Parallel and Transversal Lines

Here we discuss how the angles formed between parallel and transversal lines.

When the transversal intersects two parallel lines:

• Pairs of corresponding angles are equal.

• Pairs of alternate angles are equal

• Interior angles on the same side of transversal are supplementary.

Worked-out problems for solving parallel and transversal lines: 

1. In adjoining figure l ∥ m is cut by the transversal t. If ∠1 = 70, find the measure of ∠3, ∠5, ∠6. 

Solution:

We have ∠1 = 70°

∠1 = ∠3 (Vertically opposite angles)

Therefore, ∠3 = 70°

Now, ∠1 = ∠5 (Corresponding angles)

Therefore, ∠5 = 70°

Also, ∠3 + ∠6 = 180° (Co-interior angles)

70° + ∠6 = 180°

Therefore, ∠6 = 180° - 70° = 110°

2. In the given figure AB ∥ CD, ∠BEO = 125°, ∠CFO = 40°. Find the measure of ∠EOF.

Solution:

Draw a line XY parallel to AB and CD passing through O such that AB ∥ XY and CD ∥ XY

∠BEO + ∠YOE = 180° (Co-interior angles)

Therefore, 125° + ∠YOE = 180°

Therefore, ∠YOE = 180° - 125° = 55°

Also, ∠CFO = ∠YOF (Alternate angles)

Given ∠CFO = 40°

Therefore, ∠YOF = 40°

Then ∠EOF = ∠EOY + ∠FOY

= 55° + 40° = 95°

3. In the given figure AB ∥ CD ∥ EF and AE ⊥ AB.

Also, ∠BAE = 90°. Find the values of ∠x, ∠y and ∠z.

Solution:

y + 45° = 1800

Therefore, ∠y = 180° - 45° (Co-interior angles)

= 135°

∠y =∠x (Corresponding angles)

Therefore, ∠x = 135°

Also, 90° + ∠z + 45° = 180°

Therefore, 135° + ∠z = 180°

Therefore, ∠z = 180° - 135° = 45°

4. In the given figure, AB ∥ ED, ED ∥ FG, EF ∥ CD
Also, ∠1 = 60°, ∠3 = 55°, then find ∠2, ∠4, ∠5.

Solution:

Since, EF ∥ CD cut by transversal ED

Therefore, ∠3 = ∠5 we know, ∠3 = 55°

Therefore, ∠5 = 55°

Also, ED ∥ XY cut by transversal CD

Therefore, ∠5 = ∠x we know ∠5 = 55°

Therefore,∠x = 55°

Also, ∠x + ∠1 + ∠y = 180°

55° + 60° + ∠y = 180°

115° + ∠y = 180°

∠y = 180° - 115°

Therefore, ∠y = 65°

Now, ∠y + ∠2 = 1800 (Co-interior angles)

65° + ∠2 = 180°

∠2 = 180° - 65°

∠2 = 115°

Since, ED ∥ FG cut by transversal EF

Therefore, ∠3 + ∠4 = 180°

55° + ∠4 = 180°

Therefore, ∠4 = 180° - 55° = 125°

5. In the given figure PQ ∥ XY. Also, y : z = 4 : 5 find.

Solution:

Let the common ratio be a

Then y = 4a and z = 5a

Also, ∠z = ∠m (Alternate interior angles)

Since, z = 5a

Therefore, ∠m = 5a [RS ∥ XY cut by transversal t]

Now, ∠m = ∠x (Corresponding angles)

Since, ∠m = 5a

Therefore, ∠x = 5a [PQ ∥ RS cut by transversal t]

∠x + ∠y = 180° (Co-interior angles)

5a + 4a = 1800

9a = 180°

a = 180/9

a = 20

Since, y = 4a

Therefore, y = 4 × 20

y = 80°

z = 5a

Therefore, z = 5 × 20

z = 100°

x = 5a

Therefore, x = 5 × 20

x = 100°

Therefore, ∠x = 100°, ∠y = 80°, ∠z = 100°

● Lines and Angles

Fundamental Geometrical Concepts

Angles

Classification of Angles

Related Angles

Some Geometric Terms and Results

Complementary Angles

Supplementary Angles

Complementary and Supplementary Angles

Adjacent Angles

Linear Pair of Angles

Vertically Opposite Angles

Parallel Lines

Transversal Line

Parallel and Transversal Lines

7th Grade Math Problems

8th Grade Math Practice 

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