NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles

NCERT
SOLUTIONS
CLASS - 9th

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Book : Mathematics Ncert Solutions | Chapter-6 Maths

Class : 9th
Subject : Maths
Chapter : 6
Chapter Name : LINES AND ANGLES

Exercise 6.1

Q1 In Fig. 6.13, lines AB and CD intersect at O. If ∠ AOC + ∠ BOE = 70° and ∠ BOD = 40°,
�nd ∠ BOE and re�ex ∠ COE.

Answer. AB is a straight line, rays OC and OE stand on it.
∴ ∠AOC + ∠COE + ∠BOE = 180∘
⇒ (∠AOC + ∠BOE) + ∠COE = 180∘
⇒ 70∘ + ∠COE = 180∘
⇒ ∠COE = 180∘ − 70∘ = 110∘
Reflex ∠COE = 360∘ − 110∘ = 250∘
CD is a straight line. rays OE and 0B stand on it.
∴ ∠COE + ∠BOE + ∠BOD = 180∘
⇒ 110∘ + ∠BOE + 40∘ = 180∘
⇒ ∠BOE = 180∘ − 150∘ = 30∘

Page : 96 , Block Name : Exercise 6.1

Q2 In Fig. lines XY and MN intersect at O. If ∠ POY = 90° and a : b = 2 : 3, �nd c.

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Book : Mathematics Ncert Solutions | Chapter-6 Maths

Answer. 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.
∴ ∠XOM + ∠MOP + ∠POY = 180∘
b + a + ∠POY = 180∘
3x + 2x + 90∘ = 180∘
5x = 90∘
x = 18∘
a = 2x = 2 × 18 = 36∘
b = 3x = 3 × 18 = 54∘
MN is a straight line. Ray OX stands on it.
∴ b + c = 180∘ ( Linear Pair )
54∘ + c = 180∘
c = 180∘ − 54∘ = 126∘
∴ c = 126∘

Page : 97 , Block Name : Exercise 6.1

Q3 . In Fig. ∠ PQR = ∠ PRQ, then prove that ∠ PQS = ∠ PRT

Answer. In the given �gure, ST isa straight line and ray QP stands on it.

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Book : Mathematics Ncert Solutions | Chapter-6 Maths

∴ ∠PQS + ∠PQR = 180∘ ( Linear Pair )
∠PQR = 180∘ − ∠PQS(1)
∠PRT + ∠PRQ = 180∘ ( Linear Pair )
∠PRQ = 180∘ − ∠PRT(2)
It is given that ∠PQR = ∠PRQ .
Equating equations (1) and (2), we obtain
180∘ − ∠PQS = 180∘ − ∠PRT
∠PQS = ∠PRT

Page : 97 , Block Name : Exercise 6.1

Q4 In Fig. if x + y = w + z, then prove that AOB is a line.

Answer.
It can be observed that,
x + y + z + w = 360∘ ( Complete angle )
It is given that,
x+y=z+w
∴ x + y + x + y = 360∘
2(x + y) = 360∘
x + y = 180∘
Since x and y form a linear pair, AOB is a line.

Page : 97 , Block Name : Exercise 6.1

Q5 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).

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Book : Mathematics Ncert Solutions | Chapter-6 Maths

Answer. It is given that OR ⊥ PQ
□□POR = 90∘
□□POS + □SOR = 90∘
□ROS = 90∘ − □P OS … (1)
□QOR = 90∘ (AsOR□P Q)
On adding equations (1) and (2), we obtain
2□ROS = □QOS − □POS
□ROS = 12 (□QOS − □POS)

Page : 97 , Block Name : Exercise 6.1

Q6 It is given that ∠ XYZ = 64° and XY is produced to point P. Draw a �gure from the
given information. If ray YQ bisects ∠ ZYP, �nd ∠ XYQ and re�ex ∠ QYP.

Answer.

It is given that line YQ bisects □PYZ
Hence, □QY P = □ZY Q
It can be observed that PX is a line. Rays YQ and YZ stand on it.
□□ × Y Z + □ZY Q + □QY P = 180∘
□64∘ + 2□QY P = 180∘
□2□QY P = 180∘ − 64∘ = 116∘
□□QY P = 58∘
Also □ZY Q = □QY P = 58∘
Re�ex □QY P = 360∘ − 58∘ = 302∘
□ × Y Q = □XY Z + □ZY Q
= 64∘ + 58∘ = 122∘

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Book : Mathematics Ncert Solutions | Chapter-6 Maths

Page : 97 , Block Name : Exercise 6.1

Exercise 6.2

Q1 In Fig. 6.28, �nd the values of x and y and then show that AB || CD.

Answer. It can be observed that,
50∘ + x = 180∘ ( Linear pair )
x = 130∘ … (1)
Also, y = 130∘ ( Vertically opposite angles)
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.

Page : 103 , Block Name : Exercise 6.2

Q2 In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, �nd x.

Answer. It is given that AB II CD and CD II EF
□AB∥CD∥EF (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.
□y = 3a and z = 7a
Also, x + y = 180∘ (Co-interior angles on the same side of the transversal)

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Book : Mathematics Ncert Solutions | Chapter-6 Maths

z + y = 180∘ [ Using equation (1)]
7a + 3a = 180∘
10a = 180∘
a = 180
□x = 7a = 7 × 18∘ = 126∘

Page : 104 , Block Name : Exercise 6.2

Q3 In Fig. below, if AB || CD, EF ⊥ CD and ∠ GED = 126°, �nd ∠ AGE, ∠ GEF and ∠FGE.

Answer. It is given that
AB || CD
EF □CD
□GED = 126∘
□□GEF + □F ED = 126∘
□□GEF + 90∘ = 126∘
□□GEF = 36∘
□AGE and □GED are alternate interior angles.
□□AGE = □GED = 126∘
However, □AGE + □F GE = 180∘ ( Linear pair )
□126∘ + □F GE = 180∘
□□FGE = 180∘ − 126∘ = 54∘
□□AGE = 126∘ , □GEF = 36∘ , □FGE = 54∘

Page : 104 , Block Name : Exercise 6.2

Q4 In Fig. below, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, �nd ∠ QRS.
[Hint : Draw a line parallel to ST through point R.]

Answer.

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Book : Mathematics Ncert Solutions | Chapter-6 Maths

Let us draw a line XY parallel to ST and passing through point R.
□P QR + □QRX = 180∘ (Co-interior angles on the same side of transversal QR)
□110∘ + □QRX = 180∘
□□QRX = 70∘
Also
□RST + □SRY = 180∘ (Co-interior angles on the same side of transversal SR)
130∘ + □SRY = 180∘
□SRY = 50∘
XY is a straight line. RQ and RS stand on it.
□□QRX + □QRS + □SRY = 180∘
70∘ + □QRS + 50∘ = 180∘
□QRS = 180∘ − 120∘ = 60∘

Page : 104 , Block Name : Exercise 6.2

Q5 In Fig. 6.32 , if AB || CD, ∠ APQ = 50° and ∠ PRD = 127°, �nd x and y.

Answer. □AP R = □P RD (Alternate interior angles)
50∘ + y = 127∘
y = 127∘ − 50∘
y = 77∘
Also, □AP Q = □P QR (Alternate interior angles)
50∘ = x
□x = 50∘ and y = 77∘

Page : 104 , Block Name : Exercise 6.2

Q6 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 re�ected ray moves along the path BC and strikes the
mirror RS at C and again re�ects back along CD. Prove that AB || CD.

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Book : Mathematics Ncert Solutions | Chapter-6 Maths

Answer.

Let us draw BM□PQ and CN□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.
□□2 = □3 (Alternate interior angles)
However, □1 = □2 and □3 = □4 (By laws of re�ection)
□□1 = □2 = □3 = □4
Also, □1 + □2 = □3 + □4
□ABC = □DCB
However, these are alternate interior angles.
AB || CD

Page : 104 , Block Name : Exercise 6.2

Exercise 6.3

Q1 In Fig. below, sides QP and RQ of ∆ PQR are produced to points S and T respectively.
If ∠ SPR = 135° and ∠ PQT = 110°, �nd ∠ PRQ.

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Book : Mathematics Ncert Solutions | Chapter-6 Maths

Answer. It is given that,
□SP R = 135∘ and □P QT = 110∘
□SP R + □QP R = 180∘ ( Linear pair angles)
□135∘ + □QP R = 180∘
□□QP R = 45∘
Also, □P QT + □P QR = 180∘ (Linear pair angles)
□110∘ + □P QR = 180∘
□□P QR = 70∘
As the sum Of all interior angles of a triangle is 180∘ , therefore, for ΔPQR
□QP R + □P QR + □P RQ = 180∘
□45∘ + 70∘ + □P RQ = 180∘
□□P RQ = 180∘ − 115∘
□□P RQ = 65∘

Page : 107 , Block Name : Exercise 6.3

Q2 In Fig. 6.40, ∠ X = 62°, ∠ XYZ = 54°. If YO and ZO are the bisectors of ∠ XYZ and ∠
XZY respectively of ∆ XYZ, �nd ∠ OZY and ∠ YOZ.

Answer. As the sum of all interior angles of a triangle is 1800, therefore, for ΔXY Z ,
□X + □XY Z + □ × ZY = 180∘
62∘ + 54∘ + □XZY = 180∘
□ × ZY = 180∘ − 116∘
□XZY = 64∘
□OY Z = 64 2
= 32∘ (OZ iS the angle bisector of ΔXY Z )
54
Similarly, □OY Z = 2 = 27∘
using angle sum property for ΔOY Z , we obtain
□OY Z + □Y OZ + □OZY = 180∘
27∘ + □Y OZ + 32∘ = 180∘
□Y OZ = 180∘ − 59∘
□Y OZ = 121∘

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Book : Mathematics Ncert Solutions | Chapter-6 Maths

Page : 107 , Block Name : Exercise 6.3

Q3 In Fig. 6.41, if AB || DE, ∠ BAC = 35° and ∠ CDE = 53°, �nd ∠ DCE.

Answer. AB II DE and AE is a transversal.
□BAC = □CED (Alternate interior angles)
□□CED = 35∘
In ΔCDE, (Angle sum property of a triangle)
□CDE + □CED + □DCE = 180∘
53∘ + 35∘ + □DCE = 180∘
□DCE = 180∘ − 88∘
□DCE = 92∘

Page : 107 , Block Name : Exercise 6.3

Q4 In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠ RPT =
95° and ∠ TSQ = 75°, �nd ∠ SQT.

Answer. Using angle sum property for ΔPRT, we obtain
□P RT + □RP T + □P T R = 180∘
40∘ + 95∘ + □P T R = 180∘
□PTR = 180∘ − 135∘
□PTR = 45∘
□ST Q = □P T R = 45∘ ( Vertically opposite angles)
□ST Q = 45∘
By using angle sum property for ΔST Q, we obtain
□ST Q + □SQT + DQST = 180∘

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Book : Mathematics Ncert Solutions | Chapter-6 Maths

45∘ + □SQT + 75∘ = 180∘
□SQT = 180∘ − 120∘
□SQT = 60∘

Page : 107 , Block Name : Exercise 6.3

Q5 In Fig. 6.43, if PQ ⊥ PS, PQ || SR, ∠ SQR = 28° and ∠ QRT = 65°, then �nd the values
of x and y.

Answer. It is given that PQ II SR and QR is a transversal line.
□P QR = □QRT (Alternate interior angles)
x + 28∘ = 65∘
x = 65∘ − 28∘
x = 37∘
By using the angle sum property for ΔSPQ, we obtain
□SPQ + x + y = 180∘
90∘ + 37∘ + y = 180∘
y = 180∘ − 127∘
y = 53∘
∴ x = 37∘ and y = 53∘

Page : 108 , Block Name : Exercise 6.3

Q6 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.

Answer.

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Book : Mathematics Ncert Solutions | Chapter-6 Maths

In △QT R, □T RS is an exterior angle.
∴ □QT R + □T QR = □T RS
□QT R = □T RS − □T QR(1)
For ΔPQR, □PRS is an external angle.
∴ □QP R + □P QR = □P RS
□QP R + 2□T QR = 2□T RS(AsQT and RT are angle bisectors)
□QP R = 2(□T RS − □T QR)
□QP R = 2□QT R[By using equation (1)]
□QTR = 12 □QPR

Page : 108 , Block Name : Exercise 6.3

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