NCERT Solutions Class 9 Maths Chapter 6 Lines and Angles – Here are all the NCERT solutions for Class 9 Maths Chapter 6. This solution contains questions, answers, images, explanations of the complete Chapter 6 titled The Fun They Had of Maths taught in class 9. If you are a student of class 9 who is using NCERT Textbook to study Maths, then you must come across Chapter 6 Lines and Angles. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles in one place.

## NCERT Solutions Class 9 Maths Chapter 6 LINES AND ANGLES

Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 9 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

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 Class 9 Subject Maths Book Mathematics Chapter Number 6 Chapter Name LINES AND ANGLES

### NCERT Solutions Class 9 Maths chapter 6 LINES AND ANGLES

Class 9, Maths chapter 6, LINES AND ANGLES solutions are given below in PDF format. You can view them online or download PDF file for future use.

### LINES AND ANGLES

Q.1: In Fig. 6.13, lines AB and CD intersect at O. If ∠ AOC + ∠ BOE = 70° and ∠ BOD = 40°, find ∠ BOE and reflex ∠ COE.

Ans : AB is a straight line, rays OC and OE stand on it. $$\begin{array}{l}{\therefore \angle \mathrm{AOC}+\angle \mathrm{COE}+\angle \mathrm{BOE}=180^{\circ}} \\ {\Rightarrow(\angle \mathrm{AOC}+\angle \mathrm{BOE})+\angle \mathrm{COE}=180^{\circ}} \\ {\Rightarrow 70^{\circ}+\angle \mathrm{COE}=180^{\circ}} \\ {\Rightarrow \angle \mathrm{COE}=180^{\circ}-70^{\circ}=110^{\circ}} \\ {\text { Reflex } \angle \mathrm{COE}=360^{\circ}-110^{\circ}=250^{\circ}}\end{array}$$ CD is a straight line. rays OE and 0B stand on it. $$\begin{array}{l}{\therefore \angle \mathrm{COE}+\angle \mathrm{BOE}+\angle \mathrm{BOD}=180^{\circ}} \\ {\Rightarrow 110^{\circ}+\angle \mathrm{BOE}+40^{\circ}=180^{\circ}} \\ {\Rightarrow \angle \mathrm{BOE}=180^{\circ}-150^{\circ}=30^{\circ}}\end{array}$$

Q.2: In Fig. lines XY and MN intersect at O. If ∠ POY = 90° and a : b = 2 : 3, find c.

Ans : Let the common ratio between a and b be x. a = 2x, and b = 3x XY is a straight line, rays OM and OP stand on it. $$\begin{array}{l}{\therefore \angle \mathrm{XOM}+\angle \mathrm{MOP}+\angle \mathrm{POY}=180^{\circ}} \\ {b+a+\angle \mathrm{POY}=180^{\circ}} \\ {3 x+2 x+90^{\circ}=180^{\circ}} \\ {5 x=90^{\circ}} \\ {x=18^{\circ}} \\ {a=2 x=2 \times 18=36^{\circ}} \\ {b=3 x=3 \times 18=54^{\circ}}\end{array}$$ MN is a straight line. Ray OX stands on it. $$\begin{array}{l}{\therefore b+c=180^{\circ}(\text { Linear Pair })} \\ {54^{\circ}+c=180^{\circ}} \\ {c=180^{\circ}-54^{\circ}=126^{\circ}} \\ {\therefore c=126^{\circ}}\end{array}$$

Q.3: In Fig. ∠ PQR = ∠ PRQ, then prove that ∠ PQS = ∠ PRT

Ans : In the given figure, ST isa straight line and ray QP stands on it. $$\begin{array}{l}{\therefore \angle \mathrm{PQS}+\angle \mathrm{PQR}=180^{\circ}(\text { Linear Pair })} \\ {\angle \mathrm{PQR}=180^{\circ}-\angle \mathrm{PQS}(1)} \\ {\angle \mathrm{PRT}+\angle \mathrm{PRQ}=180^{\circ}(\text { Linear Pair })} \\ {\angle \mathrm{PRQ}=180^{\circ}-\angle \mathrm{PRT}(2)} \\ {\text { It is given that } \angle \mathrm{PQR}=\angle \mathrm{PRQ} \text { . }}\end{array}$$ Equating equations (1) and (2), we obtain $$\begin{array}{l}{180^{\circ}-\angle \mathrm{PQS}=180^{\circ}-\angle \mathrm{PRT}} \\ {\angle \mathrm{PQS}=\angle \mathrm{PRT}}\end{array}$$

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

Ans : $$\begin{array}{l}{\text { It can be observed that, }} \\ {x+y+z+w=360^{\circ}(\text { Complete angle })} \\ {\text { It is given that, }} \\ {x+y=z+w} \\ {\therefore x+y+x+y=360^{\circ}} \\ {2(x+y)=360^{\circ}} \\ {x+y=180^{\circ}}\end{array}$$ Since x and y form a linear pair, AOB is a line.

Q.5: In Fig. POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ ROS = 1/2 (∠ QOS – ∠ POS).

Ans : It is given that $$\mathrm{OR} \perp \mathrm{PQ}$$ $$\begin{array}{l}{\square \square \mathrm{POR}=90^{\circ}} \\ {\square \square \mathrm{POS}+\square \mathrm{SOR}=90^{\circ}}\end{array}$$ $$\begin{array}{l}{\square R O S=90^{\circ}-\square P O S \ldots(1)} \\ {\square Q O R=90^{\circ}(A s O R \square P Q)}\end{array}$$ On adding equations (1) and (2), we obtain $$2 \square \mathrm{ROS}=\square \mathrm{QOS}-\square \mathrm{POS}$$ $$\square \mathrm{ROS}=\frac{1}{2}(\square \mathrm{QOS}-\square \mathrm{POS})$$

Q.6: It is given that ∠ XYZ = 64° 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 : It is given that line YQ bisects $$\square \mathrm{PYZ}$$ Hence, $$\square Q Y P=\square Z Y Q$$ It can be observed that PX is a line. Rays YQ and YZ stand on it. $$\begin{array}{l}{\square \square \times Y Z+\square Z Y Q+\square Q Y P=180^{\circ}} \\ {\square 64^{\circ}+2 \square Q Y P=180^{\circ}} \\ {\square 2 \square Q Y P=180^{\circ}-64^{\circ}=116^{\circ}} \\ {\square \square Q Y P=58^{\circ}}\end{array}$$ Also $$\square Z Y Q=\square Q Y P=58^{\circ}$$ Reflex $$\square Q Y P=360^{\circ}-58^{\circ}=302^{\circ}$$ $$\begin{array}{l}{\square \times Y Q=\square X Y Z+\square Z Y Q} \\ {=64^{\circ}+58^{\circ}=122^{\circ}}\end{array}$$

Q.1: In Fig. 6.28, find the values of x and y and then show that AB || CD.

Ans : It can be observed that, $$\begin{array}{l}{50^{\circ}+x=180^{\circ}(\text { Linear pair })} \\ {x=130^{\circ} \ldots(1)} \\ {\text { Also, } y=130^{\circ}(\text { Vertically opposite angles) }}\end{array}$$ As x and y are alternate interior angles for lines AB and CD and also measures of these angles are equal to each other, therefore, line AB II CO.

Q.2: In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Ans : It is given that AB II CD and CD II EF $$\square A B\|C D\| E F$$ (Lines parallel to the same line are parallel to each other) It can be observed that x = z (Alternate interior angles) ….. (1) It is given that y: z = 3: 7 Let the common ratio between y and Z be a. $$\begin{array}{l}{\square y=3 a \text { and } z=7 a} \\ {\text { Also, } x+y=180^{\circ} \text { (Co-interior angles on the same side of the transversal) }}\end{array}$$ $$\begin{array}{l}{z+y=180^{\circ}[\text { Using equation }(1)]} \\ {7 a+3 a=180^{\circ}} \\ {10 a=180^{\circ}} \\ {a=180} \\ {\square x=7 a=7 \times 18^{\circ}=126^{\circ}}\end{array}$$

Q.3: In Fig. below, if AB || CD, EF ⊥ CD and ∠ GED = 126°, find ∠ AGE, ∠ GEF and ∠FGE.

Ans : It is given that AB || CD $$\begin{array}{l}{\text { EF } \square \mathrm{CD}} \\ {\square \mathrm{GED}=126^{\circ}}\end{array}$$ $$\begin{array}{l}{\square \square G E F+\square F E D=126^{\circ}} \\ {\square \square G E F+90^{\circ}=126^{\circ}}\end{array}$$ $$\begin{array}{l}{\square \square G E F=36^{\circ}} \\ {\square A G E \text { and } \square G E D \text { are alternate interior angles. }} \\ {\square \square A G E=\square G E D=126^{\circ}}\end{array}$$ $$\begin{array}{l}{\text { However, } \square A G E+\square F G E=180^{\circ}(\text { Linear pair })} \\ {\square 126^{\circ}+\square F G E=180^{\circ}}\end{array}$$ $$\begin{array}{l}{\square \square \mathrm{FGE}=180^{\circ}-126^{\circ}=54^{\circ}} \\ {\square \square \mathrm{AGE}=126^{\circ}, \square \mathrm{GEF}=36^{\circ}, \square \mathrm{FGE}=54^{\circ}}\end{array}$$

Q.4: In Fig. below, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠ QRS.
[Hint : Draw a line parallel to ST through point R.]

Ans : Let us draw a line XY parallel to ST and passing through point R. $$\square P Q R+\square Q R X=180^{\circ}$$ (Co-interior angles on the same side of transversal QR) $$\begin{array}{l}{\square 110^{\circ}+\square Q R X=180^{\circ}} \\ {\square \square Q R X=70^{\circ}}\end{array}$$ Also $$\square R S T+\square S R Y=180^{\circ}$$ (Co-interior angles on the same side of transversal SR) $$\begin{array}{l}{130^{\circ}+\square S R Y=180^{\circ}} \\ {\square S R Y=50^{\circ}}\end{array}$$ XY is a straight line. RQ and RS stand on it. $$\begin{array}{l}{\square \square Q R X+\square Q R S+\square S R Y=180^{\circ}} \\ {70^{\circ}+\square Q R S+50^{\circ}=180^{\circ}} \\ {\square Q R S=180^{\circ}-120^{\circ}=60^{\circ}}\end{array}$$

Q.5: In Fig. 6.32 , if AB || CD, ∠ APQ = 50° and ∠ PRD = 127°, find x and y.

Ans : $$\square A P R=\square P R D$$ (Alternate interior angles) $$\begin{array}{l}{50^{\circ}+y=127^{\circ}} \\ {y=127^{\circ}-50^{\circ}} \\ {y=77^{\circ}}\end{array}$$ Also, $$\square A P Q=\square P Q R$$ (Alternate interior angles) $$\begin{array}{l}{50^{\circ}=x} \\ {\square x=50^{\circ} \text { and } y=77^{\circ}}\end{array}$$

Q.6: 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.

Ans : Let us draw $$\mathrm{BM} \square \mathrm{PQ} \text { and } \mathrm{CN} \square \mathrm{RS}$$ As PO || RS Therefore, BM || CN Thus, BM and CN are two parallel lines and a transversal line 3C cuts them at and C respectively. $$\square \square 2=\square 3$$ (Alternate interior angles) However, $$\square 1=\square 2 \text { and } \square 3=\square 4$$ (By laws of reflection) $$\begin{array}{l}{\square \square 1=\square 2=\square 3=\square 4} \\ {\text { Also, } \square 1+\square 2=\square 3+\square 4} \\ {\square A B C=\square D C B}\end{array}$$ However, these are alternate interior angles. AB || CD

Q.1: In Fig. below, sides QP and RQ of ∆ PQR are produced to points S and T respectively. If ∠ SPR = 135° and ∠ PQT = 110°, find ∠ PRQ.

Ans : It is given that, $$\begin{array}{l}{\square S P R=135^{\circ} \text { and } \square P Q T=110^{\circ}} \\ {\square S P R+\square Q P R=180^{\circ}(\text { Linear pair angles) }} \\ {\square 135^{\circ}+\square Q P R=180^{\circ}} \\ {\square \square Q P R=45^{\circ}}\end{array}$$ Also, $$\square P Q T+\square P Q R=180^{\circ}$$ (Linear pair angles) $$\begin{array}{l}{\square 110^{\circ}+\square P Q R=180^{\circ}} \\ {\square \square P Q R=70^{\circ}}\end{array}$$ As the sum Of all interior angles of a triangle is $$180^{\circ}, \text { therefore, for } \Delta \mathrm{PQR}$$ $$\begin{array}{l}{\square Q P R+\square P Q R+\square P R Q=180^{\circ}} \\ {\square 45^{\circ}+70^{\circ}+\square P R Q=180^{\circ}} \\ {\square \square P R Q=180^{\circ}-115^{\circ}} \\ {\square \square P R Q=65^{\circ}}\end{array}$$

Q.2: In Fig.  ∠ X = 62°, ∠ XYZ = 54°. If YO and ZO are the bisectors of ∠ XYZ and ∠ XZY respectively of ∆ XYZ, find ∠ OZY and ∠ YOZ.

Ans : As the sum of all interior angles of a triangle is 1800, therefore, for $$\Delta XYZ$$, $$\begin{array}{l}{\square X+\square X Y Z+\square \times Z Y=180^{\circ}} \\ {62^{\circ}+54^{\circ}+\square X Z Y=180^{\circ}}\end{array}$$ $$\begin{array}{l}{\square \times Z Y=180^{\circ}-116^{\circ}} \\ {\square X Z Y=64^{\circ}}\end{array}$$ $$\square O Y Z=$$ $$\frac{64}{2}=32^{\circ}$$ (OZ iS the angle bisector of $$\Delta XYZ$$) Similarly, $$\square O Y Z=$$ $$\frac{54}{2}=27^{\circ}$$ using angle sum property for $$\Delta O Y Z$$, we obtain $$\begin{array}{l}{\square O Y Z+\square Y O Z+\square O Z Y=180^{\circ}} \\ {27^{\circ}+\square Y O Z+32^{\circ}=180^{\circ}} \\ {\square Y O Z=180^{\circ}-59^{\circ}} \\ {\square Y O Z=121^{\circ}}\end{array}$$

Q.3: In Fig. 6.41, if AB || DE, ∠ BAC = 35° and ∠ CDE = 53°, find ∠ DCE.

Ans : AB II DE and AE is a transversal. $$\square B A C=\square C E D$$ (Alternate interior angles) $$\begin{array}{l}{\square \square C E D=35^{\circ}} \\ {\text { In } \Delta C D E_{,}} \\ {\square C D E+\square C E D+\square D C E=180^{\circ}}\end{array}$$ (Angle sum property of a triangle) $$\begin{array}{l}{53^{\circ}+35^{\circ}+\square D C E=180^{\circ}} \\ {\square D C E=180^{\circ}-88^{\circ}} \\ {\square D C E=92^{\circ}}\end{array}$$

Q.4: In Fig, if lines PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠ RPT = 95° and ∠ TSQ = 75°, find ∠ SQT.

Ans : Using angle sum property for $$\Delta \mathrm{PRT}$$, we obtain $$\begin{array}{l}{\square P R T+\square R P T+\square P T R=180^{\circ}} \\ {40^{\circ}+95^{\circ}+\square P T R=180^{\circ}}\end{array}$$ $$\begin{array}{l}{\square \mathrm{PTR}=180^{\circ}-135^{\circ}} \\ {\square \mathrm{PTR}=45^{\circ}}\end{array}$$ $$\square S T Q=\square P T R=45^{\circ}(\text { Vertically opposite angles) }$$ $$\square S T Q=45^{\circ}$$ $$\begin{array}{l}{\text { By using angle sum property for } \Delta S T Q, \text { we obtain }} \\ {\square S T Q+\square S Q T+D Q S T=180^{\circ}}\end{array}$$ $$\begin{array}{l}{45^{\circ}+\square S Q T+75^{\circ}=180^{\circ}} \\ {\square S Q T=180^{\circ}-120^{\circ}} \\ {\square S Q T=60^{\circ}}\end{array}$$

Q.5: In Fig, if PQ ⊥ PS, PQ || SR, ∠ SQR = 28° and ∠ QRT = 65°, then find the values of x and y.

Ans : It is given that PQ II SR and QR is a transversal line. $$\square P Q R=\square Q R T$$ (Alternate interior angles) $$\begin{array}{l}{x+28^{\circ}=65^{\circ}} \\ {x=65^{\circ}-28^{\circ}} \\ {x=37^{\circ}}\end{array}$$ By using the angle sum property for $$\Delta \mathrm{SPQ}$$, we obtain $$\begin{array}{l}{\square \mathrm{SPQ}+x+y=180^{\circ}} \\ {90^{\circ}+37^{\circ}+y=180^{\circ}}\end{array}$$ $$\begin{array}{l}{y=180^{\circ}-127^{\circ}} \\ {y=53^{\circ}} \\ {\therefore x=37^{\circ} \text { and } y=53^{\circ}}\end{array}$$

Q.6: In Fig. below , the side QR of ∆ PQR is produced to a point S. If the bisectors of ∠ PQR and ∠ PRS meet at point T, then prove that ∠ QTR = 1/2 ∠ QPR.

Ans : $$\begin{array}{l}{\text { In } \triangle Q T R, \square T R S \text { is an exterior angle. }} \\ {\therefore \square Q T R+\square T Q R=\square T R S} \\ {\square Q T R=\square T R S-\square T Q R(1)}\end{array}$$ For $$\Delta \mathrm{PQR}, \square \mathrm{PRS}$$ is an external angle. $$\begin{array}{l}{\therefore \square Q P R+\square P Q R=\square P R S} \\ {\square Q P R+2 \square T Q R=2 \square T R S(A s Q T \text { and } R T \text { are angle bisectors) }} \\ {\square Q P R=2(\square T R S-\square T Q R)}\end{array}$$ $$\square Q P R=2 \square Q T R[B y \text { using equation }(1)]$$ $$\square \mathrm{QTR}=\frac{1}{2} \square \mathrm{QPR}$$

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