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Lines and Angles - Apply parallel-line theorems with transversals to deduce unknown angles

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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When two parallel lines are intersected by a transversal, the corresponding angles formed are equal in measure. For example, if line l∥ml \parallel m, then ∠1=∠5\angle 1 = \angle 5, ∠2=∠6\angle 2 = \angle 6, etc.

Diagram showing two parallel lines l and m with a transversal and marked corresponding angles 1 and 5.
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Alternate Interior Angles are equal when lines are parallel. These angles lie between the two lines and on opposite sides of the transversal (forming a 'Z' shape).

Parallel lines with alternate interior angles 3 and 5 highlighted.
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Co-interior angles (Consecutive Interior Angles) lie on the same side of the transversal and between the parallel lines. They are supplementary, meaning their sum is 180∘180^\circ.

Parallel lines showing co-interior angles 4 and 5 on the same side of the transversal.
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Transitive Property of Parallelism: Lines which are parallel to the same line are parallel to each other. If l∥ml \parallel m and m∥nm \parallel n, then l∥nl \parallel n.

📐Formulae

If l∥ml \parallel m, then ∠Corresponding1=∠Corresponding2\angle \text{Corresponding}_1 = \angle \text{Corresponding}_2

If l∥ml \parallel m, then ∠Alt. Interior1=∠Alt. Interior2\angle \text{Alt. Interior}_1 = \angle \text{Alt. Interior}_2

Sum of Co-interior Angles: ∠Int1+∠Int2=180∘\angle \text{Int}_1 + \angle \text{Int}_2 = 180^\circ (when lines are parallel)

Sum of Angles on a Straight Line: ∠a+∠b=180∘\angle a + \angle b = 180^\circ

Vertically Opposite Angles: ∠1=∠3\angle 1 = \angle 3 and ∠2=∠4\angle 2 = \angle 4 at any intersection

💡Examples

Problem 1:

In the figure, line AB∥CDAB \parallel CD and a transversal PQPQ intersects them. If one of the interior angles on the same side of the transversal is (3x+20)∘(3x + 20)^\circ and the other is (2x−10)∘(2x - 10)^\circ, find the value of xx.

Solution:

  1. Since AB∥CDAB \parallel CD, the sum of interior angles on the same side of the transversal (co-interior angles) is 180∘180^\circ.
  2. Write the equation: (3x+20)+(2x−10)=180(3x + 20) + (2x - 10) = 180.
  3. Combine like terms: 5x+10=1805x + 10 = 180.
  4. Subtract 1010 from both sides: 5x=1705x = 170.
  5. Divide by 55: x=1705=34x = \frac{170}{5} = 34.

Explanation:

The solution uses the Co-interior Angle Theorem which states that consecutive interior angles are supplementary when lines are parallel.

Problem 2:

Two parallel lines are intersected by a transversal. If a pair of alternate interior angles are given by 55∘55^\circ and (2y−5)∘(2y - 5)^\circ, find the value of yy.

Solution:

  1. Identify the relationship: Alternate interior angles are equal when lines are parallel.
  2. Set up the equation: 2y−5=552y - 5 = 55.
  3. Add 55 to both sides: 2y=602y = 60.
  4. Divide by 22: y=30y = 30.

Explanation:

This problem relies on the property that alternate interior angles (the 'Z-shape' angles) have the same measure when lines are parallel.

Problem 3:

In the given figure, line l∥ml \parallel m and tt is the transversal. If ∠1=(2x+10)∘\angle 1 = (2x + 10)^{\circ} and ∠2=(3x−20)∘\angle 2 = (3x - 20)^{\circ} represent a pair of alternate exterior angles, find the value of xx and the measure of ∠1\angle 1.

Parallel lines l and m with transversal t showing alternate exterior angles 1 and 2.

Solution:

Since l∥m, alternate exterior angles are equal.\text{Since } l \parallel m, \text{ alternate exterior angles are equal.} (2x+10)∘=(3x−20)∘(2x + 10)^{\circ} = (3x - 20)^{\circ} 10+20=3x−2x10 + 20 = 3x - 2x 30=x30 = x Now, substitute x=30 in ∠1:\text{Now, substitute } x = 30 \text{ in } \angle 1: ∠1=2(30)+10=60+10=70∘\angle 1 = 2(30) + 10 = 60 + 10 = 70^{\circ}

Explanation:

When two parallel lines are intersected by a transversal, the pair of alternate exterior angles are equal in measure. By equating the given linear expressions, we solve for the variable xx and then calculate the specific angle value.

Problem 4:

In the figure, AB∥CDAB \parallel CD and CD∥EFCD \parallel EF. Also, EA⊥ABEA \perp AB. If ∠BEF=55∘\angle BEF = 55^{\circ}, find the values of x,yx, y and zz.

Three parallel lines AB, CD, and EF with a perpendicular segment AE and a transversal through E and D/F showing angles x, y, z, and 55 degrees.

Solution:

Since AB∥CD and CD∥EF, then AB∥EF.\text{Since } AB \parallel CD \text{ and } CD \parallel EF, \text{ then } AB \parallel EF. ∠EAB+∠FEA=180∘ (Co-interior angles)\angle EAB + \angle FEA = 180^{\circ} \text{ (Co-interior angles)} 90∘+(z+55∘)=180∘90^{\circ} + (z + 55^{\circ}) = 180^{\circ} z+145∘=180∘z + 145^{\circ} = 180^{\circ} z=35∘z = 35^{\circ} Since CD∥EF, co-interior angles y+55∘=180∘\text{Since } CD \parallel EF, \text{ co-interior angles } y + 55^{\circ} = 180^{\circ} y=125∘y = 125^{\circ} Since AB∥CD, corresponding angles x=y\text{Since } AB \parallel CD, \text{ corresponding angles } x = y x=125∘x = 125^{\circ}

Explanation:

We use the properties of parallel lines: co-interior angles are supplementary (180∘180^{\circ}) and corresponding angles are equal. Since ABAB is parallel to CDCD and CDCD is parallel to EFEF, ABAB is also parallel to EFEF (transitive property).