Question

32 name the segments that are skew to (overline{ab}). (overline{eg}, overline{dh}, overline{cg}, overline{fh}) (overline{cg}, overline{ac}, overline{fh}, overline{dh}) (overline{eg}, overline{cg}, overline{cd}, overline{fh}) (overline{fh}, overline{gh}, overline{dh}, overline{cg})

Explanation:

Step1: Recall skew lines definition

Skew lines are non - parallel, non - intersecting, and lie in different planes. For segment \(\overline{AB}\) in the given polyhedron (a prism - like figure):
- \(\overline{AB}\) is in the top face (plane \(ABDC\) or \(ABFE\) - like, need to check each option's segments.
- Analyze each option:
- Option 1: \(\overline{EG}\), \(\overline{DH}\), \(\overline{CG}\), \(\overline{FH}\)
- \(\overline{AB}\) is parallel to \(\overline{CD}\), \(\overline{EF}\), \(\overline{GH}\) (since it's a prism with parallel bases). \(\overline{EG}\): Does it intersect or parallel to \(\overline{AB}\)? \(\overline{AB}\) is horizontal (assuming the figure orientation), \(\overline{EG}\) is in a different plane, non - parallel, non - intersecting. \(\overline{DH}\): In a different plane, non - parallel, non - intersecting with \(\overline{AB}\). \(\overline{CG}\): Different plane, non - parallel, non - intersecting. \(\overline{FH}\): Different plane, non - parallel, non - intersecting. Wait, no, let's re - check other options.
- Option 2: \(\overline{CG}\), \(\overline{AC}\), \(\overline{FH}\), \(\overline{DH}\)
- \(\overline{AC}\) intersects \(\overline{AB}\) at \(A\), so it's not skew. Eliminate this option.
- Option 3: \(\overline{EG}\), \(\overline{CG}\), \(\overline{CD}\), \(\overline{FH}\)
- \(\overline{CD}\) is parallel to \(\overline{AB}\) (since \(AB\) and \(CD\) are opposite sides of a parallelogram - like face), so \(\overline{CD}\) is parallel, not skew. Eliminate this option.
- Option 4: \(\overline{FH}\), \(\overline{GH}\), \(\overline{DH}\), \(\overline{CG}\)
- \(\overline{GH}\) is parallel to \(\overline{AB}\) (since \(AB\) and \(GH\) are parallel as the prism has parallel top and bottom bases), so \(\overline{GH}\) is parallel, not skew. Eliminate this option.
- Wait, maybe I made a mistake. Let's re - analyze the first option. Wait, the correct approach: Skew lines to \(\overline{AB}\) should not be parallel (so not \(\overline{CD}\), \(\overline{EF}\), \(\overline{GH}\)) and not intersecting (so not connected to \(A\) or \(B\)).
- Let's list all segments:
- Segments connected to \(A\) or \(B\) (intersecting or parallel): \(\overline{AC}\), \(\overline{AE}\), \(\overline{BD}\), \(\overline{BF}\), \(\overline{CD}\), \(\overline{EF}\), \(\overline{GH}\) are parallel or intersecting.
- Segments not connected to \(A\) or \(B\) and non - parallel: \(\overline{EG}\), \(\overline{DH}\), \(\overline{CG}\), \(\overline{FH}\) (from the first option). Wait, the fourth option has \(\overline{GH}\) which is parallel, so wrong. Third has \(\overline{CD}\) parallel. Second has \(\overline{AC}\) intersecting. So the first option: \(\overline{EG}\), \(\overline{DH}\), \(\overline{CG}\), \(\overline{FH}\) is correct? Wait, no, looking at the options again, the last option (the fourth one) is \(\overline{FH}\), \(\overline{GH}\), \(\overline{DH}\), \(\overline{CG}\). Wait, maybe the figure is a prism with bases \(ABFE\) and \(CDHG\). So \(\overline{AB}\) is in plane \(ABFE\) and \(ABDC\). Skew lines to \(\overline{AB}\) are lines that do not lie in the same plane as \(\overline{AB}\) and are not parallel.
- \(\overline{FH}\): In plane \(CDHG\) (or \(FHGD\)), not parallel to \(\overline{AB}\), non - intersecting.
- \(\overline{DH}\): In plane \(CDHG\), non - parallel, non - intersecting.
- \(\overline{CG}\): In plane \(CDHG\) (or \(CGFE\)), non - parallel, non - intersecting.
- \(\overline{GH}\): Parallel to \(\overline{AB}\), so not skew. Wait, the fourth option has \(\overline{GH}\), so wrong. The first option: \(\overline{EG}\), \(\overline{DH}\), \(\overline{CG}\), \(\overline{FH}\). Let's check \(\overline{EG}\): In plane \(AEGC\)? No, \(\overline{EG}\) is in the front - lower face. \(\overline{AB}\) is in the top face. They don't intersect and are not parallel. \(\overline{DH}\): In the right - lower face. \(\overline{CG}\): In the back - lower face. \(\overline{FH}\): In the right - lower face. Wait, maybe the correct answer is the fourth option? No, \(\overline{GH}\) is parallel. Wait, maybe I messed up the options. Let's re - read the options:

Option 1: \(\overline{EG}\), \(\overline{DH}\), \(\overline{CG}\), \(\overline{FH}\)

Option 2: \(\overline{CG}\), \(\overline{AC}\), \(\overline{FH}\), \(\overline{DH}\) ( \(\overline{AC}\) intersects \(\overline{AB}\) at \(A\), so not skew)

Option 3: \(\overline{EG}\), \(\overline{CG}\), \(\overline{CD}\), \(\overline{FH}\) ( \(\overline{CD}\) is parallel to \(\overline{AB}\), so not skew)

Option 4: \(\overline{FH}\), \(\overline{GH}\), \(\overline{DH}\), \(\overline{CG}\) ( \(\overline{GH}\) is parallel to \(\overline{AB}\), so not skew)

Wait, maybe the first option is correct. Because \(\overline{EG}\), \(\overline{DH}\), \(\overline{CG}\), \(\overline{FH}\) are all non - parallel and non - intersecting with \(\overline{AB}\).

Step2: Confirm by elimination

- Option 2: \(\overline{AC}\) intersects \(\overline{AB}\) → eliminate.
- Option 3: \(\overline{CD}\parallel\overline{AB}\) → eliminate.
- Option 4: \(\overline{GH}\parallel\overline{AB}\) → eliminate.
- Option 1: All segments (\(\overline{EG}\), \(\overline{DH}\), \(\overline{CG}\), \(\overline{FH}\)) are non - parallel and non - intersecting with \(\overline{AB}\), so they are skew.

Answer:

\(\boldsymbol{\overline{EG},\overline{DH},\overline{CG},\overline{FH}}\) (the first option: \(\overline{EG},\overline{DH},\overline{CG},\overline{FH}\))