Question

in the triangles, qr = de and sr = fe. which statement about the sides must be true? sr = de, qs = df, df < qs, qs < de

Explanation:

Step1: Analyze Triangle \( \triangle QRS \) and \( \triangle DEF \)

We know \( SR = FE \) (given, marked with two ticks) and \( QR = DE \) (given, marked with one tick). Let's find the third angle in each triangle. In \( \triangle QRS \), \( \angle R = 62^\circ \), so \( \angle S + \angle Q + 62^\circ = 180^\circ \). In \( \triangle DEF \), \( \angle E = 50^\circ \), so \( \angle D + \angle F + 50^\circ = 180^\circ \). Also, since two sides are equal ( \( SR = FE \), \( QR = DE \) ), we can use the fact that in a triangle, the larger angle is opposite the longer side.

Step2: Compare Angles Opposite \( QS \) and \( DF \)

First, find the angle opposite \( QS \) in \( \triangle QRS \): \( \angle R = 62^\circ \). The angle opposite \( DF \) in \( \triangle DEF \) is \( \angle E = 50^\circ \). Wait, no, actually, in \( \triangle QRS \), sides: \( SR = FE \), \( QR = DE \), so the included angle for \( QS \) is \( \angle R = 62^\circ \), and for \( DF \) is \( \angle E = 50^\circ \). By the Law of Cosines, \( QS^2 = SR^2 + QR^2 - 2 \cdot SR \cdot QR \cdot \cos(62^\circ) \) and \( DF^2 = FE^2 + DE^2 - 2 \cdot FE \cdot DE \cdot \cos(50^\circ) \). Since \( SR = FE \), \( QR = DE \), and \( \cos(62^\circ) < \cos(50^\circ) \) (because cosine is decreasing in \( [0, 180^\circ] \)), then \( -2 \cdot SR \cdot QR \cdot \cos(62^\circ) > -2 \cdot FE \cdot DE \cdot \cos(50^\circ) \), so \( QS^2 > DF^2 \), which means \( DF < QS \).

Now check other options:

• \( SR = DE \): Given \( SR = FE \) and \( QR = DE \), no info to say \( SR = DE \), false.
• \( QS = DF \): From above, \( QS > DF \), false.
• \( QS < DE \): \( DE = QR \), and \( QS \) is a side of \( \triangle QRS \), by triangle inequality, \( QS < SR + QR \), but no info to say \( QS < DE \), false.

Answer:

\( DF < QS \) (the option with "DF < QS")