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

NCERT Exemplar Solutions Class 9 Maths Chapter 6 – Free PDF Download

NCERT Exemplar Solutions Class 9 Mathematics Chapter 6 Lines and Angles are provided in PDF for students to prepare for the annual exam. Our subject experts have designed exemplar problems in accordance with the latest CBSE syllabus (2023-2024) for Class 9, which covers the following topics of Chapter 6, Lines and Angles:

• 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 and consecutive interior angles.
• Lines parallel to the same line
• Angle sum property of a triangle

This chapter is divided into two parts, and in the first part, the students will learn about lines, and in the second part, they 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 them understand these concepts effectively, a free PDF of the NCERT Exemplar Class 9 Maths are provided below.

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, ∠RQD = 25° and ∠CQP = 60°, then ∠QRS is equal to

(A) 85°

(B) 135°

(C) 145°

(D) 110°

Solution:

(C) 145°

Explanation:

According to the given figure, we have

AB || CD || EF

PQ || RS

∠RQD = 25°

∠CQP = 60°

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

⇒ ∠PQC = ∠BRS

We have ∠PQC = 60°

⇒ ∠BRS = 60° … 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

⇒ ∠DQR = ∠QRA

We have ∠DQR = 25°

⇒ ∠QRA = 25° … eq.(ii)

Using linear pair axiom,

We get,

∠ARS + ∠BRS = 180°

⇒ ∠ARS = 180° – ∠BRS

⇒ ∠ARS = 180° – 60° (From (i), ∠BRS = 60°)

⇒ ∠ARS = 120° … eq.(iii)

Now, ∠QRS = ∠QRA + ∠ARS

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

∠QRA = 25° and ∠ARS = 120°

Hence, the above equation can be written as:

∠QRS = 25° + 120°

⇒ ∠QRS = 145°

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 △ABC be ∠A, ∠B and ∠C

Given that ∠A= ∠B+∠C …(eq1)

But, in any △ABC,

Using the angle sum property, we have,

∠A+∠B+∠C=180^o …(eq2)

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

∠A+∠A=180^o

⇒2∠A=180^o

⇒∠A=180^o/2 = 90^o

⇒∠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 105°, and its two interior opposite angles are equal. Each of these equal angles is

(A) 37 ½^o

(B) 52 ½^o

(C) 72 ½^o

(D) 75°

Solution:

(B) 52 ½^o

Explanation:

According to the question,

The exterior angle of triangle = 105°

Let the two interior opposite angles of the triangle = x

We know that,

The exterior angle of a triangle = sum of interior opposite angles

Then, we have the equation,

105° = x + x

2x = 105°

x = 52.5°

x = 52½

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 = 5×12 = 60^o

3x = 3×12 = 36^o

7x = 7×12 = 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:

The 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

∠ABD = y

and, ∠DBC = x

We know,

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

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

Thus, for ABC to be a line,

The sum of ∠ABD and ∠DBC should be equal to 180°.

⇒ ∠ABD + ∠DBC = 180°

⇒ x + y = 180°

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

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

Solution:

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

Justification:

According to the 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 ≠ 180^o

Hence, we can say that if all the angles are less than 60^o, the measure of the angles won’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 reasons for your answer.

Solution:

No. A triangle cannot have two obtuse angles

Justification:

According to the 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 90° but less than 180°.

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 91° each.

On adding these two angles,

Sum of the two angles = 91° + 91°

⇒ Sum of the two angles = 182°

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 45°, 64° and 72°? Give reasons for your answer.

Solution:

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

Justification:

According to the 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 45°, 64° and 72°.

Sum of these angles = 45° + 64° + 72°

= 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 45°, 64° and 72°.

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

Solution:

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

Justification:

According to the 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 53°, 64°, and 63°.

Sum of these angles = 53° + 64° + 63°

= 180^o

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

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

Exercise 6.3 Page No: 58

1. In Fig. 6.9, OD is the bisector of ∠AOC, OE is the bisector of ∠BOC and OD ⊥ OE. Show that points A, O and B are collinear.

Solution:

According to the question,

In figure,

OD ⊥ OE,

OD and OE are the bisectors of ∠AOC and ∠BOC.

To prove: Points A, O and B are collinear

i.e., AOB is a straight line.

Proof:

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

∠AOC = 2∠DOC …(eq.1)

And ∠COB = 2∠COE …(eq.2)

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

∠AOC = ∠COB = 2∠DOC + 2∠COE

∠AOC +∠COB = 2(∠DOC +∠COE)

∠AOC + ∠COB = 2∠DOE

Since, OD⊥OE

We get,

∠AOC +∠COB = 2×90^o

∠AOC +∠COB =180^o

∠AOB =180^o

So, ∠AOC + ∠COB form linear pair.

Therefore, AOB is a straight line.

Hence, points A, O and B are collinear.

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

Solution:

According to the question,

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

Since, ∠1 = 60° and ∠6 = 120°

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

∠3 = ∠1 = 60°

Now, ∠3 + ∠6 = 60° + 120°

⇒ ∠3 + ∠6 = 180°

We know that,

If the sum of two interior angles on the same side of l is 180°, 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

∠MAB = ∠SBA [alternate angles]

⇒ ½ ∠MAB = ½ ∠SBA

⇒ ∠PAB = ∠QBA

⇒ ∠2 = ∠3

But, ∠2 and ∠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 ∠MAB

BQ is the bisector of ∠SBA.

Given: AP||BQ.

As AP||BQ,

We have,

So ∠2 = ∠3 [Alternate angles]

2∠2 = 2∠3

⇒ ∠2 + ∠2 = ∠3 +∠3

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

⇒ ∠1+ ∠2 = ∠3 +∠4

⇒ ∠MAB = ∠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 ∠ABC = ∠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 are the transversal,

∠DEF = ∠DPC …(eq.1) [Corresponding angles]

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

∠DPC = ∠ABC …(eq.2) [Corresponding angles]

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

∠ABC = ∠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 pair – ∠AOC and ∠BOD

2^nd pair – ∠AOD and ∠BOC

To prove:

Vertically opposite angles are equal,

i.e., ∠AOC = ∠BOD, and ∠AOD = ∠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 180°.

⇒ ∠AOC + ∠AOD = 180° (By linear pair axiom) … (i)

Similarly, the ray DO lies on line AOB.

⇒ ∠AOD + ∠BOD = 180° (By linear pair axiom) … (ii)

From equations (i) and (ii),

We have,

∠AOC + ∠AOD = ∠AOD + ∠BOD

⇒ ∠AOC = ∠BOD – – – – (iii)

Similarly, the ray BO lies on the line COD.

⇒ ∠DOB + ∠COB = 180° (By linear pair axiom) – – – – (iv)

Also, the ray CO lies on line AOB.

⇒ ∠COB + ∠AOC = 180° (By linear pair axiom) – – – – (v)

From equations (iv) and (v),

We have,

∠DOB + ∠COB = ∠COB + ∠AOC

⇒ ∠DOB = ∠AOC – – – – (vi)

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

We have,

∠AOC = ∠BOD, and ∠DOB = ∠AOC

Therefore, we get vertically opposite angles are equal.

Hence Proved.

2. Bisectors of interior ∠B and exterior ∠ACD of a Δ ABC intersect at point T.

Prove that ∠ BTC = ½ ∠ BAC.

Solution:

Given: △ ABC, produce BC to D, and the bisectors of ∠ABC and ∠ACD meet at point T.

To prove:

∠BTC = ½ ∠BAC

Proof:

In △ABC,∠ACD is an exterior angle.

We know that,

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

Then,

∠ACD = ∠ABC + ∠CAB

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

⇒ ½ ∠ACD = ½ ∠CAB + ½ ∠ABC

⇒ ∠TCD = ½ ∠CAB + ½ ∠ABC …(1)

We know that,

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

Then in △ BTC,

∠TCD = ∠BTC +∠CBT

⇒ ∠TCD = ∠BTC + ½ ∠ABC …(2)

From equations (1) and (2),

We get,

½ ∠CAB + ½ ∠ABC = ∠BTC + ½ ∠ABC

⇒ ½ ∠CAB = ∠BTC or ½ ∠BAC = ∠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,

AB ║ CD

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

PR and QS are the bisectors of ∠EPB and ∠PQD.

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

So, ∠EPB = ∠PQD

½ ∠EPB = ½ ∠PQD

∠EPR = ∠PQS

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

Since the corresponding angles are equal,

We have,

PR ║ QS

Hence Proved.

Students of Class 9 can 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 question papers are also provided at BYJU’S as study materials for students to learn and practise for their final exams.

Sample papers and previous years’ question papers will help students to know the question pattern and marks allotted for Chapter 6 of Class 9 Maths. Also, solve important questions with NCERT Exemplar for Chapter 6, Lines and Angles, by downloading the solutions in PDF, available in this article. Download BYJU’S – The Learning App to get personalised videos teaching various concepts of Maths, such as Lines and Angles and related topics, with the help of pictures and video animations.