The document NCERT Solutions Chapter 6 - Lines And Angles (I), Class 9, Maths Class 9 Notes | EduRev is a part of Class 9 category.

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**Exercise 6.1 1. In Fig. 6.13, lines AB and CD intersect at O. If âˆ AOC âˆ BOE = 70Â° and âˆ BOD = 40Â°, find âˆ BOE and reflex âˆ COE**.

**Answer** Given,

âˆ AOC + âˆ BOE = 70Â° and âˆ BOD = 40Â°

A/q,

âˆ AOC + âˆ BOE +âˆ COE = 180Â° (Forms a straight line)

â‡’ 70Â° +âˆ COE = 180Â°

â‡’ âˆ COE = 110Â°

also,

âˆ COE +âˆ BOD + âˆ BOE = 180Â° (Forms a straight line)

â‡’ 110Â° +40Â° + âˆ BOE = 180Â°

â‡’ 150Â° + âˆ BOE = 180Â°

â‡’ âˆ BOE = 30Â°

**2. In Fig. 6.14, lines XY and MN intersect at O. If âˆ POY = 90Â° and a : b = 2 : 3, find c.**

**Answer**

Given,

âˆ POY = 90Â° and a : b = 2 : 3

A/q,

âˆ POY + a + b = 180Â°

â‡’ 90Â° + a + b = 180Â°

â‡’ a + b = 90Â°

Let a be 2x then will be 3x

2x + 3x = 90Â°

â‡’ 5x = 90Â°

â‡’ x = 18Â°

âˆ´ a = 2Ã—18Â° = 36Â°

and b = 3Ã—18Â° = 54Â°

also,

b + c = 180Â° (Linear Pair)

â‡’ 54Â° + c = 180Â°

â‡’ c = 126Â°**3. In Fig. 6.15, âˆ PQR = âˆ PRQ, then prove that âˆ PQS = âˆ PRT.**

**Answer**

Given,

âˆ PQR = âˆ PRQ

To prove,

âˆ PQS = âˆ PRT

A/q,

âˆ PQR +âˆ PQS = 180Â° (Linear Pair)

â‡’ âˆ PQS = 180Â° - âˆ PQR --- (i)

also,

âˆ PRQ +âˆ PRT = 180Â° (Linear Pair)

â‡’ âˆ PRT = 180Â° - âˆ PRQ

â‡’ âˆ PRQ = 180Â° - âˆ PQR --- (ii) (âˆ PQR = âˆ PRQ)

From (i) and (ii)

âˆ PQS = âˆ PRT = 180Â° - âˆ PQR

Therefore, âˆ PQS = âˆ PRT**4. In Fig. 6.16, if x y = w z, then prove that AOB is a line.**

**Answer**

Given,

x + y = w + z

To Prove,

AOB is a line or x + y = 180Â° (linear pair.)

A/q,

x + y + w + z = 360Â° (Angles around a point.)

â‡’ (x + y) + (w + z) = 360Â°

â‡’ (x + y) + (x + y) = 360Â° (Given x + y = w + z)

â‡’ 2(x + y) = 360Â°

â‡’ (x + y) = 180Â°

Hence, x + y makes a linear pair. Therefore, AOB is a straight line.

**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 âˆ ROS = 1/2(âˆ QOS â€“ âˆ POS).**

**Answer**

Given,

OR is perpendicular to line PQ

To prove,

âˆ ROS = 1/2(âˆ QOS â€“ âˆ POS)

A/q,

âˆ POR = âˆ ROQ = 90Â° (Perpendicular)

âˆ QOS = âˆ ROQ + âˆ ROS = 90Â° + âˆ ROS --- (i)

âˆ POS = âˆ POR - âˆ ROS = 90Â° - âˆ ROS --- (ii)

Subtracting (ii) from (i)

âˆ QOS - âˆ POS = 90Â° + âˆ ROS - (90Â° - âˆ ROS)

â‡’ âˆ QOS - âˆ POS = 90Â° + âˆ ROS - 90Â° + âˆ ROS

â‡’ âˆ QOS - âˆ POS = 2âˆ ROS

â‡’ âˆ ROS = 1/2(âˆ QOS â€“ âˆ POS)

Hence, Proved.

**6. 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.****Answer**

Given,

âˆ XYZ = 64Â°

YQ bisects âˆ ZYP

âˆ XYZ +âˆ ZYP = 180Â° (Linear Pair)

â‡’ 64Â° +âˆ ZYP = 180Â°

â‡’ âˆ ZYP = 116Â°

also, âˆ ZYP = âˆ ZYQ + âˆ QYP

âˆ ZYQ = âˆ QYP (YQ bisects âˆ ZYP)

â‡’ âˆ ZYP = 2âˆ ZYQ

â‡’ 2âˆ ZYQ = 116Â°

â‡’ âˆ ZYQ = 58Â° = âˆ QYP

Now,

âˆ XYQ = âˆ XYZ + âˆ ZYQ

â‡’ âˆ XYQ = 64Â° + 58Â°

â‡’ âˆ XYQ = 122Â°

also,

reflex âˆ QYP = 180Â° + âˆ XYQ

âˆ QYP = 180Â° + 122Â°

â‡’ âˆ QYP = 302Â°**Exercise 6.2 1. In Fig. 6.28, find the values of x and y and then show that AB || CD.**

**Answer**

x + 50Â° = 180Â° (Linear pair)

â‡’ x = 130Â°

also,

y = 130Â° (Vertically opposite)

Now,

x = y = 130Â° (Alternate interior angles)

Alternate interior angles are equal.

Therefore, AB || CD.**2. In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.**

**Answer**

Given,

AB || CD and CD || EF

y : z = 3 : 7

Now,

x + y = 180Â° (Angles on the same side of transversal.)

also,

âˆ O = z (Corresponding angles)

and, y + âˆ O = 180Â° (Linear pair)

â‡’ y + z = 180Â°

A/q,

y = 3w and z = 7w

3w + 7w = 180Â°

â‡’ 10 w = 180Â°

â‡’ w = 18Â°

âˆ´ y = 3Ã—18Â° = 54Â°

and, z = 7Ã—18Â° = 126Â°

Now,

x + y = 180Â°

â‡’ x + 54Â° = 180Â°

â‡’ x = 126Â°**3. In Fig. 6.30, if AB || CD, EF âŠ¥ CD and âˆ GED = 126Â°, find âˆ AGE, âˆ GEF and âˆ FGE. **

**Answer**

Given,

AB || CD

EF âŠ¥ CD

âˆ GED = 126Â°

A/q,

âˆ FED = 90Â° (EF âŠ¥ CD)

Now,

âˆ AGE = âˆ GED (Since, AB || CD and GE is transversal. Alternate interior angles.)

âˆ´ âˆ AGE = 126Â°

Also, âˆ GEF = âˆ GED - âˆ FED

â‡’ âˆ GEF = 126Â° - 90Â°

â‡’ âˆ GEF = 36Â°

Now,

âˆ FGE +âˆ AGE = 180Â° (Linear pair)

â‡’ âˆ FGE = 180Â° - 126Â°

â‡’ âˆ FGE = 54Â°**4. In Fig. 6.31, if PQ || ST, âˆ PQR = 110Â° and âˆ RST = 130Â°, find âˆ QRS. [Hint : Draw a line parallel to ST through point R.]**

**Answer**

Given,

PQ || ST, âˆ PQR = 110Â° and âˆ RST = 130Â°

Construction,

A line XY parallel to PQ and ST is drawn.

âˆ PQR + âˆ QRX = 180Â° (Angles on the same side of transversal.)

â‡’ 110Â° + âˆ QRX = 180Â°

â‡’ âˆ QRX = 70Â°

Also,

âˆ RST + âˆ SRY = 180Â° (Angles on the same side of transversal.)

â‡’ 130Â° + âˆ SRY = 180Â°

â‡’ âˆ SRY = 50Â°

Now,

âˆ QRX +âˆ SRY + âˆ QRS = 180Â°

â‡’ 70Â° + 50Â° + âˆ QRS = 180Â°

â‡’ âˆ QRS = 60Â°**5. In Fig. 6.32, if AB || CD, âˆ APQ = 50Â° and âˆ PRD = 127Â°, find x and y.**

**Answer** Given,

AB || CD, âˆ APQ = 50Â° and âˆ PRD = 127Â°

A/q,

x = 50Â° (Alternate interior angles.)

âˆ PRD + âˆ RPB = 180Â° (Angles on the same side of transversal.)

â‡’ 127Â° + âˆ RPB = 180Â°

â‡’ âˆ RPB = 53Â°

Now,

y + 50Â° + âˆ RPB = 180Â° (AB is a straight line.)

â‡’ y + 50Â° + 53Â° = 180Â°

â‡’ y + 103Â° = 180Â°

â‡’ y = 77Â°

**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.**

**Answer**

Let us draw BE âŸ‚ PQ and CF âŸ‚ RS.

As PQ || RS

So, BE || CF

By laws of reflection we know that,

Angle of incidence = Angle of reflection

Thus, âˆ 1 = âˆ 2 and âˆ 3 = âˆ 4 --- (i)

also, âˆ 2 = âˆ 3 (alternate interior angles because BE || CF and a transversal line BC cuts them at B and C) --- (ii)

From (i) and (ii),

âˆ 1 + âˆ 2 = âˆ 3 + âˆ 4

â‡’ âˆ ABC = âˆ DCB

â‡’ AB || CD (alternate interior angles are equal)