NCERT Solutions for Exercise 6.2 Class 9 Maths Chapter 6

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.

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∠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

Q1 In Fig. 6.28, find the values of x and y and then show that AB $||$ CD.

1640149711922

Answer:

1595610584217
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)

Q2 In Fig. 6.29, if AB $||$ CD, CD $||$ EF and y : z = 3 : 7 , find x .

1640149786744

Answer:

1595610656146
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 $x=126^0$

Q3 In Fig. 6.30, if AB $||$ CD, EF $\bot$ CD and $\angle$ GED = 126°, find $\angle$ AGE, $\angle$ GEF and $\angle$ FGE.

Answer:

1595610747454
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 = $180^0-\angle AGE = 180^0 - 126^0$
$\Rightarrow \angle FGE =54^0$

Q4 In Fig. 6.31, if PQ $||$ ST, $\angle$ PQR = 110° and $\angle$ RST = 130°, find $\angle$ QRS. [ Hint : Draw a line parallel to ST through point R.]

1640149893960

Answer:

Draw a line EF parallel to the ST through R.
Since PQ || ST and ST || EF
$\Rightarrow$ EF || PQ
1595610821237
$\angle$ 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 = $110^0-50^0 = 60^0$

Q5 In Fig. 6.32, if AB $||$ CD, $\angle$ APQ = 50° and $\angle$ PRD = 127°, find x and y .

1640150269148

Answer:

1595610871500
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)
$\Rightarrow y = 77^0$

Q6 In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB $||$ CD.

1640150293845

Answer:

1595610956124
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

NCERT Solutions of Class 10 Subject Wise

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.

Also Read

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