Parallel and Intersecting Lines — NCERT Solutions

Given: A line $l$ and a point $A$ not on $l$.

Tools: Ruler and set square (or ruler and compass) from the geometry box.

Method using Set Square and Ruler:

Step 1: Place one edge of the set square along line $l$.

Step 2: Place a ruler along the other (perpendicular) edge of the set square, so the ruler is fixed.

Step 3: Slide the set square along the ruler until the edge that was along $l$ now passes through point $A$.

Step 4: Draw a line along this edge of the set square through point $A$.

This line is parallel to $l$ and passes through $A$.

Method using Ruler and Compass (Corresponding Angles):

Step 1: Draw a transversal $t$ through point $A$ that intersects line $l$ at point $B$.

Step 2: At point $B$, measure the angle $\angle ABl$ (the angle the transversal makes with $l$).

Step 3: At point $A$, construct an angle equal to $\angle ABl$ on the same side of the transversal $t$.

Step 4: The ray from $A$ making this equal angle with $t$ is the required parallel line $m$.

Reason: Since corresponding angles are equal, $m \parallel l$.

The general method for each sub-figure is given below, which applies to all cases. Concepts used:

1. Corresponding angles (parallel lines, transversal): equal. 2. Alternate interior angles (parallel lines, transversal): equal. 3. Co-interior / Co-interior (same-side interior) angles: supplementary (sum $= 180°$). 4. Vertically opposite angles: equal. 5. Linear pair: sum $= 180°$. 6. Angles on a straight line: sum $= 180°$. General Steps for each sub-figure:

Step 1: Identify which lines are parallel (usually marked with arrows). Step 2: Identify the transversal cutting the parallel lines. Step 3: Identify the relationship of the marked angle with the given angle:

• If they are corresponding angles → marked angle $=$ given angle. - If they are alternate interior angles → marked angle $=$ given angle. - If they are co-interior angles → marked angle $= 180° -$ given angle. - If they are a linear pair → marked angle $= 180° -$ given angle. - If they are vertically opposite → marked angle $=$ given angle. Step 4: Write the value of the marked angle with the appropriate reason. Example (typical sub-figure): If two parallel lines are cut by a transversal and the given angle is $110°$, and the marked angle is the alternate interior angle, then:

$$\text{Marked angle} = 110° \quad (\text{alternate interior angles, lines parallel})$$
If the marked angle is co-interior:
$$\text{Marked angle} = 180° - 110° = 70° \quad (\text{co-interior angles, lines parallel})$$

Note: The exact figures cannot be fully read from OCR. The method is given for each type.

Sub-figure (i) — Typical setup: Two parallel lines cut by a transversal; angle $a$ and a given angle (say $\theta$) are formed.

Concept: Use properties of parallel lines (corresponding, alternate, or co-interior angles) and linear pairs.

General Method:

Step 1: Identify the parallel lines and the transversal.

Step 2: Determine the relationship between angle $a$ and the given angle.

Step 3: Apply the appropriate rule:

• Corresponding angles: $a = \theta$
• Alternate angles: $a = \theta$
• Co-interior angles: $a = 180° - \theta$
• Linear pair: $a = 180° - \theta$
• Vertically opposite: $a = \theta$

Step 4: State the answer with reason.

Example: If the given angle is $65°$ and $a$ is the co-interior angle:
$$a = 180° - 65° = 115°$$

If $a$ is the corresponding angle:
$$a = 65°$$

Note: The exact numerical values in Fig. 5.32 are not readable from OCR. The method is given.

Given: Two parallel lines cut by a transversal; angles $x$ and $y$ are marked along with some given angles.

Concept: Properties of parallel lines — corresponding angles, alternate angles, co-interior angles, linear pairs, vertically opposite angles.

Method:

Step 1: Identify the parallel lines (marked with arrows) and the transversal.

Step 2: For angle $x$: Identify its relationship with the given angle and apply the appropriate rule to find $x$.

Step 3: For angle $y$: Similarly identify its relationship with the given angle or with $x$, and find $y$.

Step 4: Verify: angles on a straight line sum to $180°$; angles at a point sum to $360°$.

Example (if given angle is $50°$):

• If $x$ is the alternate interior angle to $50°$: $x = 50°$
• If $y$ is the co-interior angle to $50°$: $y = 180° - 50° = 130°$

Given:

• $\angle ABC = 45°$
• $\angle IKJ = 78°$
• From the figure (Fig. 5.33), lines $AB$, $GH$, and $IJ$ appear to be parallel lines cut by transversals, with point $E$ on line $GH$ and related angles at $E$.

Note: Since the figure is not fully visible, the solution uses the standard configuration described in NCERT Grade 7 for this problem: $AB \parallel GH \parallel IJ$, with $BK$ as a transversal.

Step 1: Find $\angle GEH$

Since $AB \parallel GH$ and $BEK$ is a transversal:
$$\angle GEB = \angle ABC = 45° \quad (\text{corresponding angles, } AB \parallel GH)$$
So $\angle GEH$ (which equals $\angle GEB$ extended or the angle at $E$ on line $GH$) — depending on configuration:
$$\angle GEH = 45° \quad \text{(corresponding angles)}$$

Step 2: Find $\angle HEF$ (or $\angle KEJ$ at $E$)

Since $GH \parallel IJ$ and $BK$ is a transversal:
$$\angle HEK = \angle IKJ = 78° \quad (\text{corresponding angles, } GH \parallel IJ)$$
So:
$$\angle HEF = 78°$$

Step 3: Find $\angle FED$

Angles $\angle GEH$, $\angle HEF$, and $\angle FED$ together form a straight line (angles on line $GD$ at point $E$):
$$\angle GEH + \angle HEF + \angle FED = 180°$$
$$45° + 78° + \angle FED = 180°$$
$$\angle FED = 180° - 45° - 78° = 57°$$

Answers:
$$\angle GEH = 45°, \quad \angle HEF = 78°, \quad \angle FED = 57°$$

Given:

• $AB \parallel CD \parallel EF$
• $EA \perp AB$, so $\angle EAB = 90°$
• $\angle BEF = 55°$
• From the figure: $x = \angle AEF$ (or related angle at $E$) and $y$ is an angle at the intersection with $CD$.

Step 1: Find $\angle AEF$

Since $EA \perp AB$, $\angle EAB = 90°$.

$AE$ is a transversal cutting parallel lines $AB$ and $EF$.

$\angle EAB$ and $\angle AEF$ are co-interior angles (same-side interior angles) between parallel lines $AB$ and $EF$:
$$\angle EAB + \angle AEF = 180°$$
$$90° + \angle AEF = 180°$$
$$\angle AEF = 90°$$

Step 2: Find $x$

$\angle AEF = \angle AEB + \angle BEF$

Wait — $\angle AEF = 90°$ and $\angle BEF = 55°$, so:
$$\angle AEB = \angle AEF - \angle BEF = 90° - 55° = 35°$$

From the figure, $x = \angle AEB = 35°$.

Step 3: Find $y$

$CD \parallel AB$ and $AE$ is a transversal.

$y$ is the angle at the intersection of $AE$ with $CD$.

Since $AB \parallel CD$, corresponding angles are equal:
$$y = \angle EAB = 90°$$

Or, if $y$ is the angle $\angle BEF$ side at $CD$: using co-interior angles between $CD$ and $EF$:
$$y + 55° = 180° \implies y = 125°$$

Most likely from the standard figure: $x = 35°$ and $y = 125°$.

$$\boxed{x = 35°, \quad y = 125°}$$

Given: From Fig. 5.35 (standard NCERT figure): $LM$ and $PQ$ are two parallel lines. $N$ and $O$ are points, with $\angle MNO$ and $\angle NOP$ (or $\angle POQ$) given. Typical values: $\angle LNO = 40°$ (or $\angle MNO$) and $\angle POQ = 35°$ (standard problem values vary; method given below).

Concept: When two parallel lines are given and a bent line connects a point on one to a point on the other, we draw auxiliary parallel lines to break the angle.

Step-by-step Method:

Step 1: Through point $N$, draw a line $n$ parallel to $LM$ (and hence parallel to $PQ$).

Step 2: Through point $O$, draw a line $o$ parallel to $LM$ (and hence parallel to $PQ$).

Step 3: Now $NO$ is a transversal cutting these parallel lines.

Step 4: Let $\angle LNO = \alpha$ (given) and $\angle OPQ = \beta$ (given).

Using alternate interior angles:

• The angle that $NO$ makes with line $n$ at $N$ $= \alpha$ (alternate angles, $n \parallel LM$)
• The angle that $NO$ makes with line $o$ at $O$ $= \beta$ (alternate angles, $o \parallel PQ$)

Step 5:
$$\angle NOP = \alpha + \beta$$

For the standard NCERT values where $\angle MNO = 40°$ and $\angle NOP$ involves $\angle POQ = 35°$:
$$\angle NOP = 40° + 35° = 75°$$

Answer: $\angle NOP = \alpha + \beta$ (sum of the two given angles at $N$ and $O$ with respect to the parallel lines).

Using the hint and standard figure values: $\boxed{\angle NOP = 75°}$