Question

consider the triangles shown. if ( mangle utv < mangle uts < mangle str ), which statement is true? ( \bigcirc vu < us < sr ) by the hinge theorem. ( \bigcirc vu = us = sr ) by the hinge theorem. ( \bigcirc mangle utv = mangle ust = mangle str ) by the converse of the hinge theorem. ( \bigcirc mangle utv > mangle uts > mangle str ) by the converse of the hinge theorem.

Explanation:

Step1: Recall the Hinge Theorem

The hinge theorem (also known as the SAS inequality theorem) states that if two sides of one triangle are congruent to two sides of another triangle, but the included angle of the first triangle is smaller than the included angle of the second triangle, then the third side of the first triangle is shorter than the third side of the second triangle. Conversely, the converse of the hinge theorem relates the lengths of sides to the measures of included angles.

Step2: Analyze the Given Triangles

From the diagram, we can see that \( \triangle UTV \) and \( \triangle STR \) have two pairs of congruent sides (the marked sides). We are given that \( m\angle UTV < m\angle UTS < m\angle STR \). Wait, actually, looking at the triangles \( \triangle UTV \), \( \triangle UTS \), and \( \triangle STR \), but more precisely, \( \triangle UTV \) and \( \triangle STR \) (and maybe \( \triangle UTS \))—wait, the sides: \( UT \) and \( ST \) are congruent (marked), \( VT \) and \( TR \) are congruent (marked as equal segments on \( VR \)). So for \( \triangle UTV \) and \( \triangle STR \), we have \( UT = ST \), \( VT = TR \), and the included angles are \( \angle UTV \) and \( \angle STR \) respectively. But also, between \( \triangle UTV \), \( \triangle UTS \), the included angles at \( T \) are \( \angle UTV \), \( \angle UTS \), and \( \angle STR \) with \( m\angle UTV < m\angle UTS < m\angle STR \).

Wait, the first option says \( VU < US < SR \) by the hinge theorem. Let's check: In \( \triangle UTV \), \( \triangle UTS \), and \( \triangle STR \), the sides opposite the angles at \( T \) are \( VU \), \( US \), and \( SR \) respectively. Since \( UT \) is a common side (or congruent sides), and \( VT = TS = TR \)? Wait, no, the segments on \( VR \) are marked equal, so \( VT = TR \), and \( UT = ST \) (marked). So for \( \triangle UTV \): sides \( UT \), \( VT \), and \( VU \); for \( \triangle UTS \): sides \( UT \), \( TS \) (but \( TS = UT \)? Wait, no, \( UT \) and \( ST \) are congruent (marked), \( VT = TR \), and \( VT = TS \)? Wait, maybe the triangles are isoceles? Wait, the key is the hinge theorem: if two sides of one triangle are congruent to two sides of another triangle, and the included angle is smaller, then the third side is shorter.

So, applying the hinge theorem: in triangles with two congruent sides, the larger the included angle, the longer the third side. So if \( m\angle UTV < m\angle UTS < m\angle STR \), then the sides opposite these angles ( \( VU \), \( US \), \( SR \) ) should satisfy \( VU < US < SR \), because the included angles are increasing, so the third sides (opposite the angles) should also be increasing. The hinge theorem (SAS inequality) tells us that if two sides are congruent, then the larger included angle gives a longer third side. So the first option, \( VU < US < SR \) by the hinge theorem, makes sense.

The second option says \( VU = US = SR \), which is wrong because the angles are not equal, so the sides can't be equal. The third option says the angles are equal, which is wrong because we are given the angles are in order of size. The fourth option reverses the angle order, which is wrong. So the first option is correct.

Answer:

\( \boldsymbol{VU < US < SR} \) by the hinge theorem. (The first option: \( \boldsymbol{\text{VU < US < SR by the hinge theorem}} \))