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

Exercise 6.1

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°

Ans.

We have, PQ || RS

Now, produce PQ to M.

We have, ∠CQP = ∠MQD                 \[\text {[Vertically opposite angles]}\]

⇒ 60° = ∠MQR + 25°

⇒ ∠MQR = 35°

Also, QM || RS and QR cut them.

∠ARQ = ∠RQD = 25°                          \[\text {[Alternate angles]}\]

Now, ∠MQR + (∠ARQ + ∠ARS) = 180°

⇒ 35° + 25° + ∠ARS = 180°

⇒ ∠ARS = 180° – 60° = 120°

∴ ∠QRS = ∠ARQ + ∠ARS = 25° + 120° = 145°

Hence, 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

Ans. Let the triangle be ∆PQR and its angles be ∠P, ∠Q and ∠R.

Now, 

∠P = ∠Q + ∠R                                          ...(1) \[\text {[Given]}\]

Also, in ∆PQR, ∠P + ∠Q + ∠R = 180º     ...(2) \[\text {[Angles sum property of triangle]}\]

Using equations (1) and (2), we have  

∠P + ∠P = 180º 

 ⇒ 2∠P = 180º 

⇒ ∠P =  = 90º

Therefore, the triangle is a right triangle

Hence, 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) $37\frac{1}{2}$ º  

(B) $52\frac{1}{2}$ º  

(C) $72\frac{1}{2}$ º  

(D) 75º

Ans. The exterior angle of a triangle is given as 105o.

Now, let each of the two interior opposite angles measure x0.

Also, exterior angle of a triangle = sum of its two interior opposite angles.

$ \Rightarrow $x0 + x0 = 105o

$ \Rightarrow $2x0 = 105o

∴ x = \[\left( {\frac{{{\text{105}}}}{{\text{2}}}} \right)\]o = ${\text{52}}\frac{{\text{1}}}{{\text{2}}}$

Hence, 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

Ans. Let the angles of the triangle be 5x°, 3x° and 7x°.

Now, the sum of all the angles of a triangle = 180°, we have

5x° + 3x° + 7x° = 180°

⇒15x° = 180°  

∴ x =  = 12°

Thus, the angles of the triangle are

5 × 12°, 3 × 12° and 7 × 12°, i.e., 60°, 36° and 84°

Since, the measure of each angle of the triangle is less than 90°, therefore the angles of the triangle are acute angles.

So, the triangle is an acute angled triangle. 

Hence, option (A) is the correct answer.

5. If one of the angles of a triangle is 130º, then the angle between the bisectors of the other two angles can be

(A) 50°            

(B) 65°  

(C) 145°          

(D) 155°

Ans. In ∆PQR, we have,

∠P = 130°

OQ and OR are the bisectors ∠Q and ∠R.

∴ ∠OQR + ∠ORQ =        

Now, ∠QOR = 180° – (∠OQR + ∠ORQ) = 180° – 25° = 155°

Hence, option (D) is the correct answer.

6. In Fig. 6.2, POQ is a line. The value of x is

(A) 200  

(B) 250  

(C) 300  

(D) 350

Ans. We have, according to the question,

3x + 4x + 40° = 180°

⇒7x + 40° = 180°  

⇒7x = 180° – 40° = 140°

∴ x =  = 20°

Hence, option (A) is the correct answer.

7. In Fig. 6.3, if OP || RS, ∠OPQ = 110º and ∠QRS = 130º, then ∠PQR is equal to 

(A) 40° 

(B) 50° 

(C) 60° 

(D) 70°

Ans:  

We have, In the given figure, produce OP such that it intersects RQ at X. 

Also, OP || RS and RX is a transversal. 

⇒ ∠RXP = ∠XRS     \[\text {[Alternate angles]}\]

∠RXP = 130º ...(1)          \[\text{[∵ ∠QRS = 130⁰]}\]

Now, RQ is a line segment. 

So, ∠PXQ + ∠RXP = 180º \[\text {[Linear pair axiom]}\]

⇒∠PXQ = 180º – ∠RXP = 180º – 130º  \[\text {[Using equation (1)]}\]

⇒∠PXQ = 50º

In ∆PQX, we have, ∠OPQ is an exterior angle,

⇒∠OPQ = ∠PXQ + ∠PQX      \[\text {[∵ exterior angle = sum of opposite interior angles]}\]

⇒110º = 50º + ∠PQX  

⇒∠PQX = 110º – 50º

⇒∠PQX = 60º 

∴ ∠PQR = 60º \[\text {[∵ ∠PQX = ∠PQR]}\]

Hence, option (C) is the correct answer.

8. Angles of a triangle are in the ratio 2 : 4 : 3. The smallest angle of the triangle is

(A) 60° 

(B) 40° 

(C) 80° 

(D) 20°

Ans. We have, the ratio of angles of the triangle is 2 : 4 : 3.

Let the angles of the triangle be 2x, 4x and 3x.

Now, we know that the sum of angles of a triangle is 180°

⇒ 2x + 4x + 3x = 180°  

⇒ 9x = 180°  

∴ x = $\left( {\frac{{{\text{180}}}}{{\text{9}}}} \right)$° = 20°

So, the angles of the triangle are 

2x = 2 × 20° = 40°

4x = 4 × 20° = 80° 

3x = 3 × 20° = 60°

Thus, the smallest angle of the triangle is 40° 

Hence, option (B) is the correct answer.

Exercise 6.2  

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

Ans. In the given figure, x and y are two adjacent angles. So, for ABC to be a straight line, the sum of two adjacent angles x and y must be 180° 

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

Ans. A triangle cannot have all angles less than 60°. Then, sum of all the angles will be less than 180° whereas sum of all the angles of a triangle is always 180°.

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

Ans. An angle whose measure is more than 90° but less than 180° is called an obtuse angle. A triangle cannot have two obtuse angles because the sum of all the angles of it cannot be more than 180°. It is always equal to 180°.

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

Ans. We cannot draw any triangle having its angles 45°, 64° and 72° because the sum of the angles (45° + 64° + 72° = 181°) cannot be 181°.

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

Ans. Sum of these angles = 53° + 64° + 63° = 180°. So, we can draw infinitely many triangles, sum of the angles of every triangle having its angles as 53°, 64° and 63° is 180°.

6. In Fig. 6.5, find the value of x for which the lines l and m are parallel.

Ans. If a transversal intersects two parallel lines, then each pair of co- interior angles are supplementary. Here, the two given lines l and m are parallel. Angles x and 44° are co interior angles on the same side of the transversal.

⇒ x + 44° = 180°

⇒ x = 180° – 44° = 136°

7. Two adjacent angles are equal. Is it necessary that each of these angles will be a right angle? Justify your answer.

Ans. No, each of these angles will be a right angle only if they form a linear pair, i.e., when the non-common arms of the given two adjacent angles are two opposite rays & common arm is perpendicular to them.

8. If one of the angles formed by two intersecting lines is a right angle, what can you say about the other three angles? Give reason for your answer.

Ans. If two intersect each other at a point, then four angles are formed. If one of these four angles is a right angle, then each of the other three angles will also be a right angle by linear pair axiom.

9. In Fig.6.6, which of the two lines are parallel and why?

Ans. For fig(i), a transversal intersects two lines such that the sum of interior angles on the same side of the transversal is 132o + 48o = 180o. Therefore, the line l and m are parallel. For fig. (ii), a transversal intersects two lines such that the sum of interior angles on the same side of the transversal is 73o + 106o = 179o. Therefore, the lines p and q are not parallel.

10. Two lines l and m are perpendicular to the same line n. Are l and m perpendicular to each other? Give reason for your answer.

Ans. When two lines l and m are perpendicular to the same line n, each of the two corresponding angles formed by these lines l and m with the line n are equal (each is equal to 90o). Hence, the line l and m are not perpendicular but parallel.

Exercise 6.3

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

Ans. Given: 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 … (1)

and ∠COB = 2∠COE … (2)

On adding equations (1) and (2), we get

∠AOC + ∠COB = 2∠DOC + 2∠COE

⇒∠AOC + ∠COB = 2 (∠DOC + ∠COE)

⇒∠AOC +∠ COB = 2∠DOE

⇒∠AOC + ∠COB = 2 × 90° [OD⊥OE]

⇒∠AOC + ∠COB = 180°

⇒∠AOB = 180°

So, ∠AOC + ∠COB are forming linear pair or 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.

Ans. We have,

∠5 + ∠6 = 180°   \[\text {[Linear pair]}\]

⇒∠5 + 120° = 180°  

⇒∠5 = 180° – 120° = 60°

Now, ∠1 = ∠5 \[\text {[Each = 60°]}\]

But, these are corresponding angles. 

Hence, the lines m and n are parallel.

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

Ans. 

∵ l || m and t is the transversal

⇒ ∠MAB = ∠SBA   \[\text {[Alternate angles]}\]

⇒ $\frac{{\text{1}}}{{\text{2}}}$ ∠MAB = $\frac{{\text{1}}}{{\text{2}}}$ ∠SBA  

⇒ ∠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.

Ans.

AP is the bisector of ∠MAB and BQ is the bisector of ∠SBA. 

We are given that

AP || BQ.

As AP || BQ, so ∠2 = ∠3 \[\text {[Alternate angles]}\]

⇒ 2∠2 = 2∠3

⇒ ∠2 + ∠2 = ∠3 + ∠3

⇒ ∠1 + ∠2 = ∠3 + ∠4 \[\text {[∵ ∠1 = ∠2 and ∠3 = ∠4]}\]

⇒ ∠MAB = ∠SBA 

But, these are alternate angles.

Thus, l || m.

5. In Fig. 6.12., BA || ED and BC || EF. Show that ∠ABC = ∠DEF.          

\[\text {[Hint: Produce DE to intersect BC at P(say)]}\]

Ans. 

Produce DE to intersect BC at P(say).

EF || BC and DP is the transversal,

⇒ ∠DEF = ∠DPC ...(1)          \[\text {[Corresponding angles]}\]

Now, AB || DP and BC is the transversal,

⇒ ∠DPC = ∠ABC ...(2)           \[\text {[Corresponding angles]}\]

From (1) and (2), we get

⇒ ∠ABC = ∠DEF

Hence, proved.

6. In Fig. 6.13, BA || ED and BC || EF. Show that ∠ABC + ∠DEF = 180°.

Ans.

Produce ED to meet BC at P(say).

Now, EF || BC and EP is the transversal.

⇒ ∠DEF + ∠EPC = 180° ...(1)      \[\text {[Co-interior angles]}\]

Again, EP || AB and BC is the transversal.

⇒ ∠EPC = ∠ABC           ...(2)      \[\text {[Corresponding angles]}\]

 From (1) and (2), we get

⇒ ∠DEF + ∠ABC = 180°

⇒ ∠ABC + ∠DEF = 180°

Hence, proved.

7. In Fig. 6.14, DE || QR and AP and BP are the bisectors of ∠EAB and ∠RBA, respectively. Find ∠APB.

Ans. DE || QR and the line n is the transversal line.

⇒ ∠EAB + ∠RBA = 180° ...(1)     \[\text {[Co-interior angles]}\]

⇒ ∠PAB + ∠PBA = 90° \[\text {[∵ AP is the bisector of ∠EAB and BP is the bisector of ∠RBA]}\]  

Now, from ∆APB, we have

⇒ ∠APB = 180° – (∠PAB + ∠PBA)

⇒ ∠APB = 180° – 90° = 90°

8. The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles of the triangle.

Ans. Given: Ratio of angles is 2 : 3 : 4.

To find: Angles of triangle.

Proof: The ratio of angles of a triangle is 2 : 3 : 4. Let the angles of a triangle be A, B and C. Therefore, A = 2x, then B = 3x and C = 4x.

In ∆ABC, ∠A + ∠B + ∠C = 180° \[\text {[∵ Sum of angles of a triangle is 180°]}\]

 ⇒ 2x + 3x + 4x = 180°  

⇒ 9x = 180°  

⇒ x = $\left( {\frac{{{\text{180}}}}{{\text{9}}}} \right)^{0}$= 20°

⇒ ∠A = 2x = 2 × 20° = 40°

⇒ ∠B = 3x = 3 × 20° = 60°

and ∠C = 4x = 4 × 20° = 80°

Hence, the angles of the triangles are 40°, 60° and 80°.

9. A triangle ABC is right-angled at A. L is a point on BC such that AL⊥BC. Prove that ∠BAL = ∠ACB.

Ans.

Given: In ∆ABC,

∠A = 90° and AL⊥BC.

To prove: ∠BAL = ∠ACB. 

Proof: In ∆ABC and ∆LAC,  

⇒∠BAC = ∠ALC … (1) \[\text {[Each 90°]}\]

and ∠ABC = ∠ABL. … (2)  \[\text {[Common angle]}\]

Adding equations (1) and (2), we get

⇒ ∠BAC + ∠ABC = ∠ALC + ∠ABC … (3)  

In ∆ABC, ∠BAC + ∠ACB + ∠ABC = 180° \[\text {[Sum of angles of triangle is 180°]}\]

⇒∠BAC + ∠ABC = 180° – ∠ACB … (4) 

In ∆ABL, ∠ABL + ∠ALB + ∠BAL = 180° \[\text {[Sum of angles of triangle is 180°]}\]

⇒∠ABL + ∠ALC = 180° – ∠BAL ... (5) \[\text {[∠ALC = ∠ALB = 90°]}\]

Substituting the value from equation (4) and (5) in equation (3), we get 

⇒180° – ∠ACB = 180° – ∠BAL  

⇒∠ACB = ∠BAL

Hence, proved.

10. Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.

Ans.

Two lines p and n are respectively perpendicular to two parallel lines l and m, i.e. p ⊥ l and n ⊥ m.

We have to show that p is parallel to n.

As n ⊥ m, so ∠1 = 90°. (1)

Again, p ⊥ l, so ∠2 = 90°. 

But, l is parallel to m, so ∠2 = ∠3 \[\text {[Corresponding angles]}\]

⇒ ∠3 = 90° ...(2)   [∠2 = 90°]

From (1) and (2), we get

⇒ ∠1 = ∠3 [Each = 90°] 

But, these are corresponding angles.

Hence, p || n.

Exercise 6.4

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

Ans.

Given: Two lines AB and CD intersect at point O.

To prove: (i) ∠AOC = ∠BOD

(ii) ∠AOD = ∠BOC

Proof: (i) Since, ray OA stands on line CD. 

⇒ ∠AOC + ∠AOD = 180° ...(1)    \[\text {[Linear pair axiom]}\]

Similarly, ray OD stands on line AB.

⇒ ∠AOD + ∠BOD = 180° ...(2)

From equations (1) and (2), we get

⇒ ∠AOC + ∠AOD = ∠AOD + ∠BOD

⇒ ∠AOC = ∠BOD 

Hence, proved.

(ii) Since, ray OD stands on line AB.

⇒ ∠AOD + ∠BOD = 180° … (3) \[\text {[Linear pair axiom]}\] 

Similarly, ray OB stands on line CD.

⇒ ∠DOB + ∠BOC = 180° … (4)

From equations (3) and (4), we get

⇒ ∠AOD + ∠BOD = ∠DOB + ∠BOC 

⇒ ∠AOD = ∠BOC

Hence, proved.

2. Bisectors of interior ∠B and exterior ∠ACD of a ∆ABD intersect at the point T. Prove that ∠BTC = $\frac{1}{2}$ ∠BAC

Ans.

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

To prove: ∠BTC = $\frac{{\text{1}}}{{\text{2}}}$ ∠BAC 

Proof: In ∆ABC, ∠ACD is an exterior angle

⇒ ∠ACD = ∠ABC + ∠CAB     \[\text {[Exterior angle of a triangle = sum of opposite angles]}\]

⇒ $\frac{{\text{1}}}{{\text{2}}}$ ∠ACD = $\frac{{\text{1}}}{{\text{2}}}$ ∠CAB + $\frac{{\text{1}}}{{\text{2}}}$ ∠ABC    \[\text {[Dividing both sides by 2]}\]

⇒ ∠TCD = $\frac{{\text{1}}}{{\text{2}}}$ ∠CAB + $\frac{{\text{1}}}{{\text{2}}}$ ∠ABC … (1)      \[\text {[ ∵ CT is a bisector of ∠ACD]}\]

In ∆BTC, ∠TCD is an exterior angle

⇒ ∠TCD = ∠BTC + ∠CBT     \[\text {[Exterior angle of a triangle = sum of opposite angles]}\]

⇒ ∠TCD = ∠BTC + $\frac{{\text{1}}}{{\text{2}}}$ ∠ABC … (2) \[\text {[∵ BT is bisector of ∠ABC]}\]

From equations (1) and (2), we get

⇒ $\frac{{\text{1}}}{{\text{2}}}$ ∠CAB + $\frac{{\text{1}}}{{\text{2}}}$ ∠ABC = ∠BTC + $\frac{{\text{1}}}{{\text{2}}}$ ∠ABC

⇒ $\frac{{\text{1}}}{{\text{2}}}$ ∠CAB = ∠BTC or $\frac{{\text{1}}}{{\text{2}}}$ ∠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.

Ans.

Given: Two lines DE and QR are parallel and are intersected by transversal at A and B respectively. Also, BP and AF are the bisectors of ∠ABR and ∠CAE respectively.

To prove: EP || FQ

Proof: Given, DE || QR  

⇒∠CAE = ∠ABR \[\text {[Corresponding angles]}\]          

⇒ $\frac{{\text{1}}}{{\text{2}}}$ ∠CAE = $\frac{{\text{1}}}{{\text{2}}}$ ∠ABR \[\text {[Dividing both sides by 2]}\] 

⇒ ∠CAF = ∠ABP  \[\text {[∵ BP and AF are the bisectors of ∠ABR and ∠CAE respectively]}\]

As these are the corresponding angles on the transversal line n and are equal.

Hence, EP || FQ. 

4. Prove that through a given point, we can draw only one perpendicular to a given line. \[\text {[Hint: Use proof by contradiction]}\]

Ans.

From the point P, a perpendicular PM is drawn to the given line AB.

⇒ ∠PMB = 90°

Let if possible, we draw another perpendicular PN to the line AB. 

Then, ∠PNB = 90°.

⇒ ∠PMB = ∠PNB, which is possible only when PM and PN coincides with each other.

Hence, through a given point, we can draw only one perpendicular to a given line.

5. Prove that two lines that are respectively perpendicular to two intersecting lines intersect each other. \[\text {[Hint: Use proof by contradiction]}\]

Ans. 

Given: Let lines l and m are two intersecting lines. Again, let n and p be another two lines which are perpendicular to the intersecting lines meet at point D.

To prove: Two lines n and p intersecting at a point.

Proof: Let us consider lines n and p are not intersecting, then it means they are parallel to each other i.e., n || p                           ... (1) 

Since, lines n and p are perpendicular to m and l respectively.

But from equation (1), n || p, it implies that l || m. It is a contradiction. Thus, our assumption is wrong. Hence, lines n and p intersect at a point.

6. Prove that a triangle must have at least two acute angles.

Ans. If the triangle is an acute angled triangle, then all its three angles are acute angles. Each of these angles is less than 90°, so they can make three angles sum equal to 180°.

If a triangle is an obtuse angled triangle, then one angle which is obtuse will be more than 90° but less than 180°, so the other two acute angles can make the three angles sum equal to 180°.

If a triangle is a right angled triangle, then one angle which is right angle will be equal to 90° and the other two acute angles can make the three angles sum equal to 180°.

Hence, we can say that a triangle must have at least two acute angles.

7. In Fig. 6.17., Q > R, PA is the bisector of ∠QPR and PM⊥QR. Prove that ∠APM = $\frac{1}{2}$(∠Q – ∠R).

Ans. 

Given: PQR, ∠Q > ∠R, PA is the bisector of ∠QPR and PM⊥QR.

To prove: ∠APM = $\frac{{\text{1}}}{{\text{2}}}$ (∠Q – ∠R)

Proof: Since, PA is the bisector of ∠QPR

⇒ ∠QPA = ∠APR                   ...(1)   

In ∆PQM, ∠Q + ∠PMQ + ∠QPM = 180°     \[\text {[Angle sum property of a triangle]}\]

⇒ ∠Q + 90° + ∠QPM = 180° \[\text {[∵ ∠PMQ = 90°]}\]

⇒ ∠Q = 90° – ∠QPM         ...(2)   

In ∆PMR, ∠PMR + ∠R + ∠RPM = 180° \[\text {[Angle sum property of a triangle]}\]

⇒ 90° + ∠R + ∠RPM = 180° \[\text {[∵ ∠PMQ = 90°]}\]

⇒ ∠R = 180° – 90° – ∠RPM 

⇒ ∠Q = 90° – ∠QPM

⇒ ∠PRM = 90° – ∠RPM … (3)

Subtracting equation (3) from equation (2), we get

⇒ ∠Q – ∠R = (90° – ∠QPM) – (90° – ∠RPM)

⇒ ∠Q – ∠R = ∠RPM – ∠QPM

⇒ ∠Q – ∠R= (∠RPA + ∠APM) – (∠QPA – ∠APM) … (4)   

⇒ ∠Q – ∠R= ∠QPA + ∠APM – ∠QPA + ∠APM       \[\text {[Using equation (1)]}\]

⇒ ∠Q – ∠R= 2∠APM

⇒ ∠APM = $\frac{{\text{1}}}{{\text{2}}}$ (∠Q – ∠R)

Hence, proved.

NCERT Exemplar Solutions for Class 9 Maths Chapter 6 'Lines and Angles', urge you to work on your thinking abilities. This part has 4 activities and 35 altogether. In practice 1 there are 8 questions, in practice 2 there are 10 questions altogether, practice 3 has 10 problems and exercise 4 has 7 questions altogether. From these NCERT Exemplar problems for Chapter 6, you will figure out how to find and make various types of points. Students will figure out how to demonstrate proclamations identified with the given theorems.