Class 9 Maths Chapter 6 (Lines and Angles) Important Questions with solutions are given here, which can be easily accessed. These questions have been prepared by our experts for students of standard 9 to make them prepare for final **exam 2020**. All the questions are based on **CBSE **syllabus and taken in reference from **NCERT** book. Students can do their revision by praticing the questions here and score good marks.

To revise the important questions chapter-wise for 9th Maths, reach us at BYJUâ€™S. The chapter lines and angles will consist of topics such as angles formed after the intersection of two lines, linear pair of angles, complementary angles, etc. Let us solve the questions here to understand all the concepts.

**Also Check:**

- Important 2 Marks Questions for CBSE 9th Maths
- Important 3 Marks Questions for CBSE 9th Maths
- Important 4 Marks Questions for CBSE 9th Maths

## Important Questions & Solutions For Class 9 Chapter 6 (Lines and Angles)

**Q.1: In the figure, lines AB and CD intersect at O. If âˆ AOC + âˆ BOE = 70Â° and âˆ BOD = 40Â°, find âˆ BOE and reflex âˆ COE. **

Solution:

From the given figure, we can see;

âˆ AOC + âˆ BOE +âˆ COE and âˆ COE +âˆ BOD + âˆ BOE form a straight line.

So, âˆ AOC + âˆ BOE +âˆ COE = âˆ COE +âˆ BOD + âˆ BOE = 180Â°

Now, by substituting the values of âˆ AOC + âˆ BOE = 70Â° and âˆ BOD = 40Â° we get

âˆ COE = 110Â° and

âˆ BOE = 30Â°

**Q.2: In the Figure, lines XY and MN intersect at O. If âˆ POY = 90Â° and a : b = 2 : 3, find c.**

Solution:

As we know, the sum of linear pair are always equal to 180Â°

So,

âˆ POY + a + b = 180Â°

substituting the value of âˆ POY = 90Â° (as given in the question) we get,

a + b = 90Â°

Now, it is given that a : b = 2 : 3 so,

Let a be 2x and b be 3x

âˆ´ 2x + 3x = 90Â°

Solving this we get

5x = 90Â°

So, x = 18Â°

âˆ´ a = 2 Ã— 18Â° = 36Â°

Similarly b can be calculated and the value will be

b = 3 Ã— 18Â° = 54Â°

From the diagram, b + c also forms a straight angle so,

b + c = 180Â°

=> c + 54Â° = 180Â°

âˆ´ c = 126Â°

**Q.3: In the Figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that âˆ ROS = 1/2(âˆ QOS â€“ âˆ POS).**

Solution:

In the question, it is given that (OR âŠ¥ PQ) and âˆ POQ = 180Â°

So, âˆ POS + âˆ ROS + âˆ ROQ = 180Â°

Now, âˆ POS + âˆ ROS = 180Â° â€“ 90Â° (Since âˆ POR = âˆ ROQ = 90Â°)

âˆ´ âˆ POS + âˆ ROS = 90Â°

Now, âˆ QOS = âˆ ROQ + âˆ ROS

It is given that âˆ ROQ = 90Â°,

âˆ´ âˆ QOS = 90Â° + âˆ ROS

Or, âˆ QOS – âˆ ROS = 90Â°

As âˆ POS + âˆ ROS = 90Â° and âˆ QOS – âˆ ROS = 90Â°, we get

âˆ POS + âˆ ROS = âˆ QOS – âˆ ROS

=>2 âˆ ROS + âˆ POS = âˆ QOS

Or, âˆ ROS = Â½ (âˆ QOS â€“ âˆ POS) (Hence proved).

**Q.4: It is given that âˆ XYZ = 64Â° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects âˆ ZYP, find âˆ XYQ and reflex âˆ QYP.**

Solution:

Here, XP is a straight line

So, âˆ XYZ +âˆ ZYP = 180Â°

substituting the value of âˆ XYZ = 64Â° we get,

64Â° +âˆ ZYP = 180Â°

âˆ´ âˆ ZYP = 116Â°

From the diagram, we also know that âˆ ZYP = âˆ ZYQ + âˆ QYP

Now, as YQ bisects âˆ ZYP,

âˆ ZYQ = âˆ QYP

Or, âˆ ZYP = 2âˆ ZYQ

âˆ´ âˆ ZYQ = âˆ QYP = 58Â°

Again, âˆ XYQ = âˆ XYZ + âˆ ZYQ

By substituting the value of âˆ XYZ = 64Â° and âˆ ZYQ = 58Â° we get.

âˆ XYQ = 64Â° + 58Â°

Or, âˆ XYQ = 122Â°

Now, reflex âˆ QYP = 180Â° + âˆ XYQ

We computed that the value of âˆ XYQ = 122Â°. So,

âˆ QYP = 180Â° + 122Â°

âˆ´ âˆ QYP = 302Â°

**Q.5: In the Figure, if AB || CD, EF âŠ¥ CD and âˆ GED = 126Â°, find âˆ AGE, âˆ GEF and âˆ FGE**.

Solution:

Since AB || CD GE is a transversal.

It is given that âˆ GED = 126Â°

So, âˆ GED = âˆ AGE = 126Â° (As they are alternate interior angles)

Also,

âˆ GED = âˆ GEF + âˆ FED

As

EF âŠ¥ CD, âˆ FED = 90Â°

âˆ´ âˆ GED = âˆ GEF + 90Â°

Or, âˆ GEF = 126 â€“ 90Â° = 36Â°

Again, âˆ FGE + âˆ GED = 180Â° (Transversal)

substituting the value of âˆ GED = 126Â° we get,

âˆ FGE = 54Â°

So,

âˆ AGE = 126Â°

âˆ GEF = 36Â° and

âˆ FGE = 54Â°

**Q.6: In the Figure, if PQ || ST, âˆ PQR = 110Â° and âˆ RST = 130Â°, find âˆ QRS.**

Solution: First, construct a line XY parallel to PQ.

As we know, the angles on the same side of transversal is equal to 180Â°.

So, âˆ PQR + âˆ QRX = 180Â°

Or,âˆ QRX = 180Â° â€“ 110Â°

âˆ´ âˆ QRX = 70Â°

Similarly,

âˆ RST + âˆ SRY = 180Â°

Or, âˆ SRY = 180Â° â€“ 130Â°

âˆ´ âˆ SRY = 50Â°

Now, for the linear pairs on the line XY-

âˆ QRX + âˆ QRS + âˆ SRY = 180Â°

substituting their respective values we get,

âˆ QRS = 180Â° â€“ 70Â° â€“ 50Â°

Or, âˆ QRS = 60Â°

**Q.7: 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 âŸ‚ PQ and CF âŸ‚ RS.

Now, since PQ || RS,

So, BE || CF

As we know,,

Angle of incidence = Angle of reflection (By the law of reflection)

So,

âˆ 1 = âˆ 2 and

âˆ 3 = âˆ 4

We also know that alternate interior angles are equal. Here, BE âŠ¥ CF and the transversal line BC cuts them at B and C

So, âˆ 2 = âˆ 3 (As they are alternate interior angles)

Now, âˆ 1 + âˆ 2 = âˆ 3 + âˆ 4

Or, âˆ ABC = âˆ DCB

So, AB âˆ¥ CD (alternate interior angles are equal)

**Q.8: In Fig. 6.40, âˆ X = 62Â°, âˆ XYZ = 54Â°. If YO and ZO are the bisectors of âˆ XYZ and âˆ XZY respectively of Î” XYZ, find âˆ OZY and âˆ YOZ.**

Solution:

As we know, the sum of the interior angles of the triangle.

So, âˆ X +âˆ XYZ + âˆ XZY = 180Â°

substituting the values as given in the question we get,

62Â° + 54Â° + âˆ XZY = 180Â°

Or, âˆ XZY = 64Â°

Now, As we know, ZO is the bisector so,

âˆ OZY = Â½ âˆ XZY

âˆ´ âˆ OZY = 32Â°

Similarly, YO is a bisector and so,

âˆ OYZ = Â½ âˆ XYZ

Or, âˆ OYZ = 27Â° (As âˆ XYZ = 54Â°)

Now, as the sum of the interior angles of the triangle,

âˆ OZY +âˆ OYZ + âˆ O = 180Â°

substituting their respective values we get,

âˆ O = 180Â° â€“ 32Â° â€“ 27Â°

Or, âˆ O = 121Â°

**Q.9: In the figure, if AB || CD || EF, PQ || RS, âˆ RQD = 25Â° and âˆ CQP = 60Â°, then find âˆ QRS?**

Solution:

According to the given figure, we have

AB || CD || EF

PQ || RS

âˆ RQD = 25Â°

âˆ CQP = 60Â°

PQ || RS.

As we know,

If a transversal intersects two parallel lines, then each pair of alternate exterior angles is equal.

Now, since, PQ || RS

â‡’ âˆ PQC = âˆ BRS

We have âˆ PQC = 60Â°

â‡’ âˆ BRS = 60Â° â€¦ eq.(i)

We also know that,

If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.

Now again, since, AB || CD

â‡’ âˆ DQR = âˆ QRA

We have âˆ DQR = 25Â°

â‡’ âˆ QRA = 25Â° â€¦ eq.(ii)

Using linear pair axiom,

We get,

âˆ ARS + âˆ BRS = 180Â°

â‡’ âˆ ARS = 180Â° â€“ âˆ BRS

â‡’ âˆ ARS = 180Â° â€“ 60Â° (From (i), âˆ BRS = 60Â°)

â‡’ âˆ ARS = 120Â° â€¦ eq.(iii)

Now, âˆ QRS = âˆ QRA + âˆ ARS

From equations (ii) and (iii), we have,

âˆ QRA = 25Â° and âˆ ARS = 120Â°

Hence, the above equation can be written as:

âˆ QRS = 25Â° + 120Â°

â‡’ âˆ QRS = 145Â°

### Class 9 Maths Chapter 6 Extra Questions

- If two lines intersect, prove that the vertically opposite angles are equal.
- Bisectors of interior âˆ B and exterior âˆ ACD of a Î” ABC intersect at the point T.Prove that âˆ BTC = Â½ âˆ BAC.
- A transversal intersects two parallel lines. Prove that the bisectors of any pair of corresponding angles so formed are parallel.
- In the figure, OD is the bisector of âˆ AOC, OE is the bisector of âˆ BOC and OD âŠ¥ OE. Show that the points A, O and B are collinear.

5. The angles of a triangle are in the ratio 5 : 3: 7. The triangle is

- An acute-angled triangle
- An obtuse-angled triangle
- A right triangle
- An isosceles triangle

6. Can a triangle have all angles less than 60Â°? Give reason for your answer.

7. Can a triangle have two obtuse angles? Give reason for your answer.

8. How many triangles can be drawn having its angles as 45Â°, 64Â° and 72Â°? Give reason for your answer.

9. How many triangles can be drawn having its angles as 53Â°, 64Â° and 63Â°? Give reason for your answer.