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**

**Some Geometric Terms and Results**

**Complementary and Supplementary Angles**

**Parallel and Transversal Lines**

**8th Grade Math Practice**** ****From Parallel and Transversal Lines to HOME PAGE**

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