Question

if \\( \overline{tu} \cong \overline{uv} \\) and \\( sv = 35 \\), what is \\( tv \\)?
\\( tv = \square \\)

Explanation:

Step1: Identify Segment Relationships

Since \( \overline{TU} \cong \overline{UV} \), \( U \) is the midpoint of \( TV \). Also, the right angle at \( S \) (perpendicular bisector) implies \( S \) is the midpoint of \( TV \)? Wait, no—wait, the diagram shows a perpendicular bisector (the right angle at \( S \) on the line through \( S \) and \( U \)? Wait, actually, since \( TU \cong UV \), triangle \( TUV \) is isoceles with \( TU = UV \). But also, the segment \( SV = 35 \), and if \( S \) is the midpoint (because the line through \( S \) is a perpendicular bisector, so \( TS = SV \))? Wait, no—wait, the key is: if \( TU \cong UV \), and the line through \( S \) is a perpendicular bisector, then \( S \) is the midpoint of \( TV \), so \( TS = SV \). But also, since \( U \) is the midpoint (from \( TU \cong UV \)), but actually, the problem is that \( SV = 35 \), and since the line is a perpendicular bisector, \( TV = 2 \times SV \)? Wait, no—wait, let's re-express.

Wait, the diagram: there's a segment \( TV \), with a point \( U \) such that \( TU = UV \) (so \( U \) is midpoint of \( TV \)), and a line through \( S \) (with a right angle) intersecting \( TV \) at \( S \) and \( UV \) at \( U \)? Wait, no, the problem says \( SV = 35 \), and we need to find \( TV \). Wait, actually, the perpendicular bisector: if the line through \( S \) is perpendicular to \( TV \) and bisects it, then \( TS = SV \). But also, since \( TU = UV \), \( U \) is the midpoint, so \( TV = 2 \times UV \). Wait, maybe the key is that \( S \) is the midpoint, so \( TS = SV = 35 \), but no—wait, no, the problem is that \( TU \cong UV \), so \( U \) is midpoint, and the line through \( S \) is a perpendicular bisector, so \( S \) is midpoint, so \( TV = 2 \times SV \)? Wait, no, wait: if \( SV = 35 \), and \( S \) is the midpoint, then \( TV = 2 \times SV = 70 \). Wait, that makes sense. Because the perpendicular bisector implies \( TS = SV \), so \( TV = TS + SV = 35 + 35 = 70 \). Also, since \( TU \cong UV \), \( U \) is the midpoint, but the perpendicular bisector at \( S \) (so \( S \) is midpoint) gives \( TS = SV \). So \( TV = 2 \times SV \).

Step2: Calculate \( TV \)

Given \( SV = 35 \), and \( TV = 2 \times SV \) (because \( S \) is the midpoint, so \( TS = SV \)), then \( TV = 2 \times 35 = 70 \).

Answer:

\( 70 \)