Q1: Find the angles marked below.

To Find: The value of the marked angles in each figure. (a) Given: Two parallel lines are intersected by a transversal. An angle is given as 60^°. We need to find angle 'a'. Solution: Angle 'a' and the given 60^° angle are in corresponding positions. Since the lines are parallel, corresponding…

Q4: In Fig. 5.33, angle ABC=45^° and angle IKJ=78^°. Find angles angle GEH, angle HEF, angle FED

Given: Three parallel horizontal lines: AB DE IJ. Three parallel slanted lines: AC GH IK. angle ABC = 45^°. angle IKJ = 78^°. To Find: angle GEH, angle HEF, angle FED. Solution: 1. Finding angle GEH: The lines forming angle GEH are GE (which is on line DE) and EH (which is on line GH). The lines…

Q6: What is the measure of angle angle NOP in Fig. 5.35? [Hint: Draw lines parallel to LM and PQ through points N and O.]

Given: Line LM is parallel to line PQ (LM PQ). A zigzag line connects M to N, N to O, and O to Q. angle LMN = 25^°. angle PQO = 45^°. Assuming the question asks for angle NOQ as P is a point on the line PQ. To Find: The measure of angle NOQ. Solution: Let's use the hint and properties of parallel…

Parallel and Intersecting Lines Class 7 Mathematics NCERT Solutions

Q1Figure it Out (Section 5.1)

List all the linear pairs and vertically opposite angles you observe in Fig. 5.3: Linear Pairs $\angle a$ and $\angle b, \ldots$ Pairs of Vertically Opposite Angles $\angle b$ and $\angle d, \ldots$

Solution

Given: Two lines intersecting, forming four angles labeled a, b, c, and d in a clockwise or counter-clockwise manner.

To Find: All linear pairs and pairs of vertically opposite angles.

Solution:

Linear Pairs: A linear pair consists of two adjacent angles whose non-common sides are opposite rays. They add up to

180^\circ

. The linear pairs are:

$\angle a$ and $\angle b$
$\angle b$ and $\angle c$
$\angle c$ and $\angle d$
$\angle d$ and $\angle a$

Pairs of Vertically Opposite Angles: Vertically opposite angles are the angles opposite each other when two lines cross. They are equal. The pairs of vertically opposite angles are:

$\angle a$ and $\angle c$
$\angle b$ and $\angle d$

Final Answer:

Linear Pairs(

\angle a, \angle b

), (

\angle b, \angle c

), (

\angle c, \angle d

), (

\angle d, \angle a

) Pairs of Vertically Opposite Angles(

\angle a, \angle c

), (

\angle b, \angle d

)

Q1Figure it Out (Section 5.4)

Draw some lines perpendicular to the lines given on the dot paper in Fig. 5.10.

Solution

To Do: Draw lines perpendicular to given lines on a dot paper.

Method: A perpendicular line forms a right angle (

90^\circ

) with the original line. On a dot paper, we can use the grid of dots to draw perpendicular lines.

For a horizontal line: A perpendicular line will be a vertical line. We can draw this by connecting dots that are vertically aligned.
For a vertical line: A perpendicular line will be a horizontal line. We can draw this by connecting dots that are horizontally aligned.
For a diagonal line (e.g., one that goes 1 unit right and 1 unit up for every step): A perpendicular line will have a slope that is the negative reciprocal. On a dot paper, this means drawing a line that goes 1 unit right and 1 unit down (or 1 unit left and 1 unit up) for every step. The new line will form a right angle with the original diagonal line.

Q2Figure it Out (Section 5.4)

In Fig. 5.11, mark the parallel lines using the notation given above (single arrow, double arrow etc.). Mark the angle between perpendicular lines with a square symbol.

(a)

How did you spot the perpendicular lines?

(b)

How did you spot the parallel lines?

Solution

Given: A figure containing various lines and shapes.

To Do: Identify and mark parallel and perpendicular lines.

Solution:

(a) How to spot perpendicular lines: Perpendicular lines are lines that intersect at a right angle (

90^\circ

).

Visually, they form a perfect 'L' or 'T' shape at their intersection.
In many geometric figures like squares and rectangles, adjacent sides are perpendicular.
We can confirm by placing the corner of a set square or a protractor at the intersection to check if the angle is exactly $90^\circ$ .
The symbol for a right angle (a small square at the vertex) indicates perpendicular lines.

(b) How to spot parallel lines: Parallel lines are lines on the same plane that never intersect, no matter how far they are extended. They always maintain the same distance from each other.

Visually, they look like railway tracks or the opposite edges of a ruler.
In geometric figures like parallelograms, rectangles, and squares, opposite sides are parallel.
We can check if lines are parallel by measuring the perpendicular distance between them at several points. If the distance is constant, the lines are parallel.
Another way is to draw a transversal line cutting across the pair of lines and measure the corresponding or alternate interior angles. If these angles are equal, the lines are parallel.

Q3Figure it Out (Section 5.4)

In the dot paper following, draw different sets of parallel lines. The line segments can be of different lengths but should have dots as endpoints.

Solution

To Do: Draw different sets of parallel lines on dot paper.

Method: To draw parallel lines on a dot paper, we can use the concept of slope or 'rise over run'. Parallel lines have the same slope.

Draw the first line segment: Connect any two dots on the paper. Let's say this line goes 'a' units horizontally and 'b' units vertically between its endpoints.
Draw a parallel line: To draw a line parallel to the first one, start from a different dot and move the same number of units horizontally and vertically. For example, if the first line connects a dot to another dot that is 3 dots to the right and 2 dots up, any other line that connects a dot to another dot 3 dots to the right and 2 dots up will be parallel to the first line.

Examples:

Horizontal parallel lines: Connect dots on the same horizontal row. All such lines will be parallel to each other.
Vertical parallel lines: Connect dots on the same vertical column. All such lines will be parallel to each other.
Diagonal parallel lines: Draw a line segment by moving 2 dots right and 1 dot up. Draw another line segment anywhere else on the paper by also moving 2 dots right and 1 dot up. These two segments will be parallel.

Q4Figure it Out (Section 5.4)

Using your sense of how parallel lines look, try to draw lines parallel to the line segments on this dot paper.

(a)

Did you find it challenging to draw some of them?

(b)

Which ones?

(c)

How did you do it?

Solution

To Do: Draw lines parallel to given segments on dot paper using intuition.

Solution:

(a) Did you find it challenging to draw some of them? Yes, it can be challenging to draw parallel lines for segments that have a complex slope (e.g., moving 3 units right and 4 units up), especially when relying only on visual sense without counting the dots.

(b) Which ones?

Lines that are horizontal or vertical are the easiest to draw parallel lines for.
Lines with a simple diagonal slope (like 1 unit right, 1 unit up) are also relatively easy.
Lines with a steeper or more irregular slope (like 4 units right, 1 unit up, or 2 units right, 3 units up) are the most challenging to draw parallel to by just looking.

(c) How did you do it?

Visual Method: I tried to keep my ruler at the same angle as the original line and then slid it across the paper to a new position to draw the parallel line.
Dot Counting Method (more accurate): For a more precise way, I determined the 'rise' and 'run' of the original line segment by counting the dots. For example, if the original line goes from a starting dot to an ending dot by moving 3 dots to the right and 2 dots up, I drew the parallel line by picking a new starting dot and finding its corresponding endpoint by also moving 3 dots to the right and 2 dots up.

Q5Figure it Out (Section 5.4)

In Fig. 5.13, which line is parallel to line a - line b or line c? How do you decide this?

Solution

Given: A figure showing three lines: line 'a', line 'b', and line 'c'. Line 'b' appears to maintain a constant distance from line 'a', while line 'c' appears to converge towards line 'a'.

To Find: Which line ('b' or 'c') is parallel to line 'a' and the reason for it.

Solution:

Line b is parallel to line a.

Reasoning: Parallel lines are defined as lines that lie in the same plane and never intersect, no matter how far they are extended. This means the perpendicular distance between them is always constant.

By observing the figure, line 'b' appears to run alongside line 'a' at a constant distance. If we were to extend both lines 'a' and 'b' indefinitely, they would never meet.
On the other hand, line 'c' is drawn at a slight angle to line 'a'. If we extend both lines 'a' and 'c', they will eventually intersect at some point. Therefore, line 'c' is not parallel to line 'a'.

Final Answer: Line b is parallel to line a. We can decide this because the distance between line a and line b appears to be constant, while the distance between line a and line c is changing, indicating that line c will eventually intersect line a.

Q1Figure it Out (Section 5.7)

Can you draw a line parallel to l, that goes through point A? How will you do it with the tools from your geometry box? Describe your method.

Solution

To Construct: A line parallel to a given line 'l' and passing through a given external point 'A'.

Tools: Ruler and set square.

Method using a Ruler and Set Square:

Step 1: Place the ruler along the given line 'l'.
Step 2: Place one of the edges of the set square (the one that forms a right angle) along the ruler.
Step 3: Slide the set square along the edge of the ruler until the other perpendicular edge of the set square touches the point 'A'.
Step 4: Hold the set square firmly and draw a line along this edge through point 'A'. Let's call this new line 'm'.
Step 5: This line 'm' is perpendicular to the line drawn in step 3. Let's try another method.

A better method using Ruler and Set Square (based on corresponding angles):

Step 1: Place one edge of the set square along the given line 'l'.
Step 2: Place the ruler along another edge of the set square (for example, the hypotenuse).
Step 3: Hold the ruler firmly in place and slide the set square along the ruler's edge until the edge that was on line 'l' passes through point 'A'.
Step 4: Draw a line along this edge through point 'A'. This new line will be parallel to the original line 'l' because the corresponding angles formed by the ruler (acting as a transversal) are equal.

Method using a Ruler and Compass:

Step 1: Take any point B on the line 'l' and join A to B.
Step 2: With B as the center and a convenient radius, draw an arc cutting 'l' at C and AB at D.
Step 3: With A as the center and the same radius as in Step 2, draw an arc EF cutting AB at G.
Step 4: Place the compass pointer at C and open it to the length of the arc CD.
Step 5: With G as the center and the same compass opening, draw an arc cutting the arc EF at H.
Step 6: Join A and H and extend the line. This line AH is parallel to 'l'. (This works by constructing equal alternate interior or corresponding angles).

Q1Figure it Out (Section 5.8)

Find the angles marked below.

Solution

To Find: The value of the marked angles in each figure.

(a) Given: Two parallel lines are intersected by a transversal. An angle is given as

60^\circ

. We need to find angle 'a'. Solution: Angle 'a' and the given

60^\circ

angle are in corresponding positions. Since the lines are parallel, corresponding angles are equal. Final Answer:

a = 60^\circ

.

(b) Given: Two parallel lines are intersected by a transversal. An angle is given as

110^\circ

. We need to find angle 'b'. Solution: Angle 'b' and the given

110^\circ

angle are interior angles on the same side of the transversal. Their sum must be

180^\circ

.

b + 110^\circ = 180^\circ

b = 180^\circ - 110^\circ = 70^\circ

Final Answer:

b = 70^\circ

.

(c) Given: Two lines intersect. An angle is given as

75^\circ

. We need to find angles 'c' and 'd'. Solution: Angle 'c' and the

75^\circ

angle are vertically opposite angles. Vertically opposite angles are equal. So,

c = 75^\circ

. Angle 'd' and the

75^\circ

angle form a linear pair. Their sum is

180^\circ

.

d + 75^\circ = 180^\circ

d = 180^\circ - 75^\circ = 105^\circ

Final Answer:

c = 75^\circ

,

d = 105^\circ

.

(d) Given: Two parallel lines are intersected by a transversal. An angle is given as

55^\circ

. We need to find angle 'e'. Solution: Angle 'e' and the given

55^\circ

angle are alternate interior angles. Since the lines are parallel, alternate interior angles are equal. Final Answer:

e = 55^\circ

.

(e) Given: Two parallel lines with a third line intersecting them, forming a shape with a vertex between the parallel lines. Two angles,

60^\circ

and

70^\circ

, are given. We need to find angle 'f'. Solution: Draw a line through the vertex of angle 'f' parallel to the other two lines. This divides angle 'f' into two parts,

f_1

and

f_2

.

$f_1$ is an alternate interior angle to the $60^\circ$ angle, so $f_1 = 60^\circ$ .
$f_2$ is an alternate interior angle to the $70^\circ$ angle, so $f_2 = 70^\circ$ . Angle 'f' is the sum of these two parts. $f = f_1 + f_2 = 60^\circ + 70^\circ = 130^\circ$ Final Answer: $f = 130^\circ$ .

(f) Given: Two parallel lines are intersected by a transversal. Two interior angles on the same side are given as

2x

and

3x

. We need to find 'x'. Solution: The interior angles on the same side of the transversal are supplementary (add up to

180^\circ

).

2x + 3x = 180^\circ

5x = 180^\circ

x = \frac{180^\circ}{5} = 36^\circ

Final Answer:

x = 36^\circ

.

Q2Figure it Out (Section 5.8)

Find the angle represented by a.

Solution

Given: Two sets of parallel lines intersecting each other. Two angles,

60^\circ

and

70^\circ

, are given at two of the intersection points. We need to find angle 'a'.

Solution: Let's extend the lines forming the angle 'a' to form a triangle with the other given lines.

The angle given as $70^\circ$ has a corresponding angle inside the central parallelogram, which is also $70^\circ$ . Let's call this angle inside the parallelogram angle P.
The angle given as $60^\circ$ and an adjacent angle inside the parallelogram are supplementary. So the angle inside is $180^\circ - 60^\circ = 120^\circ$ . This is not the easiest way.

Alternate Method (using a triangle):

Consider the transversal line associated with the $70^\circ$ angle. Extend it.
The angle corresponding to the given $60^\circ$ angle is also $60^\circ$ . Let's call this angle 1.
The angle vertically opposite to the given $70^\circ$ angle is also $70^\circ$ . Let's call this angle 2.
Angle 1, Angle 2, and Angle 'a' form a triangle. The sum of angles in a triangle is $180^\circ$ .
Let's trace the lines more carefully. Let's form a triangle using the transversal that forms the top side of the angle 'a' and the two other intersecting lines.
- The angle corresponding to the given $60^\circ$ is an interior angle of this triangle. So, one angle is $60^\circ$ .
- The angle alternate interior to the given $70^\circ$ is another interior angle of this triangle. So, the second angle is $70^\circ$ .
- The third angle of this triangle is vertically opposite to angle 'a'. Let's call it 'a''.
The sum of angles in this triangle is $180^\circ$ . $60^\circ + 70^\circ + a' = 180^\circ$ $130^\circ + a' = 180^\circ$ $a' = 180^\circ - 130^\circ = 50^\circ$
Since 'a' and 'a'' are vertically opposite angles, they are equal. $a = a' = 50^\circ$

Final Answer:

a = 50^\circ

.

Q3Figure it Out (Section 5.8)

In the figures below, what angles do x and y stand for?

Solution

To Find: The values of x and y in each figure.

Figure 1 (Left): Given: Two parallel lines are intersected by a transversal. An angle is given as

130^\circ

. We need to find angles x and y. Solution:

The angle given as $130^\circ$ and angle 'x' are alternate exterior angles. Since the lines are parallel, alternate exterior angles are equal. Therefore, $x = 130^\circ$ .
The angle corresponding to the $130^\circ$ angle is also $130^\circ$ . This corresponding angle and angle 'y' form a linear pair, so their sum is $180^\circ$ . $130^\circ + y = 180^\circ$ $y = 180^\circ - 130^\circ = 50^\circ$ Final Answer (Figure 1): $x = 130^\circ$ , $y = 50^\circ$ .

Figure 2 (Right): Given: A triangle with two interior angles

50^\circ

and

70^\circ

. The third interior angle is 'y'. An exterior angle is 'x'. Solution:

The sum of the angles in a triangle is $180^\circ$ . $50^\circ + 70^\circ + y = 180^\circ$ $120^\circ + y = 180^\circ$ $y = 180^\circ - 120^\circ = 60^\circ$
The exterior angle of a triangle is equal to the sum of the two opposite interior angles. $x = 50^\circ + 70^\circ = 120^\circ$ Alternatively, x and y form a linear pair, so $x + y = 180^\circ$ . Since $y=60^\circ$ , $x = 180^\circ - 60^\circ = 120^\circ$ . Final Answer (Figure 2): $x = 120^\circ$ , $y = 60^\circ$ .

Q4Figure it Out (Section 5.8)

In Fig. 5.33, $\angle \mathrm{ABC}=45^{\circ}$ and $\angle \mathrm{IKJ}=78^{\circ}$ . Find angles $\angle \mathrm{GEH}, \angle \mathrm{HEF}$ , $\angle \mathrm{FED}$

Solution

Given:

Three parallel horizontal lines: AB || DE || IJ.
Three parallel slanted lines: AC || GH || IK.
$\angle ABC = 45^\circ$ .
$\angle IKJ = 78^\circ$ .

To Find:

\angle GEH

,

\angle HEF

,

\angle FED

.

Solution:

Finding $\angle GEH$ :
- The lines forming $\angle GEH$ are GE (which is on line DE) and EH (which is on line GH).
- The lines forming $\angle IKJ$ are IK and KJ (which is on line IJ).
- We are given that DE || IJ and GH || IK.
- When two pairs of lines are parallel to each other, the angle formed by the intersection of the first pair is equal to the angle formed by the intersection of the second pair.
- Therefore, $\angle GEH = \angle IKJ = 78^\circ$ .
Finding $\angle HEF$ and $\angle FED$ :
- Let's assume from the diagram that the points D, E, and some point F lie on the same straight line, so we are dealing with angles around point E on the line DF. Let's also assume B, C, F are collinear, making BF a transversal.
- Let's find $\angle GEF$ first. The lines forming this angle are GE and EF. We are given AB || GE (part of DE). If we assume BF is a transversal cutting these parallel lines, then $\angle ABC$ and $\angle GEF$ are corresponding angles.
- So, $\angle GEF = \angle ABC = 45^\circ$ .
- Now we can find $\angle HEF$ . From the figure, it appears that ray EF is between rays EG and EH.
- $\angle GEH = \angle GEF + \angle HEF$ .
- $78^\circ = 45^\circ + \angle HEF$ .
- $\angle HEF = 78^\circ - 45^\circ = 33^\circ$ .
- Now, to find $\angle FED$ . Since D, E, F are on a straight line, $\angle FED$ represents the angle of the straight line, which is $180^\circ$ . However, this is likely not what is being asked. It is more probable that 'F' is not on the line DE. Assuming the question intended to find the angles around point E based on the given transversals.
- Let's assume the question labels are $\angle GEH$ , $\angle DEC$ (assuming the transversal is BC), and $\angle FED$ where F is a point on transversal BC.
- Let's stick to the most plausible interpretation based on the question's labels:
  - $\angle GEH = 78^\circ$ .
  - $\angle HEF$ can be found if we assume a transversal creates $\angle GEF$ . As calculated, $\angle GEF=45^\circ$ . This leads to $\angle HEF = 33^\circ$ .
  - $\angle FED$ : If D-E-G is a straight line, then $\angle FED$ and $\angle GEF$ are supplementary angles. $\angle FED = 180^\circ - \angle GEF = 180^\circ - 45^\circ = 135^\circ$ .

Final Answer: Based on the most common interpretation of such diagrams:

$\angle \mathrm{GEH} = 78^\circ$
$\angle \mathrm{HEF} = 33^\circ$
$\angle \mathrm{FED} = 135^\circ$

Q5Figure it Out (Section 5.8)

In Fig. 5.34, $A B$ is parallel to CD and CD is parallel to EF . Also, EA is perpendicular to AB . If $\angle \mathrm{BEF}=55^{\circ}$ , find the values of $x$ and $y$ .

Solution

Given:

AB || CD || EF.
EA ⊥ AB, which means $\angle EAB = 90^\circ$ .
$\angle BEF = 55^\circ$ .
Angle 'x' is $\angle AEB$ .
Angle 'y' is $\angle BCD$ .

To Find: The values of x and y.

Solution:

Finding x:
- Since AB || EF and AE is a transversal, the sum of the interior angles on the same side is $180^\circ$ .
- $\angle EAB + \angle AEF = 180^\circ$ .
- $90^\circ + \angle AEF = 180^\circ$ .
- $\angle AEF = 180^\circ - 90^\circ = 90^\circ$ .
- The angle $\angle AEF$ is composed of two angles, $\angle AEB$ (which is x) and $\angle BEF$ .
- $\angle AEF = \angle AEB + \angle BEF$ .
- $90^\circ = x + 55^\circ$ .
- $x = 90^\circ - 55^\circ = 35^\circ$ .
Finding y:
- Since AB || CD and BC is a transversal, the sum of the interior angles on the same side is $180^\circ$ .
- $\angle ABC + \angle BCD = 180^\circ$ .
- We need to find $\angle ABC$ first.
- Consider the parallel lines AB and EF with BE as the transversal.
- $\angle ABE$ and $\angle BEF$ are alternate interior angles.
- Therefore, $\angle ABE = \angle BEF = 55^\circ$ .
- Since $\angle ABC$ is the same as $\angle ABE$ , we have $\angle ABC = 55^\circ$ .
- Now, using the relationship between interior angles for AB || CD:
- $\angle ABC + \angle BCD = 180^\circ$ .
- $55^\circ + y = 180^\circ$ .
- $y = 180^\circ - 55^\circ = 125^\circ$ .

Final Answer:

x = 35^\circ

and

y = 125^\circ

.

Q6Figure it Out (Section 5.8)

What is the measure of angle $\angle \mathrm{NOP}$ in Fig. 5.35? [Hint: Draw lines parallel to LM and PQ through points N and O.]

Solution

Given:

Line LM is parallel to line PQ (LM || PQ).
A zigzag line connects M to N, N to O, and O to Q.
$\angle LMN = 25^\circ$ .
$\angle PQO = 45^\circ$ .
Assuming the question asks for $\angle NOQ$ as P is a point on the line PQ.

To Find: The measure of

\angle NOQ

.

Solution: Let's use the hint and properties of parallel lines. Let

\angle MNO = n

and

\angle NOQ = o

.

Method 1: Using Alternate Interior Angles

Draw a line XNY through N parallel to LM and PQ.
Draw a line UOV through O parallel to LM and PQ.
Since LM || XNY, $\angle LMN$ and $\angle MNX$ are alternate interior angles. So, $\angle MNX = 25^\circ$ .
Since PQ || UOV, $\angle PQO$ and $\angle QOV$ are alternate interior angles. So, $\angle QOV = 45^\circ$ .
We can express the angles at N and O as:
- $n = \angle MNO = \angle MNX + \angle ONX = 25^\circ + \angle ONX$ .
- $o = \angle NOQ = \angle NOV + \angle QOV = \angle NOV + 45^\circ$ .
Now consider the parallel lines XNY and UOV with transversal NO. The angles $\angle ONX$ $∠ ONX$ and $\angle NOV$ $∠ NO V$ are alternate interior angles, so they are equal.
- $\angle ONX = \angle NOV$ .
From step 5, we have $\angle ONX = n - 25^\circ$ and $\angle NOV = o - 45^\circ$ .
Equating them: $n - 25^\circ = o - 45^\circ$ , which gives the relation $o - n = 20^\circ$ .

Method 2: Using Consecutive Interior Angles

Using the same parallel lines XNY and UOV.
Since LM || XNY, $\angle LMN$ and $\angle MNY$ are consecutive interior angles. Their sum is $180^\circ$ . So, $\angle MNY = 180^\circ - 25^\circ = 155^\circ$ .
Since PQ || UOV, $\angle PQO$ and $\angle QOU$ are consecutive interior angles. Their sum is $180^\circ$ . So, $\angle QOU = 180^\circ - 45^\circ = 135^\circ$ .
We can express the angles around N and O as:
- $\angle ONY = \angle MNY - \angle MNO = 155^\circ - n$ .
- $\angle NOU = \angle QOU - \angle NOQ = 135^\circ - o$ .
Now consider the parallel lines XNY and UOV with transversal NO. The angles $\angle ONY$ $∠ ON Y$ and $\angle NOU$ $∠ NO U$ are consecutive interior angles, so their sum is $180^\circ$ $18 0^{\circ}$ .
- $\angle ONY + \angle NOU = 180^\circ$ .
Substitute the expressions from step 4:
- $(155^\circ - n) + (135^\circ - o) = 180^\circ$ .
- $290^\circ - (n + o) = 180^\circ$ .
- $n + o = 290^\circ - 180^\circ = 110^\circ$ .

Solving the System of Equations We now have a system of two linear equations:

$o - n = 20^\circ$
$o + n = 110^\circ$

Add the two equations: $(o - n) + (o + n) = 20^\circ + 110^\circ$ $2o = 130^\circ$ $o = 65^\circ$
Substitute $o = 65^\circ$ into the second equation: $n + 65^\circ = 110^\circ$ $n = 110^\circ - 65^\circ = 45^\circ$

The question asks for

\angle NOP

, which we assume is

\angle NOQ

. The value of this angle is 'o'.

Final Answer: The measure of angle

\angle NOP

(interpreted as

\angle NOQ

) is

65^\circ

.

Parallel and Intersecting LinesClass 7 Mathematics NCERT Solutions

List all the linear pairs and vertically opposite angles you observe in Fig. 5.3: Linear Pairs∠a\angle a∠a and ∠b,…\angle b, \ldots∠b,… Pairs of Vertically Opposite Angles∠b\angle b∠b and ∠d,…\angle d, \ldots∠d,…

Solution

Draw some lines perpendicular to the lines given on the dot paper in Fig. 5.10.

Solution

In Fig. 5.11, mark the parallel lines using the notation given above (single arrow, double arrow etc.). Mark the angle between perpendicular lines with a square symbol.

Solution

In the dot paper following, draw different sets of parallel lines. The line segments can be of different lengths but should have dots as endpoints.

Solution

Using your sense of how parallel lines look, try to draw lines parallel to the line segments on this dot paper.

Solution

In Fig. 5.13, which line is parallel to line a - line b or line c? How do you decide this?

Solution

Can you draw a line parallel to l, that goes through point A? How will you do it with the tools from your geometry box? Describe your method.

Solution

Find the angles marked below.

Solution

Find the angle represented by a.

Solution

In the figures below, what angles do x and y stand for?

Solution

In Fig. 5.33, ∠ABC=45∘\angle \mathrm{ABC}=45^{\circ}∠ABC=45∘ and ∠IKJ=78∘\angle \mathrm{IKJ}=78^{\circ}∠IKJ=78∘. Find angles ∠GEH,∠HEF\angle \mathrm{GEH}, \angle \mathrm{HEF}∠GEH,∠HEF, ∠FED\angle \mathrm{FED}∠FED

Solution

In Fig. 5.34, ABA BAB is parallel to CD and CD is parallel to EF . Also, EA is perpendicular to AB . If ∠BEF=55∘\angle \mathrm{BEF}=55^{\circ}∠BEF=55∘, find the values of xxx and yyy.

Solution

What is the measure of angle ∠NOP\angle \mathrm{NOP}∠NOP in Fig. 5.35? [Hint: Draw lines parallel to LM and PQ through points N and O.]

Solution

List all the linear pairs and vertically opposite angles you observe in Fig. 5.3: Linear Pairs $\angle a$ and $\angle b, \ldots$ Pairs of Vertically Opposite Angles $\angle b$ and $\angle d, \ldots$

In Fig. 5.33, $\angle \mathrm{ABC}=45^{\circ}$ and $\angle \mathrm{IKJ}=78^{\circ}$ . Find angles $\angle \mathrm{GEH}, \angle \mathrm{HEF}$ , $\angle \mathrm{FED}$

In Fig. 5.34, $A B$ is parallel to CD and CD is parallel to EF . Also, EA is perpendicular to AB . If $\angle \mathrm{BEF}=55^{\circ}$ , find the values of $x$ and $y$ .

What is the measure of angle $\angle \mathrm{NOP}$ in Fig. 5.35? [Hint: Draw lines parallel to LM and PQ through points N and O.]