \( UV \) to \( VW \) should be \( 3:1 \)? Wait, maybe I mixed up the ratio. Let's use the Basic Proportionality Theorem (Thales' theorem): 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 \( TUW \), \( SV \) is parallel to \( TU \), intersecting \( TW \) at \( S \) and \( UW \) at \( V \). So \( \frac{TS}{SW} = \frac{UV}{VW} \). Wait, \( TS = 12 \), \( SW = 36 \), so \( \frac{12}{36} = \frac{UV}{VW} \), so \( \frac{UV}{VW} = \frac{1}{3} \). Let \( UV = x \), \( VW = 3x \). Then \( x + 3x = 60 \) (since \( UW = 60 \)), so \( 4x = 60 \), \( x = 15 \). Wait, but that seems small. Wait, maybe the top length is not \( UW \) but \( U \) to \( W \) as 60, and the side from \( T \) to \( W \) is 36 + 12 = 48. Then the ratio of \( UW \) to \( VW \) would be \( 60:VW = 48:36 \), so \( 60/VW = 48/36 \), \( VW = (60 \times 36)/48 = 45 \), so \( UV = 60 - 45 = 15 \). Wait, that's the same. But maybe I made a mistake. Wait, let's check the ratio again. \( TS = 12 \), \( SW = 36 \), so \( TS/SW = 12/36 = 1/3 \). So by Thales' theorem, \( UV/VW = TS/SW = 1/3 \), so \( UV = (1/3)VW \). Then \( UV + VW = 60 \), so \( (1/3)VW + VW = 60 \), \( (4/3)VW = 60 \), \( VW = 45 \), \( UV = 15 \). Wait, but that seems correct? Wait, no, maybe the other way. If \( TU \parallel SV \), then the ratio of \( TW \) to \( TS \) is \( (12 + 36)/12 = 48/12 = 4 \), so the ratio of \( UW \) to \( UV \) is 4, so \( UW = 4 \times UV \), so \( 60 = 4 \times UV \), so \( UV = 15 \). Yes, that makes sense. So \( UV = 15 \)? Wait, no, wait, \( UW \) is 60, and \( UW = UV + VW \), and \( VW = 3 \times UV \), so \( UV + 3UV = 60 \), \( 4UV = 60 \), \( UV = 15 \). Wait, but let's confirm with the diagram. The top segment is 60 (from \( U \) to \( W \)), \( V \) is between \( U \) and \( W \). The side from \( T \) to \( S \) is 12, and from \( S \) to the end (with tick mark) is 36. So the ratio of \( TS \) to \( SW \) (36) is 1:3, so \( UV \) to \( VW \) is 1:3, so \( UV = 15 \), \( VW = 45 \), and \( 15 + 45 = 60 \), which matches the top length. So that's correct.

Step2: Solve for \( UV \)

Let \( UV = x \) and \( VW = 60 - x \). Since \( TU \parallel SV \), by the Basic Proportionality Theorem, \(\frac{TS}{SW} = \frac{UV}{VW}\). We know \( TS = 12 \), \( SW = 36 \), so:

\[ \frac{12}{36} = \frac{x}{60 - x} \]

Simplify \(\frac{12}{36} = \frac{1}{3}\), so:

\[ \frac{1}{3} = \frac{x}{60 - x} \]

Cross-multiplying:

\[ 60 - x = 3x \]

Add \( x \) to both sides:

\[ 60 = 4x \]

Divide both sides by 4:

\[ x = 15 \]

Answer:

\( UV = \boxed{15} \)