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Parallel and Intersecting Lines - Between Lines

Grade 7CBSE

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

🔑Concepts

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When a transversal tt intersects two lines ll and mm, several pairs of angles are formed including interior angles, exterior angles, and corresponding angles.

A transversal t intersecting two lines l and m with angles 1 through 8 labeled.
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If two parallel lines are cut by a transversal, each pair of alternate interior angles are equal (e.g., ∠3=∠6\angle 3 = \angle 6 and ∠4=∠5\angle 4 = \angle 5).

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If two parallel lines are cut by a transversal, each pair of interior angles on the same side of the transversal (co-interior angles) are supplementary (e.g., ∠3+∠5=180∘\angle 3 + \angle 5 = 180^\circ).

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If two parallel lines are cut by a transversal, each pair of corresponding angles are equal (e.g., ∠1=∠5\angle 1 = \angle 5, ∠2=∠6\angle 2 = \angle 6, etc.).

📐Formulae

Linear Pair: ∠a+∠b=180∘\text{Linear Pair: } \angle a + \angle b = 180^\circ

Vertically Opposite Angles: ∠1=∠3\text{Vertically Opposite Angles: } \angle 1 = \angle 3

If l∥m, then Alternate Interior Angles ∠3=∠5\text{If } l \parallel m, \text{ then Alternate Interior Angles } \angle 3 = \angle 5

If l∥m, then Co-interior Angles: ∠4+∠5=180∘\text{If } l \parallel m, \text{ then Co-interior Angles: } \angle 4 + \angle 5 = 180^\circ

💡Examples

Problem 1:

In the given figure, line ll is parallel to line mm (l∥ml \parallel m) and a transversal tt cuts them. If one of the interior angles is 125∘125^\circ, find the measure of its adjacent interior angle on the same side of the transversal.

Solution:

Let the given interior angle be ∠1=125∘\angle 1 = 125^\circ and the required angle be xx. Since the lines are parallel, the sum of interior angles on the same side of the transversal is 180∘180^\circ. Therefore, 125∘+x=180∘125^\circ + x = 180^\circ. To find xx: x=180∘−125∘x = 180^\circ - 125^\circ 180−12555\begin{array}{r} 180 \\ - 125 \\ \hline 55 \end{array} Thus, x=55∘x = 55^\circ.

Explanation:

We use the property of co-interior angles (interior angles on the same side of the transversal), which states that they are supplementary when the lines are parallel.

Problem 2:

Two lines pp and qq are intersected by a transversal nn. If the alternate interior angles are (3x−10)∘(3x - 10)^\circ and (2x+15)∘(2x + 15)^\circ, find the value of xx for which p∥qp \parallel q.

Solution:

For lines pp and qq to be parallel, the alternate interior angles must be equal. Therefore: (3x−10)=(2x+15)(3x - 10) = (2x + 15) Subtract 2x2x from both sides: 3x−2x−10=153x - 2x - 10 = 15 x−10=15x - 10 = 15 Add 1010 to both sides: x=15+10x = 15 + 10 x=25x = 25

Explanation:

Parallel lines require alternate interior angles to be equal. We set up an algebraic equation based on this property and solve for the unknown variable xx.

Problem 3:

In the following figure, l∥ml \parallel m and tt is the transversal. If ∠1=70∘\angle 1 = 70^\circ, find the value of ∠7\angle 7.

Two parallel lines l and m with transversal and angles 1 and 7 indicated.

Solution:

Given: ∠1=70∘\angle 1 = 70^\circ Since ∠1\angle 1 and ∠3\angle 3 are vertically opposite angles: ∠3=∠1=70∘\angle 3 = \angle 1 = 70^\circ Since l∥ml \parallel m, alternate interior angles ∠3\angle 3 and ∠5\angle 5 are equal: ∠5=∠3=70∘\angle 5 = \angle 3 = 70^\circ Now, ∠5\angle 5 and ∠7\angle 7 are vertically opposite angles: ∠7=∠5=70∘\angle 7 = \angle 5 = 70^\circ Alternatively, ∠1\angle 1 and ∠5\angle 5 are corresponding angles, so ∠5=70∘\angle 5 = 70^\circ, and then ∠7=70∘\angle 7 = 70^\circ by vertically opposite angles property.

Explanation:

The problem uses properties of parallel lines (corresponding angles) and the property of vertically opposite angles to find the unknown angle.

Problem 4:

In the figure, line ABAB is parallel to line CDCD. Find the value of yy if the co-interior angles are (2y+10)∘(2y + 10)^\circ and (3y−20)∘(3y - 20)^\circ.

Parallel lines AB and CD with co-interior angles labeled with algebraic expressions.

Solution:

Given AB∥CDAB \parallel CD. The sum of co-interior angles is 180∘180^\circ. (2y+10)+(3y−20)=180(2y + 10) + (3y - 20) = 180 5y−10=1805y - 10 = 180 5y=1905y = 190 y=1905y = \frac{190}{5} y=38y = 38 Thus, the value of yy is 3838.

Explanation:

Since the lines are parallel, the interior angles on the same side of the transversal are supplementary, allowing us to set up and solve a linear equation.