NCERT Solutions for Exercise 6.2 Class 9 Maths Chapter 6 - Lines and Angles

# NCERT Solutions for Exercise 6.2 Class 9 Maths Chapter 6 - Lines and Angles

Edited By Safeer PP | Updated on Jul 18, 2022 02:18 PM IST

NCERT Solutions for Class 9 Maths exercise 6.2 – A transversal is a line that intersects two or more lines at distinct points. Line l crosses lines m and n at P and Q, respectively. Line l is thus a transversal for lines m and n.

• ∠1=∠5, ∠2=∠6, ∠4=∠8 and ∠3=∠7(Corresponding angles)

• ∠3=∠5, ∠4=∠6 (Alternate interior angles)

• ∠1=∠7, ∠2=∠8 (Alternate exterior angles)

Consecutive interior angles, also known as associated angles or co-interior angles, are interior angles on the same side of the transversal.

The corresponding angles are identical when a transversal connects two parallel lines.

The two lines are parallel if a transversal intersects them in such a way that a pair of corresponding angles are equal.

NCERT solutions Class 9 Maths exercise 6.2 – this exercise includes some theorems which are very important from the examination point of view. So, the theorems and their proof given in NCERT are:

When a transversal intersects two parallel lines, each pair of alternate interior angles is equal.

The lines are parallel if a transversal intersects two lines in such a way that their inner angles are equal.

If a transversal overlaps two parallel lines, each pair of internal angles on the same side is additional.

Along with Class 9 Maths, chapter 6 exericse 6.2 the following exercises are also present.

## Lines and Angles Class 9 Maths Chapter 6 Exercise: 6.1

Answer:

Given that,
In the figure, CD and PQ intersect at F
Therefore, $y =130^0$ (vertically opposite angles)

PQ is a straight line. So,
$\\x + 50^0 =180^0\\ x = 130^0$

Hence AB || CD (since $x$ and are $y$ alternate interior angles)

Answer:

Given AB || CD and CD || EF and $y:z = 3:7 \Rightarrow z = \frac{7}{3}y$
therefore, AB || EF and $x =z$ (alternate interior angles)..............(i)

Again, CD || AB
$\\\Rightarrow x+y =180^0\\ \Rightarrow z+y =180^0$ .............(ii)

Put the value of $z$ in equation (ii), we get

$\frac{10}{3}y =180^0 \Rightarrow y =54^0$
Then $z=180^0-54^0=126^0$

By equation (i), we get the value of $x=126^0$

Answer:

Given AB || CD, EF $\perp$ CD and $\angle$ GED = $126^0$
In the above figure,
GE is transversal. So, that $\angle$ AGE = $\angle$ GED = $126^0$ [Alternate interior angles]
Also, $\angle$ GEF = $\angle$ GED - $\angle$ FED
= $126^0-90^0 = 36^0$
$\angle GEF = 36^0$

Since AB is a straight line
Therefore, $\angle$ AGE + $\angle$ FGE = $180^0$
So, $\angle$ FGE = $180^0-\angle AGE = 180^0 - 126^0$
$\Rightarrow \angle FGE =54^0$

Answer:

Draw a line EF parallel to the ST through R.
Since PQ || ST and ST || EF
$\Rightarrow$ EF || PQ

$\angle$ PQR = $\angle$ QRF = $110^0$ (Alternate interior angles)
$\angle$ QRF = $\angle$ QRS + $\angle$ SRF .............(i)

Again, $\angle$ RST + $\angle$ SRF = $180^0$ (Interior angles of two parallels ST and RF)
$\Rightarrow \angle SRF =180^0-130^0 = 50^0$ ( $\angle$ RST = $130^0$ , given)

Thus, $\angle$ QRS = $110^0-50^0 = 60^0$

Answer:

Given, AB || CD, $\angle$ APQ = $50^0$ and $\angle$ PRD = $127^0$
PQ is a transversal. So,
$\angle$ APQ = $\angle$ PQR= $50^0$ (alternate interior angles)
$\therefore x = 50^0$

Again, PR is a transversal. So,
$\angle$ y + $50^0$ = $127^0$ (Alternate interior angles)
$\Rightarrow y = 77^0$

Answer:

Draw a ray BL $\perp$ PQ and CM $\perp$ RS
Since PQ || RS (Given)
So, BL || CM and BC is a transversal
$\therefore$ $\angle$ LBC = $\angle$ MCB (Alternate interior angles).............(i)

It is known that, angle of incidence = angle of reflection
So, $\angle$ ABL = $\angle$ LBC and $\angle$ MCB = $\angle$ MCD
$\Rightarrow \angle ABL =\angle MCD$ ..................(ii)

Adding eq (i) and eq (ii), we get

$\angle$ ABC = $\angle$ DCB
Both the interior angles are equal
Hence AB || CD

## More About NCERT Solutions for Class 9 Maths Exercise 6.2

When a transversal intersects two lines and a pair of interior angles on the same side of the transversal are supplementary, the two lines are parallel. Lines parallel to the same line are parallel to one another. If a transversal intersects two lines in such a way that the bisectors of two corresponding angles are parallel, the two lines are parallel.

NCERT book Class 9 Maths chapter 6 exercise 6.2 Question 2 is based on ratio and proportion. We know that for solving this question we will apply the property of ratio and proportion.

Also Read| Lines And Angles Class 9 Notes

Benefits of NCERT Solutions for Class 9 Maths Exercise 6.2

• NCERT Exercise 6.2 Class 9 Maths, is based on the basic concepts of parallel lines and transversal

• From Class 9 Maths chapter 6 exercise 6.2 we learn the different types of angles and the relationship between them if two parallel lines are cut by a transverse.

• Understanding the concepts from Class 9 Maths chapter 6 exercise 6.2 will allow us to understand the higher concepts which include triangles are quadrilaterals related to PARALLEL LINES AND A TRANSVERSAL

Also, See

## Subject Wise NCERT Exemplar Solutions

### Frequently Asked Question (FAQs)

1. What is the main concept of NCERT solutions for Class 9 Maths exercise 6.2?

This exercise is about Parallel Lines and a Transversal, types of angles like vertically opposite angles and axioms that are formed due to this intersection.

2. What are parallel lines?

Parallel lines are coplanar straight lines that never intersect with each other at any point.

3. What do you understand by consecutive interior angle?

Consecutive angles are the angles formed by a transversal that intersects two parallel lines. Each angle forms a pair of consecutive angles which lies on each of the parallel lines on any one side of the transversal which can be either be interior or exterior.

4. How many angles are formed when a transverse cuts a pair of parallel lines?

Four angles are formed on each of the parallel lines. Thus, making the total sum eight.

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