This answer key gives a full worked solution for every problem on the parallel lines and transversals worksheet. Check each reason against the theorem you cited, and compare your algebra step by step.
Problem 1: Name the angle pairs
- \(\angle 4\) and \(\angle 5\) lie between the lines on opposite sides of the transversal: alternate interior angles.
- \(\angle 2\) and \(\angle 6\) are in the same position at the two intersections: corresponding angles.
- \(\angle 1\) and \(\angle 8\) lie outside the lines on opposite sides of the transversal: alternate exterior angles.
- \(\angle 3\) and \(\angle 5\) lie between the lines on the same side of the transversal: same-side interior angles.
Problem 2: Corresponding angles
\(\angle 1\) and \(\angle 5\) are corresponding angles. By the Corresponding Angles Theorem they are equal.
\(m\angle 5\) = \(118^{\circ}\).
Problem 3: Alternate interior angles
\(\angle 3\) and \(\angle 6\) are alternate interior angles, so they are equal.
\(m\angle 6\) = \(72^{\circ}\).
Problem 4: Same-side interior angles
\(\angle 3\) and \(\angle 5\) are same-side interior angles, so they are supplementary.
\(180-53=127\). \(m\angle 5\) = \(127^{\circ}\).
Problem 5: Alternate exterior angles
\(\angle 2\) and \(\angle 7\) are alternate exterior angles. For parallel lines they are equal.
\(m\angle 7\) = \(109^{\circ}\).
Problem 6: Find all eight angles
Vertical angles are equal: \(m\angle 4\) = \(124^{\circ}\). Corresponding angles are equal: \(m\angle 5\) = \(m\angle 8\) = \(124^{\circ}\). Each other angle forms a linear pair with a \(124^{\circ}\) angle: \(180-124=56\).
\(m\angle 1\) = \(m\angle 4\) = \(m\angle 5\) = \(m\angle 8\) = \(124^{\circ}\); \(m\angle 2\) = \(m\angle 3\) = \(m\angle 6\) = \(m\angle 7\) = \(56^{\circ}\).
Problem 7: Corresponding angles with algebra
Corresponding angles are equal: \(3x+15=5x-25\).
Subtract \(3x\) and add 25: \(40=2x\), so \(x=20\).
\(3(20)+15=75\) and \(5(20)-25=75\). Each angle measures \(75^{\circ}\).
Problem 8: Alternate interior with algebra
\(\angle 4\) and \(\angle 5\) are alternate interior angles, so \(4x-12=2x+30\).
Subtract \(2x\), add 12: \(2x=42\), so \(x=21\).
\(m\angle 4\) = \(4(21)-12=72\), so \(m\angle 4\) = \(72^{\circ}\). Check: \(2(21)+30=72\).
Problem 9: Same-side interior with algebra
Same-side interior angles are supplementary: \((2x+20)+(3x-10)=180\).
Combine: \(5x+10=180\), so \(5x=170\) and \(x=34\).
\(2(34)+20=88\) and \(3(34)-10=92\). The angles are \(88^{\circ}\) and \(92^{\circ}\); \(88+92=180\).
Problem 10: Same-side exterior with algebra
Same-side exterior angles between parallel lines are supplementary: \((4x+8)+(2x+22)=180\).
\(6x+30=180\), so \(6x=150\) and \(x=25\).
\(m\angle 1\) = \(4(25)+8=108\) and \(m\angle 7\) = \(2(25)+22=72\). So \(108^{\circ}\) and \(72^{\circ}\).
Problem 11: Are the lines parallel? (alternate interior)
The alternate interior angles are equal (\(64^{\circ}\) = \(64^{\circ}\)).
Yes, \(a \parallel b\) by the Alternate Interior Angles Converse.
Problem 12: Are the lines parallel? (same-side interior)
Add the angles: \(112+78=190\). The sum is not \(180^{\circ}\).
No, the lines are not parallel, because same-side interior angles of parallel lines must be supplementary.
Problem 13: Value of x that makes lines parallel (corresponding)
The lines are parallel exactly when corresponding angles are equal: \(5x+4=7x-20\).
\(24=2x\), so \(x=12\).
\(5(12)+4=64\) and \(7(12)-20=64\). Each angle is \(64^{\circ}\).
Problem 14: Value of x that makes lines parallel (alternate exterior)
Set the alternate exterior angles equal: \(9x-7=6x+23\).
\(3x=30\), so \(x=10\).
\(9(10)-7=83\) and \(6(10)+23=83\). Each angle is \(83^{\circ}\).
Problem 15: Complete the proof
Each missing reason is shown in the table.
| Statements | Reasons |
|---|---|
| \(a \parallel b\) | Given |
| \(m\angle 4\) = \(m\angle 8\) | Corresponding Angles Theorem |
| \(m\angle 8\) = \(m\angle 5\) | Vertical Angles Theorem |
| \(m\angle 4\) = \(m\angle 5\) | Transitive Property of Equality |
Problem 16: Write a proof of the converse
| Statements | Reasons |
|---|---|
| \(m\angle 1\) = \(m\angle 8\) | Given |
| \(m\angle 1\) = \(m\angle 4\) | Vertical Angles Theorem |
| \(m\angle 4\) = \(m\angle 8\) | Substitution Property of Equality |
| \(a \parallel b\) | Corresponding Angles Converse (\(\angle 4\) and \(\angle 8\) are corresponding) |
Problem 17: Perpendicular bisector with algebra
By the Perpendicular Bisector Theorem, \(PA=PB\): \(3x+4=5x-8\).
\(12=2x\), so \(x=6\).
\(PA=3(6)+4=22\) and \(PB=5(6)-8=22\). \(PA=22\).
Problem 18: Distance to endpoints
Right triangle \(\triangle PMA\) has legs 12 and 9: \(PA^{2}=12^{2}+9^{2}=144+81=225\), so \(PA=15\).
By the Perpendicular Bisector Theorem, \(PB=PA\). \(PA=PB=15\).
Problem 19: Streets problem
- The northeast angles are corresponding angles, so the angle at Oak Street is \(71^{\circ}\).
- The northwest and northeast angles at Oak Street form a linear pair: \(180-71=109\). The northwest angle is \(109^{\circ}\).
Problem 20: Railroad crossing
- Alternate interior angles are equal: \(3x+2=2x+37\), so \(x=35\).
- \(3(35)+2=107\) and \(2(35)+37=107\). Each is \(107^{\circ}\).
- Same-side interior angles are supplementary: \(180-107=73\). The angle is \(73^{\circ}\).
Problem 21: Bookshelf brace
Parallel lines have equal corresponding angles. Here \(83 \neq 85\).
No, the shelves are not parallel. They are \(2^{\circ}\) off, so one shelf must be adjusted.
Problem 22: Bus stop
\(M\) is the midpoint, so \(AM=16 : 2=8\). Right triangle \(\triangle PMA\): \(PA^{2}=8^{2}+6^{2}=64+36=100\), so \(PA=10\).
By the Perpendicular Bisector Theorem, \(PB=PA\). The stop is 10 blocks from each house.
Problem 23: Multi-step angle problem
\(\angle 1\) and \(\angle 5\) are corresponding: \(6x-12=4x+18\), so \(2x=30\) and \(x=15\). Then \(m\angle 1\) = \(6(15)-12=78\) and \(m\angle 5\) = \(78^{\circ}\).
\(\angle 4\) and \(\angle 1\) are vertical angles: \(m\angle 4\) = \(78^{\circ}\).
\(\angle 5\) and \(\angle 6\) form a linear pair: \(180-78=102\), so \(m\angle 6\) = \(102^{\circ}\).
Problem 24: Notebook paper
Same-side interior angles are supplementary: \((6y+4)+(3y+14)=180\).
\(9y+18=180\), so \(9y=162\) and \(y=18\).
\(6(18)+4=112\) and \(3(18)+14=68\). The angles are \(112^{\circ}\) and \(68^{\circ}\).
See every chapter on the Geometry math page.