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Chapter 4 Lines and Angles

NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1

Question 1.
In figure, lines AB and CD intersect at 0. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 img 1
Solution:
Here, ∠AOC and ∠BOD are vertically opposite angles.
∴ ∠AOC = ∠BOD
⇒ ∠AOC = 40° [∵ ∠BOD = 40°(Given)] …(i)
We have, ∠AOC + ∠ BOE = 70° (Given)
40°+ ∠BOE = 70° [From Eq. (i)]
⇒ ∠BOE = 30°
Also, ∠AOC + ∠COE + ∠BOE = 180° (Linear pair axiom)
⇒ 40° + ∠COE + 30° = 180°
⇒ ∠COE = 110°
Now, ∠COE + reflex ∠COE = 360° (Angles at a point)
110°+reflex ∠COE = 360°
⇒ Reflex ∠COE = 250°

Question 2.
In figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. find c.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 img 2
Solution:
We have, ∠POY = 90°
⇒ ∠POY + ∠POX = 180° (Linear pair axiom)
⇒ ∠POX = 90°
⇒ a+b = 90°
Also, a : b = 2 : 3 (Given)
⇒ Let a = 2k,b = 3k
Now, from Eq. (j), we get
2k + 3k = 90°
⇒ 5k = 90°
⇒ k = 18°
∴ a = 2 x 18°=36°
and b=3 x 18°=54°
Now, ∠MOX + ∠XON = 1800 (Linear pair axiom)
b+ c = 180°
⇒ 540 + c= 180°
⇒ c = 126°

Question 3.
In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 img 3
Solution:
∵ ∠PQS+ ∠PQR = 180° (Linear pair axiom) ,..(i)
and ∠PRT + ∠PRQ = 180° (Linear pair axiom).. .(ii)
From Eqs. (i) and (ii), we get
∠PQS + ∠PQR =∠PRT + ∠PRQ
∠PQS + ∠PRQ =∠PRT + ∠PRQ
[Given, ∠PQR = ∠PRQ]
⇒ ∠PQS = ∠PRT

Question 4.
In figure, if x + y = w + z, then prove that AOB is a line.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 img 4
Solution:
∵ x+ y+w+ z = 360° (Angle at a point)
x + y = w + z (Given)…(i)
∴ x+ y+ x+ y = 360° [From Eq. (i)]
2(x + y) = 360°
⇒ x + y = 180° (Linear pair axiom)
Hence, AOB is a straight line.

Question 5.
In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 img 5
Solution:
We have,
∠POR = ∠ROQ = 90° (∵ Given that, OR is perpendicular to PQ)
∴ ∠POS + ∠ROS = 90°
⇒ ∠ROS = 90° – ∠POS
On adding ∠ROS both sides, we get
2 ∠ROS = 90° – ∠POS + ∠ROS
⇒ 2 ∠ROS = (90° + ∠ROS) – ∠POS
⇒ 2∠ROS = ∠QOS – ∠POS (∵ ∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROS)
⇒ ∠ROS = 12 (∠QOS – ∠POS)
Hence proved.

Question 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.
Solution:
Here, YQ bisects ∠ZYP.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 img 6
Hence, ∠ZYQ = ∠QYP = 12 ∠ZYP ……..(i)
Given, ∠XYZ = 64° ….(ii)
∵ ∠XYZ + ∠ZYQ + ZQYP = 180° (Linear pair axiom)
⇒ 64° + ∠ZYQ + ∠ZYQ = 180° [From Eqs. (i) and (ii)]
⇒ 2 ∠ZYQ = 180° – 64°
⇒ ∠ZYQ = 12 x 116°
⇒ ∠ZYQ = 58°
∴ ∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58° = 122°
Now, ∠QYP + reflex ∠QYP = 360°
58° + reflex ∠QYP = 360°
⇒ reflex ∠ QYP = 302°

NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2

Question 1.
In figure, find the values of x and y and then show that AB || CD.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 1
Solution:
∵ x + 50° = 180° (Linear pair)
⇒ x = 130°
∴ y = 130° (Vertically opposite angle)
Here, ∠x = ∠COD = 130°
These are corresponding angles for lines AB and CD.
Hence, AB || CD

Question 2.
In figure, if AB || CD, CD || EF and y: z = 3:7, find x.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 2
Solution:
Given
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 3
⇒ Let y = 3k, z = 7k
x = ∠CHG (Corresponding angles)…(i)
∠CHG = z (Alternate angles)…(ii)
From Eqs. (i) and (ii), we get
x = z …….(iii)
Now, x+y = 180°
(Internal angles on the same side of the transversal)
⇒ z+y = 180° [From Eq. (iii)]
7k + 3k = 180°
⇒ 10k = 180°
⇒ k = 18
∴ y = 3 x 18° = 54°
and z = 7x 18°= 126°
∴ x = z
Now, x + y = 180°
(Internal angles on the same side of the transversal)
⇒ z + y = 180° [From Eq. (iii)]
7k + 3k = 180°
⇒ 10k = 180°
⇒ k = 18
∴ y = 3 x 18° = 54°
and z = 7x 18°= 126°
x = z
⇒ x = 126°

Question 3.
In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 4
Solution:
∵ ∠AGE = ∠GED (Alternate interior angles)
But ∠GED = 126°
⇒ ∠AGE = 126° ….(i)
∴ ∠GEF + ∠FED= 126°
⇒ ∠GEF + 90° =126° (∵ EF ⊥ CD)
⇒ ∠GEF = 36°
Also, ∠AGE + ∠FGE = 180° (Linear pair axiom)
⇒ 126° + ∠FGE =180°
⇒ ∠FGE = 54°

Question 4.
In figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠QRS.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 5
Solution:
Drawing a tine parallel to ST through R.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 6
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 7

Question 5.
In figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 8
Solution:
We have, AB || CD
⇒ ∠APQ = ∠PQR (Alternate interior angles)
⇒ 50° = x
⇒ x = 50°
Now, ∠PQR + ∠QPR = 127°
(Exterior angle is equal to sum of interior opposite angles of a triangle)
⇒ 50°+ ∠QPR = 127°
⇒ y = 77°.

Question 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 again reflects back along CD. Prove that AB || CD.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 9
Solution:
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 10

NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3

Question 1.
In figure, sides QP and RQ of APQR are produced to points S and T, respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 1
Solution:
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 2
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 3

Question 2.
In figure, ∠X – 62°, ∠XYZ = 54°, if YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OZY and ∠YOZ.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 4
Solution:
In ∆XYZ,
∵ ∠X+ ∠Y+ ∠Z = 180°
(Sum of all angles of triangle is equal to
∴ 62° + ∠Y + ∠Z = 180° [YZX = 62° (Given)]
⇒ ∠Y + ∠Z = 118° .
⇒ 12∠Y + 12 ∠Z = 12 x 118°
⇒ ∠OYZ + ∠OZY = 59°
(∵ YO and ZO are the bisectors of ∠XYZ and ∠XZY)
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 5
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 6

Question 3.
In figure, if AB || DE, ∠BAC = 35° and ∠CDE = 539 , find ∠DCE.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 7
Solution:
We have AB || DE
⇒ ∠AED= ∠ BAE (Alternate interior angles)
Now, ∠ BAE = ∠ BAC
⇒ ∠ BAE = 35° [ ∵ ∠ BAC = 35° (Given) ]
∴ ∠ AED = 35°
In ∆DCE,
∵ ∠DCE + ∠CED+ ∠EDC= 180° ( ∵Sum of all angles of triangle is equal to 180°)
⇒ ∠ DCE + 35°+ 53° = 180° ( ∵∠ AED = ∠ CED = 35°)
⇒ ∠ DCE = 180° – (35° + 53°)
⇒ ∠ DCE = 92°

Question 4.
In figure, if lines PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠ RPT = 95° and ∠TSQ = 75°, find ∠ SQT.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 8
Solution:
∵∠PTS = ∠RPT + ∠PRT (Exterior angle = Sum of interior opposite angles)
∠ PTS = 95° + 40° [∵ ∠PPT = 95° (Given)]
⇒ ∠ PTS = 135° [and ∠PRT = 40°]
Also, ∠ TSQ + ∠ SQT = ∠ PTS (Exterior angle = Sum of interior opposite angles)
⇒ 75°+ ∠ SQT = 135°
⇒ ∠ SQT = 60° [∵ ∠ TSQ = 75° (Given)]

Question 5.
In figure, if PQ ⊥ PS, PQ||SR, ∠SQR = 2S° and ∠QRT = 65°, then find the values of x and y.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 9
Solution:
Here, PQ || SR
⇒ ∠ PQR = ∠ OPT (Alternate interior angles)
⇒ x + 28° = 65° ⇒ x = 37°
Now, in right angled ASPQ, we have ∠P = 90°
∴ ∠P + x + y = 180° (∵ Sum of all angles of a triangle is equal to 180°)
⇒ 90°+ 37°+ y= 180°
⇒ 127°+ y=180°
⇒ y = 53°

Question 6.
In figure, the side QR of A PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 10
Solution:
In ∆ PQP,
∵ ∠QPR + ∠PQR = ∠PRS …(i)
(∵ Sum of interior opposite angles = Exterior angle)
Now, in ∆ TOR,
∵ ∠QTR + ∠TQR = ∠TRS …..(ii)
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 11

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