SEBA Class 9 Mathematics Chapter 6 Lines and Angles

SEBA Class 9 Mathematics Chapter 6 Lines and Angles Solutions, SEBA Class 9 Maths Textbook Notes in English Medium, SEBA Class 9 Mathematics Chapter 6 Lines and Angles Solutions in English to each chapter is provided in the list so that you can easily browse throughout different chapter Assam Board SEBA Class 9 Mathematics Chapter 6 Lines and Angles Notes and select needs one.

SEBA Class 9 Mathematics Chapter 6 Lines and Angles

Join Telegram channel

Follow us:

Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. SEBA Class 9 Mathematics Chapter 6 Lines and Angles Question Answer. These solutions are part of SCERT All Subject Solutions. Here we have given SEBA Class 9 Mathematics Chapter 6 Lines and Angles Solutions for All Subject, You can practice these here.

Lines and Angles

Chapter – 6

Q.1. In figure, lines AB and CD intersect at 0. If ∠ AOC+ ∠ BOE = 70⁰and ∠ BOD = 40⁰ , find ∠ BOE and reflex ∠COE.

Ans: ∴ Lines AB and CD intersect at O 

∴ ∠ AOC + ∠ BOD 

| Vertically Opposite Angles 

But ∠ BOD = 40⁰ , …. (1) | Given 

∠ AOC = 40⁰ …(2) 

Now, ∠ AOC + ∠BOE = 70⁰ 

⇒ 40⁰ + ∠ BOE = 70⁰   | Using (2)

⇒ ∠ BOE = 70⁰ – 40⁰ 

⇒ ∠ BOE = 30⁰

Again, Reflex  ∠COE =∠COD + ∠ BOD + ∠BOE 

= ∠ COD + 40⁰ + 30⁰

 | Using (1) and (2)

= 180⁰ + 40⁰ + 30⁰

 ∴ Ray OA stands on line CD 

= 250⁰ (Linear Pair Axiom)

Q.2. In figure, lines XT and MN intersect at 0. If⇒∠POY=90⁰

and a:b = 2:3, find c.

Ans: Ray OP stands on line XY

∴ ∠POX + ∠POY = 180⁰ | Linear Pair Axioms

⇒ ∠POX +90⁰ = 180⁰ |∵∠POY=90⁰(Given)

⇒ ∠POX = 180⁰-90⁰

⇒ ∠POX  = 90⁰

⇒ ∠POM + ∠XOM = 90⁰

⇒ a+b=90  ….(1)

A:b = 2:3

⇒ a = 2k

B = 3k

Putting the values of a and b in (1), we get

2k + 3k = 90⁰

⇒ k=18⁰

…(2)

∴ Ray OX stands on line MN

∴ ∠XOM + ∠ XON = 180⁰ | Linear Pair Axioms

⇒b+c=180⁰ ⇒54⁰ +c=180⁰ | Using (2)

⇒c = 180⁰-54⁰⇒c = 126⁰

Q.3. In figure,∠PQR=∠PRQ then prove that ∠PQS = ZPRT.

Ans: Ray QP stands on line ST

∴ ∠ PQS+ ∠ PQR = 180⁰   …(1)| Linear Pair Axiom 

∴ Ray RP stands on line ST 

∴ ∠ PRQ + ∠ PRT = 180⁰

…(2) Linear Pair Axiom 

From (1) and (2), we obtain 

∠ PQS + ∠ PQR=∠PRQ+ ∠ PRT 

⇒ ∠ PQS = ∠PRT 

|∵  ∠ PQR = ∠ PRQ(Given)

Q.4. In figure, if x + y = w + z then prove that AOB is a line.

Ans: x + y = w + z    …(1)| Given 

∴ The sum of all the angles around a point is equal to 360⁰ 

∴ x + y + w + z = 360⁰  

⇒ x + y + x + y = 360⁰ | Using (1) 

⇒ 2(x + y) = 360⁰

⇒ x + y = 180⁰ 

∴ AOB is a line If the sum of two adjacent angles is 180⁰, then the non-common arms of the angles form a line

Q.5. In the figure, POQ is a line. Ray OR is perpendicular to line PQ. 

OS is another ray lying between rays OP and OR. Prove that

Ans:  Ray OR is perpendicular to the line PQ

From (2) and (3),

∴ ∠ QOS – ∠ POS =(∠QOR- ∠POR) +2 ∠ROS=2∠ROS

| Using (1)

Q.6. It is given that ∠ XYZ = 64⁰ and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ ZYP find ∠ XYQ and reflex ∠ QYP

Ans:  Ray YZ stands on line PX 

∴ ∠ XYZ+ ∠ ZYP = 180⁰ l Linear Pair Axiom

⇒ 64^ 0 + ∠ ZYP = 180⁰

| ∵ ∠ XYZ = 64⁰ Given) 

⇒ ∠ ZYP = 180⁰- 64⁰

⇒∠ZYP = 116⁰     …(1)

∵ Ray YQ bisects ∠ ZYP

=58⁰    …(2)

∴ Reflex ∠ QYP = 360⁰ – 58⁰=302⁰ 

∵ The sum of all the angles round a point is equal to 360⁰ 

Again, ∠ XYQ = ∠ XYZ+ ∠ ZYQ 

= 64⁰ + 58⁰

= 122⁰

Q.1. In figure, find the values of x and y and then show that AB||CD

Ans: Let transversal l intersects AB and CD at P and Q respectively.

According to given fig., 

50⁰ + x = 180⁰ (linear pair) 

⇒ x = 180⁰ – 50⁰ 

⇒ x= 130⁰  …(i) 

y = 130⁰

(vertically opposite angles) … (ii) 

From (i) and (ii), we observe that: x = y

It shows that interior, alternate interior angles are equal.

As we know that if a transversal (say l) intersects two lines in such a way that a pair of alternate interior angles are equal, then the two lines are parallel.

Therefore, ABIICD

Q.2. In figure, if AB|| CD, CD || EF and y:z = 3:7 find x

Ans:

∴ AB|| CD and CD ED

∴ AB|| EF | Lines parallel to the same line are parallel to each other

∴ x = z    …(1)

x+y=180⁰   …(2)

Alternate Interior Angles 

Consecutive interior angles on the same side of the transversal GH to parallel lines AB and CD

From (1) and (2), 

z + y = 180⁰ 

y:z = 3:7 

Sum of the ratios =3+7=10

∴ x = z = 126⁰ 

Q.3. In figure, if AB II CD, EF⊥CD and ∠ GED = 126 deg , find∠AGE, GEF and ∠ FGE.

Ans: (i) ∠ AGE= ∠ GED = 126⁰

| Alternate Interior Angles 

(ii) ∠ GED = 126⁰ 

⇒ ∠GEF+ ∠ FED = 126⁰ 

⇒ ∠GEF + 90⁰ = 126⁰

| ∵  EF⊥CD ∠ FED = 90⁰ 

⇒ ∠GEF = 126⁰- 90⁰= 36⁰

(iii) ∠ GEC+ ∠GEF+ ∠ FED = 180⁰ |∵ CD is a line

Now, ∠FGE = ∠ GEC = 54⁰| Alternate Interior Angles

Q.4. In figure, if PQ || ST, ∠PQR = 110⁰ and ∠RST=130⁰, find ∠QRS

 [Hint. Draw a line parallel to ST through point R ]

Ans: Construction: Draw a line RU parallel to ST through Point R

∠RST + SRU=180⁰ Sum of the consecutive

interior angles on the

⇒130⁰+SRU=180⁰ same side of the

 transversal is 180⁰

⇒ ∠SRU=180⁰-130⁰ 50⁰ ….(1)

∠QRU = ∠ PQR = 110⁰ | Alternate Interior Angles

⇒ ∠QRS +∠SRU=110⁰

⇒ ∠QRS+50⁰ 110⁰    | Using (1)

⇒∠QRS=110⁰-50⁰ = 60⁰

Q.5. In figure, If AB|| CD, ∠APQ = 50⁰ and PRD=127⁰, find x and y.

Ans: x = APQ = 50⁰

| Alternate Interior Angles

x=∠PRD=127⁰

| Alternate Interior Angles

Q.6. In figure, 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 againreflects back along CD. Prove that AB|| CD.

Or

In figure, m and n are two plane mirrors parallel to each other. Prove that incident ray CA is parallel to reflected ray BD.

[Hint. Draw perpendiculars (normals) at A and B to the plane mirrors. Recall that angle of incidence is equal to angle of reflection.]

Ans: Construction: Draw ray BL ⊥ PQ and ray CM⊥RS

Proof: ∴ BL⊥PQ, CM⊥RS and PQ II RS 

∴ BL | CM

= Angle of reflection

From (1), (2) and (3), we get

 ∠ ABL = ∠ MCD   …(4)

Adding (1) and (4), we get

∠ LBC + ∠ ABL=∠MCB+ ∠ MCD 

⇒ ∠ ABC =∠BCD

But these are alternate interior angles and they are equal. 

So, AB || CD

Q.1. In figure, sides QP and RQ of ∆PQR are produced to points S and respectively. If ∠SPR = 135⁰ and∠PQT = 110⁰, find ∠PRQ.

Ans: TR is a line

∴ ∠PQT+∠PQR = 180⁰

⇒ 110⁰+∠PQR = 180⁰

⇒ ∠PQR = 180⁰-110⁰= 70⁰

∴ QS is a line

∴ ∠SPR+∠QPR = 180⁰

⇒ 135⁰ +∠QPR = 180⁰

⇒ ∠QPR =180⁰-135⁰= 45υ   …(2)

In ∆ PQR,

∠ PQR +∠ QPR + ∠PRQ=180⁰ | ∵ The sum of all the angles of a triangles is 180⁰

⇒ 70⁰ +45⁰ + PRQ=180⁰ | Using (1) and (2)

⇒ 115⁰ + ∠PRQ = 180⁰

⇒∠PRO=180⁰-115⁰ = 65⁰.

Q.2. In figure, ∠XYZ=54⁰. IfYO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OZY and∠YOZ

Ans: In ∆ XYZ

∠XYZ + ∠YZX + ∠ZXY = 180⁰

| ∵ The sum of all the angles of a triangle is 180⁰

⇒ 54⁰+ ∠ YZX + 62⁰ = 180⁰

⇒116⁰+ ∠ YZX = 180⁰

⇒ ∠YZX=180⁰-116⁰ = 64⁰     …(1)

∵ YZ is the bisector of ∠XYZ

∵ YO is the bisector of ZYZX

| Using (1)

∠OZ+∠OZ∠YOZ = 180⁰ 

 | ∵ The sum of all the angles of triangle is 180 

⇒ 27⁰ +32⁰+ ∠ YOZ = 180⁰ | Using (2) and (3)

⇒ 59⁰ +∠YOZ=180⁰⇒∠ YOZ=180⁰- 59⁰= 121⁰

Q.3. In figure, if AB || DE, ∠ BAC = 35⁰ and ∠ CDE = 35⁰, find∠DCE.

Ans: ∠DEC=∠BAC = 35⁰ …(1) 

| Alternate Interior Angles 

∠CDE = 53⁰ …(2) | Given 

In∆CDE, 

∠CDE+∠DEC∠DCE = 180⁰  

| ∵ The sum of all the angles of triangle is 180⁰ 

⇒ 53⁰ +35⁰+ ∠ DCE = 180⁰ | Using (1) and (2) 

⇒88⁰ +∠DCE = 180⁰

⇒∠DCE = 180⁰ – 88⁰ = 92⁰

Q.4. In figure, if lines PQ and RS intersect at point T, such that ∠ PRT = 40⁰, ∠ RPT = 95⁰ and ∠ TSQ = 75⁰, find ∠ SQT

Ans: In ∆ PRT,

∠PTR+ ∠PRT+ ∠RPT = 180⁰

| ∵ The sum of all the ∠ of a triangle is 180⁰ 

⇒ ∠PTR+ ∠40⁰ + 95⁰= 180⁰

⇒ ∠PTR + 135⁰= 180⁰

⇒ ∠PTR = 45⁰ 

⇒ ∠QTS= ∠ PTR = 45⁰ 

Vertically Opposite Angles

In∆TSQ,

∠QTS + ∠TSQ+∠SQT = 180⁰

| ∵ The sum of all the angles of a triangle is 180⁰

 ⇒ 45⁰ +75⁰+ ∠ SQT = 180⁰ 

⇒ 120⁰ +∠SQT = 180⁰

⇒ ∠ SQT = 180⁰ – 120⁰= 60⁰

Q.5. In figure, if PQ⊥PS, PQIISR, ∠ SQR = 28⁰ and ∠ QRT = 65⁰ , then find the values of x and y.

Ans: ∠QRT =∠RQS+ ∠ QSR

| ∵ The Exterior angle is equal to the sum of the two interior opposite angles

⇒ 65⁰=28⁰+∠QSR 

⇒ ∠ QSR = 65⁰ – 28⁰ = 37⁰ 

∵ PQ⊥SP 

∴ ∠QPS = 90⁰ 

∵ PQ||SR 

∴ ∠QPS+∠PSR= 180⁰

∴ The sum of consecutive interior angles on the same side of the transversal is 180⁰

⇒ 90⁰ +∠ PSR = 180⁰ 

⇒ ∠ PSR = 180⁰ – 90⁰ = 90⁰ 

⇒ ∠ PSR+ ∠ QSR = 90⁰

⇒ y + 37⁰- 37⁰ = 53⁰ 

In ∆PQS, 

∠ PQS+ ∠ QSP+ ∠ QPS = 180⁰

∵ The sum of all the angles of a triangle is 180⁰

⇒ x + y + 90⁰ = 180⁰

⇒ x + 53⁰ + 90⁰ = 180⁰ 

⇒ x + 143⁰= 180⁰

⇒ x = 180⁰ – 143⁰ = 37⁰

Q.6. In figure, the side QR of triangle PQR is produced to a point S. If the bisectors of ∆ PQR and ∠ PRS meet at point T, then that 

Ans: ∵ ∠ TRS is an exterior angle of ∆ TQR 

∴ ∠ TRS = ∠ TQR+ ∠ QTR …(1) 

∵ The exterior angle is equal to sum of the two interior opposite angles 

∵ PRS is an exterior angle of ∆ PQR 

∴ ∠ PRS = ∠ PQR+ ∠ QPR …(2) 

∵ The exterior angle is equal to sum of

the two interior opposite angles 

⇒ 2 ∠ TRS =2 ∠ TQR+ ∠ QPR |∵ QT is the bisector of ∠ PQR and RT is the bisector of ∠ PRS 

⇒ 2( ∠ TRS – ∠ TQR)= ∠ QPR …(3) 

From (1), 

∠TRS -∠TQR=∠QTR …(4) 

From (3) and (4), we obtain