NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles
Neha Tanna7 Aug, 2024
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NCERT Solutions for Class 9 Maths Chapter 6: The questions and answers about the topic Lines and Angles are included in the NCERT Solutions for Class 9 Maths topic 6. This topic serves as an introduction to fundamental geometry, with a particular emphasis on the characteristics of the angles that are created when two lines meet (i) and when a line crosses two or more parallel lines at different points.
This chapter is covered in the CBSE Syllabus for Class 9 Maths and is part of the Unit - Geometry. NCERT Solutions are available here, and they can assist students do well on their final exams.NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Overview
Chapter 6 of NCERT Class 9 Maths, titled "Lines and Angles," delves into the fundamental concepts of geometry involving lines and angles. This chapter builds upon the foundational knowledge of Euclidean geometry introduced earlier and expands students' understanding of how lines and angles interact and relate to one another. The chapter begins by defining essential terms such as intersecting lines, parallel lines, and various types of angles, including complementary, supplementary, adjacent, and vertically opposite angles. It then explores the properties of these angles, providing students with the tools to analyze and solve problems involving them.NCERT Solutions for Class 9 Maths Chapter 6 PDF
This chapter is the concept of parallel lines and their corresponding angles when intersected by a transversal. Students learn about alternate interior angles, alternate exterior angles, and corresponding angles, along with their properties and relationships. These concepts are vital for solving geometric problems and understanding the principles of congruence and similarity in triangles. Here we have provided the NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles -NCERT Solutions for Class 9 Maths Chapter 6 PDF
NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1
Below we have provided NCERT Solutions for Class 9 Maths Chapter 6 -1. In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC +∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
Solution:
From the diagram, we have (∠AOC +∠BOE +∠COE) and (∠COE +∠BOD +∠BOE) forms a straight line. So, ∠AOC+∠BOE +∠COE = ∠COE +∠BOD+∠BOE = 180° Now, by putting the values of ∠AOC + ∠BOE = 70° and ∠BOD = 40° we get ∠COE = 110° and ∠BOE = 30° So, reflex ∠COE = 360 o – 110 o = 250 o2. In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
Solution:
We know that the sum of linear pair is always equal to 180° So, ∠POY +a +b = 180° Putting the value of ∠POY = 90° (as given in the question), we get, a+b = 90° Now, it is given that a:b = 2:3, so Let a be 2x and b be 3x ∴ 2x+3x = 90° Solving this, we get 5x = 90° So, x = 18° ∴ a = 2×18° = 36° Similarly, b can be calculated, and the value will be b = 3×18° = 54° From the diagram, b+c also forms a straight angle, so b+c = 180° c+54° = 180° ∴ c = 126°3. In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
Solution:
Since ST is a straight line, so ∠ PQS+ ∠ PQR = 180° (linear pair) and ∠ PRT+ ∠ PRQ = 180° (linear pair) Now, ∠ PQS + ∠ PQR = ∠ PRT+ ∠ PRQ = 180° Since ∠ PQR = ∠ PRQ (as given in the question) ∠ PQS = ∠ PRT. (Hence proved).4. In Fig. 6.16, if x+y = w+z, then prove that AOB is a line.
Solution:
To prove AOB is a straight line, we will have to prove x+y is a linear pair i.e. x+y = 180° We know that the angles around a point are 360°, so x+y+w+z = 360° In the question, it is given that, x+y = w+z So, (x+y)+(x+y) = 360° 2(x+y) = 360° ∴ (x+y) = 180° (Hence proved).5. In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = ½ (∠QOS – ∠POS).
Solution:
In the question, it is given that (OR ⊥ PQ) and ∠POQ = 180° We can write it as ∠ROP = ∠ROQ = 90 0 We know that ∠ROP = ∠ROQ It can be written as ∠POS + ∠ROS = ∠ROQ ∠POS + ∠ROS = ∠QOS – ∠ROS ∠SOR + ∠ROS = ∠QOS – ∠POS So we get 2∠ROS = ∠QOS – ∠POS Or, ∠ROS = 1/2 (∠QOS – ∠POS)(Hence proved).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, XP is a straight line
So, ∠XYZ +∠ZYP = 180°
Putting the value of ∠XYZ = 64°, we get
64° +∠ZYP = 180°
∴ ∠ZYP = 116°
From the diagram, we also know that ∠ZYP = ∠ZYQ + ∠QYP
Now, as YQ bisects ∠ZYP,
∠ZYQ = ∠QYP
Or, ∠ZYP = 2∠ZYQ
∴ ∠ZYQ = ∠QYP = 58°
Again, ∠XYQ = ∠XYZ + ∠ZYQ
By putting the value of ∠XYZ = 64° and ∠ZYQ = 58°, we get.
∠XYQ = 64°+58°
Or, ∠XYQ = 122°
Now, reflex ∠QYP = 180°+XYQ
We computed that the value of ∠XYQ = 122°.
So,
∠QYP = 180°+122°
∴ ∠QYP = 302°
NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2
1. In Fig. 6.28, find the values of x and y and then show that AB || CD.
Solution:
We know that a linear pair is equal to 180°. So, x+50° = 180° ∴ x = 130° We also know that vertically opposite angles are equal. So, y = 130° In two parallel lines, the alternate interior angles are equal. In this, x = y = 130° This proves that alternate interior angles are equal, so AB || CD.2. In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.
Solution:
It is known that AB || CD and CD||EF As the angles on the same side of a transversal line sum up to 180°, x + y = 180° —–(i) Also, ∠O = z (Since they are corresponding angles) and, y +∠O = 180° (Since they are a linear pair) So, y+z = 180° Now, let y = 3w and hence, z = 7w (As y : z = 3 : 7) ∴ 3w+7w = 180° Or, 10 w = 180° So, w = 18° Now, y = 3×18° = 54° and, z = 7×18° = 126° Now, angle x can be calculated from equation (i) x+y = 180° Or, x+54° = 180° ∴ x = 126°3. In Fig. 6.30, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
Solution:
Since AB || CD, GE is a transversal. It is given that ∠GED = 126° So, ∠GED = ∠AGE = 126° (As they are alternate interior angles) Also, ∠GED = ∠GEF +∠FED As EF⊥ CD, ∠FED = 90° ∴ ∠GED = ∠GEF+90° Or, ∠GEF = 126° – 90° = 36° Again, ∠FGE +∠GED = 180° (Transversal) Putting the value of ∠GED = 126°, we get ∠FGE = 54° So, ∠AGE = 126° ∠GEF = 36° and ∠FGE = 54°4. In Fig. 6.31, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
[Hint : Draw a line parallel to ST through point R.]
Solution:
First, construct a line XY parallel to PQ.
We know that the angles on the same side of the transversal is equal to 180°.
So, ∠PQR+∠QRX = 180°
Or, ∠QRX = 180°-110°
∴ ∠QRX = 70°
Similarly,
∠RST +∠SRY = 180°
Or, ∠SRY = 180°- 130°
∴ ∠SRY = 50°
Now, for the linear pairs on the line XY-
∠QRX+∠QRS+∠SRY = 180°
Putting their respective values, we get
∠QRS = 180° – 70° – 50°
Hence, ∠QRS = 60°
5. In Fig. 6.32, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
Solution:
From the diagram, ∠APQ = ∠PQR (Alternate interior angles) Now, putting the value of ∠APQ = 50° and ∠PQR = x, we get x = 50° Also, ∠APR = ∠PRD (Alternate interior angles) Or, ∠APR = 127° (As it is given that ∠PRD = 127°) We know that ∠APR = ∠APQ+∠QPR Now, putting values of ∠QPR = y and ∠APR = 127°, we get 127° = 50°+ y Or, y = 77° Thus, the values of x and y are calculated as: x = 50° and y = 77°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.
Solution:
First, draw two lines, BE and CF, such that BE ⊥ PQ and CF ⊥ RS. Now, since PQ || RS, So, BE || CF
NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.3
1. In Fig. 6.39, sides QP and RQ of ΔPQR are produced to points S and T,respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
Solution:
It is given the TQR is a straight line, and so, the linear pairs (i.e. ∠TQP and ∠PQR) will add up to 180° So, ∠TQP +∠PQR = 180° Now, putting the value of ∠TQP = 110°, we get ∠PQR = 70° Consider the ΔPQR, Here, the side QP is extended to S, and so ∠SPR forms the exterior angle. Thus, ∠SPR (∠SPR = 135°) is equal to the sum of interior opposite angles. (Triangle property) Or, ∠PQR +∠PRQ = 135° Now, putting the value of ∠PQR = 70°, we get ∠PRQ = 135°-70° Hence, ∠PRQ = 65°2. In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY, respectively of Δ XYZ, find ∠OZY and ∠YOZ.
Solution:
We know that the sum of the interior angles of the triangle. So, ∠X +∠XYZ +∠XZY = 180° Putting the values as given in the question, we get 62°+54° +∠XZY = 180° Or, ∠XZY = 64° Now, we know that ZO is the bisector, so ∠OZY = ½ ∠XZY ∴ ∠OZY = 32° Similarly, YO is a bisector, so ∠OYZ = ½ ∠XYZ Or, ∠OYZ = 27° (As ∠XYZ = 54°) Now, as the sum of the interior angles of the triangle, ∠OZY +∠OYZ +∠O = 180° Putting their respective values, we get ∠O = 180°-32°-27° Hence, ∠O = 121°3. In Fig. 6.41, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
Solution:
We know that AE is a transversal since AB || DE Here ∠BAC and ∠AED are alternate interior angles. Hence, ∠BAC = ∠AED It is given that ∠BAC = 35° ∠AED = 35° Now consider the triangle CDE. We know that the sum of the interior angles of a triangle is 180°. ∴ ∠DCE+∠CED+∠CDE = 180° Putting the values, we get ∠DCE+35°+53° = 180° Hence, ∠DCE = 92°4. In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
Solution:
Consider triangle PRT. ∠PRT +∠RPT + ∠PTR = 180° So, ∠PTR = 45° Now ∠PTR will be equal to ∠STQ as they are vertically opposite angles. So, ∠PTR = ∠STQ = 45° Again, in triangle STQ, ∠TSQ +∠PTR + ∠SQT = 180° Solving this, we get 74° + 45° + ∠SQT = 180° ∠SQT = 60°5. In Fig. 6.43, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.
Solution:
x +∠SQR = ∠QRT (As they are alternate angles since QR is transversal) So, x+28° = 65° ∴ x = 37° It is also known that alternate interior angles are the same, and so ∠QSR = x = 37° Also, now ∠QRS +∠QRT = 180° (As they are a Linear pair) Or, ∠QRS+65° = 180° So, ∠QRS = 115° Using the angle sum property in Δ SPQ, ∠SPQ + x + y = 180° Putting their respective values, we get 90°+37° + y = 180° y = 180 0 – 127 0 = 53 0 Hence, y = 53°6. In Fig. 6.44, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = ½ ∠QPR.
Solution:
Consider the ΔPQR. ∠PRS is the exterior angle, and ∠QPR and ∠PQR are the interior angles. So, ∠PRS = ∠QPR+∠PQR (According to triangle property) Or, ∠PRS -∠PQR = ∠QPR ———–(i) Now, consider the ΔQRT, ∠TRS = ∠TQR+∠QTR Or, ∠QTR = ∠TRS-∠TQR We know that QT and RT bisect ∠PQR and ∠PRS, respectively. So, ∠PRS = 2 ∠TRS and ∠PQR = 2∠TQR Now, ∠QTR = ½ ∠PRS – ½∠PQR Or, ∠QTR = ½ (∠PRS -∠PQR) From (i), we know that ∠PRS -∠PQR = ∠QPR So, ∠QTR = ½ ∠QPR (hence proved).Benefits of NCERT Solutions for Class 9 Maths Chapter 6
The NCERT Solutions for Class 9 Maths Chapter 6: Lines and Angles offer several benefits that aid in building a strong understanding of geometry:Clear Understanding of Concepts : The solutions provide detailed explanations of key concepts like intersecting lines, parallel lines, and various angles, ensuring students grasp the foundational principles of geometry.
Enhanced Problem-Solving Skills : By working through a variety of problems, students develop their ability to analyze and solve geometric problems involving lines and angles, improving their overall problem-solving skills.
Logical Reasoning : The chapter emphasizes the properties and relationships between angles and lines, fostering logical reasoning and analytical thinking that are essential for mathematics and other subjects.
Exam Preparation : NCERT Solutions align with the CBSE curriculum, ensuring comprehensive coverage of all topics and helping students prepare effectively for exams with step-by-step solutions to textbook problems.
Application of Theorems : The solutions guide students in applying theorems related to lines and angles, such as alternate interior angles and corresponding angles, reinforcing their understanding of geometric principles.
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