Lines and Angles is a core and conceptual chapter in Class 9 Mathematics that builds the foundation of geometrical reasoning. This chapter covers important concepts such as intersecting lines, transversal, angles formed by a transversal, parallel lines, corresponding angles, alternate interior angles, interior angles on the same side of a transversal, and angle sum properties—all of which are frequently asked in school exams.
NCERT Solutions for Class 9 Maths Chapter 6 – Lines and Angles are prepared strictly according to the latest CBSE syllabus and exam pattern. The solutions are written in simple, step-by-step language with clear diagrams and logical explanations, helping students understand proofs easily and score well in Class 9 examinations.
NCERT Solutions For Class 9 Maths Chapter 6 Lines and Angles
Q.
In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠ BOD = 40°, find ∠BOE and reflex ∠COE.
Q.
In Fig. 6.32, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find
x and
y.
Q.
In Fig. 6.43, if PQ ⊥ PS, PQ||SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.
Q.
In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
Q.
In Fig. 6.41, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
Q.
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.
Q.
In Fig. 6.39, sides QP and RQ of Δ PQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
Q.
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.
Q.
In Fig. 6.31, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
Q.
In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
Q.
In Fig. 6.30, if AB||CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
Q.
In Fig. 6.29, if AB || CD, CD || EF and
y :
z = 3 : 7, find
x.
Q.
In Fig. 6.28, find the values of x and y and then show that AB || CD.
Q.
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.
Q.
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 = (1/2)(∠QOS – ∠POS).
Q.
In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.
Q.
In Fig. 6.15, ∠ PQR = ∠ PRQ, then prove that ∠PQS = ∠PRT.
Q.
In Fig. 6.44, the side QR of Δ 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.
1. Lines AB and CD intersect at point O. Find the required angle and its reflex angle.
Answer:
Line AB is a straight line. OC and OE are rays originating from point O.
We know that a straight line forms an angle of 180 degrees.
Using the linear pair property, the sum of the angles on line AB is 180°.
Substituting the given values and simplifying, we obtain the required angle.
The reflex angle is found by subtracting the angle from 360 degrees.
Similarly, CD is also a straight line and OB and OE are rays from point O.
Using the same linear pair property, the value of the required angle and its reflex angle can be calculated.
2. Lines XY and MN intersect at point O. If a : b = 2 : 3, find angle c.
Answer:
Let the common ratio between angles a and b be x.
So,
a = 2x
b = 3x
Since XY is a straight line, the angles formed on it add up to 180 degrees.
Therefore,
2x + 3x = 180°
5x = 180°
x = 36°
Substituting the value of x, we find the values of a and b.
Using the linear pair property again, the value of angle c is obtained.
3. In the given figure, prove that the given angles are equal.
Answer:
ST is a straight line and QP is a transversal.
By the linear pair property, the sum of adjacent angles on a straight line is 180°.
Writing equations for both sides and using the given condition, we equate the two expressions.
On simplification, both sides turn out to be equal.
Hence, the required result is proved.
4. If the given angle relations hold, prove that AOB is a straight line.
Answer:
We know that the sum of angles around a point is 360 degrees.
Using the given relations and simplifying the expression, we find that angles x and y form a linear pair.
Since angles forming a linear pair lie on a straight line,
AOB is a straight line.
Hence proved.
5. POQ is a straight line. OR is perpendicular to PQ. OS lies between OP and OR. Prove the given relation.
Answer:
Since OR is perpendicular to PQ, the angle formed is 90 degrees.
Using the linear pair property for adjacent angles and writing equations for each part, we add the equations.
On simplification, the unwanted terms cancel out.
The final relation is obtained easily.
Hence proved.
6. A line is produced and a ray bisects the given angle. Find the angle and its reflex angle.
Answer:
Since the ray bisects the angle, both parts are equal.
A straight line forms an angle of 180 degrees.
Dividing this equally, we find the value of the required angle.
The reflex angle is calculated as:
360° − given angle.
Thus, both the angle and its reflex angle are obtained.
EXERCISE 6.2
1. If AB ∥ CD and CD ∥ EF, and y : z = 3 : 7, find x.
Answer:
Since AB is parallel to CD and CD is parallel to EF,
AB ∥ CD ∥ EF (lines parallel to the same line are parallel to each other).
Let y = 3k and z = 7k.
Using the property that co-interior angles on the same side of a transversal sum to 180 degrees,
we form an equation and solve for k.
Substituting the value of k, we obtain the value of x.
2. If AB ∥ CD and EF ∥ CD, find the required angles.
Answer:
Using the properties of alternate interior angles and linear pairs,
the angles are calculated step by step.
Thus, all the required angle values are obtained.
3. If PQ ∥ ST, find the required angle. (Hint: Draw a line parallel to ST through point R.)
Answer:
Draw a line XY through point R such that XY ∥ ST.
Now, XY, PQ and ST are parallel lines.
Using the property that co-interior angles on the same side of a transversal sum to 180 degrees,
we form equations and simplify.
Hence, the required angle is obtained.
4. If AB ∥ CD, find the values of x and y.
Answer:
Since AB is parallel to CD,
alternate interior angles are equal.
Using this property, we write equations for x and y.
Solving these equations gives the required values of x and y.
5. PQ and RS are two parallel mirrors. An incident ray reflects twice. Prove that AB ∥ CD.
Answer:
Draw perpendiculars from the points of incidence to the mirrors.
Since PQ ∥ RS, the perpendiculars drawn are also parallel.
Using the laws of reflection, the angles of incidence and reflection are equal.
Thus, corresponding angles formed are equal, which implies
AB ∥ CD.
Hence proved.
FAQs: Class 9 Maths Chapter 6 – Lines and Angles
Q1. Is Lines and Angles important for Class 9 exams?
Yes, it is a high-weightage geometry chapter.
Q2. Which topics are most important in this chapter?
Angles formed by a transversal, parallel lines, and angle relationships.
Q3. Are proof-based questions asked from this chapter?
Yes, theorem and reasoning-based questions are common.
Q4. Is this chapter useful for higher classes?
Yes, it forms the base for triangles and circles in Class 9 & 10.
Q5. How do NCERT Solutions help?
They provide NCERT-aligned, exam-ready explanations with clear diagrams and logic.