Geometry Parallel Lines and Transversals Worksheet Solutions (PDF)

Geometry math worksheet solutions: Parallel Lines and Transversals – Maths PDF

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This parallel lines answers for Geometry covers parallel lines and transversals (Common Core G-CO.C.9), with a free PDF to print.

Use this answer key to check the 20 problems on parallel lines. Every solution shows the formula or theorem used, each step of the work, and the final answer in bold. If an answer differs from yours, compare the steps line by line to find where the work changed.

Geometry · Common Core State Standards: G-CO.C.9

Parallel lines answers: exercises 1 to 10

Exercise 1 – Vertical angles

Vertical angles are congruent.

68°

Exercise 2 – Linear pair

\(180^{\circ} – 112^{\circ} = 68^{\circ}\)

68°

Exercise 3 – Corresponding angles

Corresponding angles are equal.

Angles 2 and 6 are corresponding angles, so they are congruent.

65°

Exercise 4 – Alternate interior angles

Angles 4 and 5 are alternate interior angles, so they are congruent.

125°

Exercise 5 – Same-side interior angles

Same-side interior angles are supplementary.

\(180^{\circ} – 125^{\circ} = 55^{\circ}\)

55°

Exercise 6 – Name the pair

Two parallel lines cut by a transversal form eight angles.

3 and 6 are on opposite sides of the transversal between the lines: alternate interior.

2 and 6 are in the same position at the two intersections: corresponding.

4 and 6 are between the lines on the same side: same-side interior.

1 and 8 are outside the lines on opposite sides: alternate exterior.

Exercise 7 – Alternate exterior angles

Angles 1 and 8 are alternate exterior angles, so they are congruent.

118°

Exercise 8 – Corresponding angles with x

\(3x + 10 = 5x – 30\)

\(40 = 2x\)

\(x = 20\)

\(3(20) + 10 = 70^{\circ}\)

x = 20; both angles measure 70°

Exercise 9 – Alternate interior with x

\(4x – 7 = 2x + 25\)

\(2x = 32\)

\(x = 16\)

\(4(16) – 7 = 57^{\circ}\)

x = 16; 57°

Exercise 10 – Same-side interior with x

\((6x + 10) + (4x + 20) = 180\)

\(10x + 30 = 180\)

\(x = 15\)

\(6(15) + 10 = 100^{\circ}\) and \(4(15) + 20 = 80^{\circ}\)

x = 15; 100° and 80°

Exercises 11 and up

Exercise 11 – Vertical angles with x

\(7x – 12 = 4x + 21\)

\(3x = 33\)

\(x = 11\)

\(7(11) – 12 = 65^{\circ}\)

x = 11; 65° each

Exercise 12 – Are the lines parallel? (no)

Parallel lines would make corresponding angles congruent. \(74^{\circ} \neq 76^{\circ}\).

No, the lines are not parallel

Exercise 13 – Are the lines parallel? (yes)

Congruent alternate interior angles prove that the lines are parallel (converse).

Yes, the lines are parallel

Exercise 14 – All eight angles

Each angle is 57° or 123°.

Angles that equal 57°: the angle itself, its vertical angle, and the corresponding and alternate angles of those.

The rest form a linear pair with a 57° angle: \(180^{\circ} – 57^{\circ} = 123^{\circ}\).

Four angles measure 57° and four measure 123°

Exercise 15 – Perpendicular bisector

Points on the perpendicular bisector are equidistant from A and B.

\(2x + 5 = 4x – 11\)

\(16 = 2x\)

\(x = 8\)

\(PA = 2(8) + 5 = 21\)

x = 8; PA = 21

Exercise 16 – Perpendicular bisector equation

Midpoint: \(\left( \frac{-2 + 6}{2}, \frac{1 + 5}{2} \right) = (2, 3)\).

Slope of AB: \(\frac{5 – 1}{6 – (-2)} = \frac{1}{2}\), so the perpendicular slope is \(-2\).

\(y – 3 = -2(x – 2)\) gives \(y = -2x + 7\).

y = −2x + 7

Exercise 17 – Crossing streets

Maple (top), Oak (bottom) and the avenue.

Corresponding angles are congruent, so the angle at Oak Street is 52°.

The obtuse angle forms a linear pair with it: \(180^{\circ} – 52^{\circ} = 128^{\circ}\).

52° and 128°

Exercise 18 – Railing

Same-side interior angles are supplementary.

\(180^{\circ} – 64^{\circ} = 116^{\circ}\)

116°

Exercise 19 – Shelf bracket

\(3x + 12 = 5x – 8\)

\(20 = 2x\)

\(x = 10\)

\(3(10) + 12 = 42^{\circ}\)

x = 10; 42°

Exercise 20 – Equal distance

PA = PB = 13, AM = 5, PM = 12.

Points on the perpendicular bisector are equidistant from A and B, so PB = 13 miles.

Triangle PMA is a right triangle with AM = 5 and PA = 13: \(PM = \sqrt{13^{2} – 5^{2}} = \sqrt{144} = 12\).

PB = 13 miles; PM = 12 miles

Key facts behind this parallel lines answers

Key facts

Vertical angles are congruent, and a linear pair adds to 180 degrees.
Parallel lines: corresponding and alternate interior angles are congruent.
Parallel lines: same-side interior angles are supplementary.
If those angle conditions hold for a transversal, the two lines are parallel.
Points on the perpendicular bisector of a segment are equidistant from its endpoints.

How to use this parallel lines answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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