NCERT Exemplar Class 9 Mathematics Chapter 6 Lines and Angles, is provided in pdf format for students to prepare for exams. Our experts have designed exemplar problems in accordance with **CBSE syllabus (2020-2021)** for 9th standard, which covers the following topics of chapter Lines and Angles given below;

- Basic terms like line-segment, collinear points, non-collinear points, right angle, straight angle, acute angle, reflex angle, complementary angles, etc.
- Lines which are intersecting and non-intersecting
- Linear pair of angles such as adjacent angles, vertically opposite angles, etc.
- Transversal intersecting Parallel lines forming angles such as exterior angles, consecutive interior angle.
- Lines parallel to the same line
- Angle Sum Property of a Triangle

**Also, check**: NCERT Exemplar Class 9 Maths

This chapter is divided into two parts in the first part the students will learn about lines and in the second part, the students will learn about different angles. Learning the concepts of lines and angles is very much important to understand the concepts of geometry in Class 9 as well as in Class 10. To make understand these concepts, free NCERT exemplars are provided here.

Students of Class 9 can also use these exemplar solutions as a reference tool while practising the **NCERT book** exercise questions, which can also be downloaded in PDF form. Exemplar books, NCERT solutions, notes and questions papers are also provided here, in BYJU’S as study materials for students to learn and practice for their final exams.

Sample papers and previous year question papers will help to know the question pattern and marks contained by chapter 6 in Maths exam for Class 9. Also, solve important questions with NCERT exemplar for chapter lines and angles. To download pdf click on the link below.

## Download PDF of NCERT Exemplar Solutions for Class 9 Maths Chapter 6 Lines and Angles

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Exercise 6.1 Page No: 55

**Write the correct answer in each of the following:**

**1. In Fig. 6.1, if AB || CD || EF, PQ || RS, âˆ RQD = 25Â° and âˆ CQP = 60Â°, then âˆ QRS is equal to**

**(A) 85Â° **

**(B) 135Â°**

**(C) 145Â° **

**(D) 110Â°**

**Solution:**

**(C) 145Â° **

Explanation:

According to the given figure, we have

AB || CD || EF

PQ || RS

âˆ RQD = 25Â°

âˆ CQP = 60Â°

PQ || RS.

We know that,

If a transversal intersects two parallel lines, then each pair of alternate exterior angles is equal.

Now, since, PQ || RS

â‡’Â âˆ PQC =Â âˆ BRS

We haveÂ âˆ PQC = 60Â°

â‡’Â âˆ BRS = 60Â° â€¦ eq.(i)

We also know that,

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

Now again, since, AB || CD

â‡’Â âˆ DQR =Â âˆ QRA

We haveÂ âˆ DQR = 25Â°

â‡’Â âˆ QRA = 25Â° â€¦ eq.(ii)

Using linear pair axiom,

We get,

âˆ ARS +Â âˆ BRS = 180Â°

â‡’Â âˆ ARS = 180Â° –Â âˆ BRS

â‡’Â âˆ ARS = 180Â° – 60Â° (From (i),Â âˆ BRS = 60Â°)

â‡’Â âˆ ARS = 120Â° â€¦ eq.(iii)

Now,Â âˆ QRS =Â âˆ QRA +Â âˆ ARS

From equations (ii) and (iii), we have,

âˆ QRA = 25Â° andÂ âˆ ARS = 120Â°

Hence, the above equation can be written as:

âˆ QRS = 25Â° + 120Â°

â‡’Â âˆ QRS = 145Â°

Therefore, option (C) is the correct answer.

**2. If one angle of a triangle is equal to the sum of the other two angles, then the triangle is**

**(A) An isosceles triangle**

**(B) An obtuse triangle**

**(C) An equilateral triangle**

**(D) A right triangle**

**Solution:**

**(D) A right triangle**

Explanation:

Let the angles of â–³ABC be âˆ A, âˆ B and âˆ C

Given that âˆ A= âˆ B+âˆ C â€¦(eq1)

But, in any â–³ABC,

Using angle sum property, we have,

âˆ A+âˆ B+âˆ C=180^{o} â€¦(eq2)

From equations (eq1) and (eq2), we get

âˆ A+âˆ A=180^{o}

â‡’2âˆ A=180^{o}

â‡’âˆ A=180^{o}/2 = 90^{o}

â‡’âˆ A = 90^{o}

Hence, we get that the triangle is a right triangle

Therefore, option (D) is the correct answer.

**3. An exterior angle of a triangle is 105Â° and its two interior opposite angles are equal. Each of these equal angles is**

**(A) 37 Â½ ^{o}**

**(B) 52 Â½ ^{o}**

**(C) 72 Â½ ^{o} **

**(D) 75Â°**

**Solution:**

**(B) 52 Â½ ^{o}**

Explanation:

According to the question,

Exterior angle of triangle= 105Â°

Let the two interior opposite angles of the triangle = x

We know that,

Exterior angle of a triangle = sum of interior opposite angles

Then, we have the equation,

105Â° = x + x

2x = 105Â°

x = 52.5Â°

x = 52Â½

Therefore, option (B) is the correct answer.

**4. The angles of a triangle are in the ratio 5 : 3 : 7. The triangle is**

**(A) An acute angled triangle **

**(B) An obtuse angled triangle**

**(C) A right triangle **

**(D) An isosceles triangle**

**Solution:**

**(A) An acute angled triangle **

Explanation:

According to the question,

The angles of a triangle are of the ratio 5 : 3 : 7

Let 5:3:7 be 5x, 3x and 7x

Using the angle sum property of a triangle,

5x + 3x +7x =180

15x=180

x=12

Substituting the value of x, x = 12, in 5x, 3x and 7x we get,

5x = 5Ã—12 = 60^{o}

3x = 3Ã—12 = 36^{o}

7x = 7Ã—12 = 84^{o}

Since all the angles are less than 90^{o}, the triangle is an acute angled triangle.

Therefore, option (A) is the correct answer.

Exercise 6.2 Page No: 56

**1. For what value of x + y in Fig. 6.4 will ABC be a line? Justify your answer.**

**Solution:**

Value of x + y should be 180^{o} for ABC to be a line.

Justification:

From the figure we can say that,

BD is a ray that intersects AB and BC at the point B which results in

âˆ ABD = y

and,Â âˆ DBC = x

We know,

If a ray stands on a line, then the sum of two adjacent angles so formed is 180Â°.

â‡’Â If the sum of two adjacent angles is 180Â°, then a ray stands on a line.

Thus, for ABC to be a line,

The sum ofÂ âˆ ABD andÂ âˆ DBC should be equal to 180Â°.

â‡’Â âˆ ABD +Â âˆ DBC = 180Â°

â‡’Â x + y = 180Â°

Therefore, the value of x + y should be equal to 180Â° for ABC to be a line.

**2. Can a triangle have all angles less than 60Â°? Give reason for your answer.**

**Solution:**

No. A triangle cannot have all angles less than 60Â°

Justification:

According to angle sum property,

We know that the sum of all the interior angles of a triangle should be = 180^{o}.

Suppose, all the angles are 60^{o},

Then we get, 60^{o} + 60^{o} + 60^{o} = 180^{o}.

Now, considering angles less than 60^{o},

Let us take 59^{o}, which is the highest natural number less than 60^{o}.

Then we have,

59^{o} +59^{o} + 59^{o} = 177^{o} â‰ 180^{o}

Hence, we can say that if all the angles are less that 60^{o}, the measure of the angles wonâ€™t satisfy the angle sum property.

Therefore, a triangle cannot have all angles less than 60^{o}.

**3. Can a triangle have two obtuse angles? Give reason for your answer.**

**Solution:**

No. A triangle cannot have two obtuse angles

Justification:

According to angle sum property,

We know that the sum of all the interior angles of a triangle should be = 180^{o}.

An obtuse angle is one whose value is greater than 90Â° but less than 180Â°.

Considering two angles to be equal to the lowest natural number greater than 90^{o}, i.e., 91^{o}.

According to the question,

If the triangle has two obtuse angles, then there are two angles which are at least 91Â° each.

On adding these two angles,

Sum of the two angles = 91Â° + 91Â°

â‡’ Sum of the two angles = 182Â°

The sum of these two angles already exceeds the sum of three angles of the triangle, even without considering the third angle.

Therefore, a triangle cannot have two obtuse angles.

**4. How many triangles can be drawn having its angles as 45Â°, 64Â° and 72Â°? Give reason for your answer.**

**Solution:**

No triangle can be drawn having its angles 45Â°, 64Â° and 72Â°.

Justification:

According to angle sum property,

We know that the sum of all the interior angles of a triangle should be = 180^{o}.

But, according to the question,

We have the angles 45Â°, 64Â° and 72Â°.

Sum of these angles = 45Â° + 64Â° + 72Â°

= 181^{o}, which is greater than 180^{o}.

Hence, the angles do not satisfy the angle sum property of a triangle.

Therefore, no triangle can be drawn having its angles 45Â°, 64Â° and 72Â°.

**5. How many triangles can be drawn having its angles as 53Â°, 64Â° and 63Â°? Give reason for your answer.**

**Solution:**

Infinitely many triangles can be drawn having its angles as 53Â°, 64Â° and 63Â°.

Justification:

According to angle sum property,

We know that the sum of all the interior angles of a triangle should be = 180^{o}.

According to the question,

We have the angles 53Â°, 64Â°, and 63Â°.

Sum of these angles = 53Â° + 64Â° + 63Â°

= 180^{o}

Hence, the angles satisfy the angle sum property of a triangle.

Therefore, infinitely many triangles can be drawn having its angles as 53Â°, 64Â° and 63Â°.

Exercise 6.3 Page No: 58

**1. In Fig. 6.9, OD is the bisector of âˆ AOC, OE is the bisector of âˆ BOC and OD âŠ¥ OE. Show that the points A, O and B are collinear.**

**Solution:**

According to the question,

In figure,

OD âŠ¥ OE,

OD and OE are the bisector of âˆ AOC and âˆ BOC.

To prove: Points A, O and B are collinear

i.e., AOB is a straight line.

Proof:

Since, OD and OE bisect angles âˆ AOC and âˆ BOC respectively.

âˆ AOC = 2âˆ DOC â€¦(eq.1)

And âˆ COB = 2âˆ COE â€¦(eq.2)

Adding (eq.1) and (eq.2), we get

âˆ AOC = âˆ COB = 2âˆ DOC + 2âˆ COE

âˆ AOC +âˆ COB = 2(âˆ DOC +âˆ COE)

âˆ AOC + âˆ COB = 2âˆ DOE

Since, ODâŠ¥OE

We get,

âˆ AOC +âˆ COB = 2Ã—90^{o}

âˆ AOC +âˆ COB =180^{o}

âˆ AOB =180^{o}

So, âˆ AOC + âˆ COB are forming linear pair.

Therefore, AOB is a straight line.

Hence, points A, O and B are collinear.

**2. In Fig. 6.10, âˆ 1 = 60Â° and âˆ 6 = 120Â°. Show that the lines m and n are parallel.**

**Solution:**

According to the question,

We have from the figure âˆ 1 = 60Â° and âˆ 6 = 120Â°

Since, âˆ 1 = 60Â° and âˆ 6 = 120Â°

Here, âˆ 1 = âˆ 3 [since they are vertically opposite angles]

âˆ 3 = âˆ 1 = 60Â°

Now, âˆ 3 + âˆ 6 = 60Â° + 120Â°

â‡’ âˆ 3 + âˆ 6 = 180Â°

We know that,

If the sum of two interior angles on same side of *l* is 180Â°, then the lines are parallel.

Therefore, m || n

**3. AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (Fig. 6.11). Show that AP || BQ.**

**Solution:**

*l *|| m and t is the transversal

âˆ MAB = âˆ SBA [alternate angles]

â‡’ Â½ âˆ MAB = Â½ âˆ SBA

â‡’ âˆ PAB = âˆ QBA

â‡’ âˆ 2 = âˆ 3

But, âˆ 2 and âˆ 3 are alternate angles.

Hence, AP||BQ.

**4. If in Fig. 6.11, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m.**

**Solution:**

AP is the bisector of âˆ MAB

BQ is the bisector of âˆ SBA.

Given: AP||BQ.

As AP||BQ,

We have,

So âˆ 2 = âˆ 3 [Alternate angles]

2âˆ 2 = 2âˆ 3

â‡’ âˆ 2 + âˆ 2 = âˆ 3 +âˆ 3

From figure, we have âˆ 1= âˆ 2and âˆ 3 = âˆ 4

â‡’ âˆ 1+ âˆ 2 = âˆ 3 +âˆ 4

â‡’ âˆ MAB = âˆ SBA

But, we know that these are alternate angles.

Hence, the lines *l* and m are parallel, i.e., *l* ||m.

**5. In Fig. 6.12, BA || ED and BC || EF. Show that âˆ ABC = âˆ DEF [Hint: Produce DE to intersect BC at P (say)].**

**Solution:**

Construction:

Extend DE to intersect BC at point, P.

Given, EF||BC and DP is the transversal,

âˆ *DEF *= âˆ *DPC *â€¦(eq.1) [Corresponding angles]

Also given, AB||DP and BC is the transversal,

âˆ *DPC *= âˆ *ABC *â€¦(eq.2) [Corresponding angles]

From (eq.1) and (eq.2), we get

âˆ ABC = âˆ DEF

Hence, Proved.

Exercise 6.4 Page No: 61

**1. If two lines intersect, prove that the vertically opposite angles are equal.**

**Solution:**

From the figure, we know that,

AB and CD intersect each other at point O.

Let the two pairs of vertically opposite angles be,

1^{st}Â pair –Â âˆ AOC andÂ âˆ BOD

2^{nd}Â pair –Â âˆ AOD andÂ âˆ BOC

To prove:

Vertically opposite angles are equal,

i.e., âˆ AOC =Â âˆ BOD, andÂ âˆ AOD =Â âˆ BOC

From the figure,

The ray AO stands on the line CD.

We know that,

If a ray lies on a line then the sum of the adjacent angles is equal to 180Â°.

â‡’Â âˆ AOC +Â âˆ AOD = 180Â° (By linear pair axiom) â€¦ (i)

Similarly, the ray DO lies on the line AOB.

â‡’Â âˆ AOD +Â âˆ BOD = 180Â° (By linear pair axiom) â€¦ (ii)

From equations (i) and (ii),

We have,

âˆ AOC +Â âˆ AOD =Â âˆ AOD +Â âˆ BOD

â‡’Â âˆ AOC =Â âˆ BOD – – – – (iii)

Similarly, the ray BO lies on the line COD.

â‡’Â âˆ DOB +Â âˆ COB = 180Â° (By linear pair axiom) – – – – (iv)

Also, the ray CO lies on the line AOB.

â‡’Â âˆ COB +Â âˆ AOC = 180Â° (By linear pair axiom) – – – – (v)

From equations (iv) and (v),

We have,

âˆ DOB +Â âˆ COB =Â âˆ COB +Â âˆ AOC

â‡’Â âˆ DOB =Â âˆ AOC – – – – (vi)

Thus, from equation (iii) and equation (vi),

We have,

âˆ AOC =Â âˆ BOD, andÂ âˆ DOB =Â âˆ AOC

Therefore, we get, vertically opposite angles are equal.

Hence Proved.

**2. Bisectors of interior âˆ B and exterior âˆ ACD of a Î” ABC intersect at the point T.**

**Prove that âˆ BTC = Â½ âˆ BAC.**

**Solution:**

Given: â–³ ABC, produce BC to D and the bisectors of âˆ ABC and âˆ ACD meet at point T.

To prove:

âˆ BTC = Â½ âˆ BAC

Proof:

In â–³ABC,âˆ ACD is an exterior angle.

We know that,

Exterior angle of a triangle is equal to the sum of two opposite angles,

Then,

âˆ ACD = âˆ ABC + âˆ CAB

Dividing L.H.S and R.H.S by 2,

â‡’ Â½ âˆ ACD = Â½ âˆ CAB + Â½ âˆ ABC

â‡’ âˆ TCD = Â½ âˆ CAB + Â½ âˆ ABC â€¦(1)

[âˆµCT is a bisector of âˆ ACDâ‡’ Â½ âˆ ACD = âˆ TCD]We know that,

Exterior angle of a triangle is equal to the sum of two opposite angles,

Then in â–³ BTC,

âˆ TCD = âˆ BTC +âˆ CBT

â‡’ âˆ TCD = âˆ BTC + Â½ âˆ ABC â€¦(2)

[âˆµBT is bisector of â–³ ABC â‡’âˆ CBT = Â½ âˆ ABC ]From equation (1) and (2),

We get,

Â½ âˆ CAB + Â½ âˆ ABC = âˆ BTC + Â½ âˆ ABC

â‡’ Â½ âˆ CAB = âˆ BTC or Â½ âˆ BAC = âˆ BTC

Hence, proved.

**3. A transversal intersects two parallel lines. Prove that the bisectors of any pair of corresponding angles so formed are parallel.**

**Solution:**

Let,

ABÂ â•‘ CD

EF be the transversal passing through the two parallel lines at P and Q respectively.

PR and QS are the bisectors ofÂ âˆ EPB and âˆ PQD.

We know that the corresponding angles of parallel lines are equal,

So, âˆ EPB = âˆ PQD

Â½ âˆ EPB = Â½ âˆ PQD

âˆ EPR = âˆ PQS

But, we also know that they are corresponding angles of PR and QS

Since the corresponding angles are equal,

We have,

PRÂ â•‘ QS

Hence Proved.

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