In this exercise, students will introduce some new angles made by transversal with two given lines. RD Sharma Class 9 Solutions Chapter 8 exercise 8.4 also contains answers related to chapters like alternate interior angles, consecutive interior angles and corresponding angles axiom.

## Download PDF of RD Sharma Solutions for Class 9 Maths Chapter 8 Lines and Angles Exercise 8.4

### Access Answers to Maths RD Sharma Class 9 Chapter 8 Lines and Angles Exercise 8.4 Page number 8.38

**Question 1: In figure, AB, CD and âˆ 1 and âˆ 2 are in the ratio 3 : 2. Determine all angles from 1 to 8.**

**Solution: **

Let âˆ 1 = 3x and âˆ 2 = 2x

From figure: âˆ 1 and âˆ 2 are linear pair of angles

Therefore, âˆ 1 + âˆ 2 = 180

3x + 2x = 180

5x = 180

x = 180 / 5

=> x = 36

So, âˆ 1 = 3x = 108^{0} and âˆ 2 = 2x = 72^{0}

As we know, vertically opposite angles are equal.

Pairs of vertically opposite angles are:

(âˆ 1 = âˆ 3); (âˆ 2 = âˆ 4) ; (âˆ 5, âˆ 7) and (âˆ 6 , âˆ 8)

âˆ 1 = âˆ 3 = 108Â°

âˆ 2 = âˆ 4 = 72Â°

âˆ 5 = âˆ 7

âˆ 6 = âˆ 8

We also know, if a transversal intersects any parallel lines, then the corresponding angles are equal

âˆ 1 = âˆ 5 = âˆ 7 = 108Â°

âˆ 2 = âˆ 6 = âˆ 8 = 72Â°

Answer: âˆ 1 = 108Â°, âˆ 2 = 72Â°, âˆ 3 = 108Â°, âˆ 4 = 72Â°, âˆ 5 = 108Â°, âˆ 6 = 72Â°, âˆ 7 = 108Â° and âˆ 8 = 72Â°

**Question 2: In figure, I, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find âˆ 1, âˆ 2 and âˆ 3.**

**Solution:** From figure:

âˆ Y = 120Â° [Vertical opposite angles]

âˆ 3 + âˆ Y = 180Â° [Linear pair angles theorem]

=> âˆ 3= 180 â€“ 120

=> âˆ 3= 60Â°

Line l is parallel to line m,

âˆ 1 = âˆ 3 [ Corresponding angles]

âˆ 1 = 60Â°

Also, line m is parallel to line n,

âˆ 2 = âˆ Y [Alternate interior angles are equal]

âˆ 2 = 120Â°

Answer: âˆ 1 = 60Â°, âˆ 2 = 120Â° and âˆ 3 = 60Â°.

**Question 3: In figure, AB || CD || EF and GH || KL. Find âˆ HKL.**

**Solution:**

Extend LK to meet line GF at point P.

From figure, CD || GF, so, alternate angles are equal.

âˆ CHG =âˆ HGP = 60Â°

âˆ HGP =âˆ KPF = 60Â° [Corresponding angles of parallel lines are equal]

Hence, âˆ KPG =180 â€“ 60 = 120Â°

=> âˆ GPK = âˆ AKL= 120Â° [Corresponding angles of parallel lines are equal]

âˆ AKH = âˆ KHD = 25Â° [alternate angles of parallel lines]

Therefore, âˆ HKL = âˆ AKH + âˆ AKL = 25 + 120 = 145Â°

**Question 4: In figure, show that AB || EF.**

**Solution: **Produce EF to intersect AC at point N.

From figure, âˆ BAC = 57Â° and

âˆ ACD = 22Â°+35Â° = 57Â°

Alternative angles of parallel lines are equal

=> BA || EF â€¦..(1)

Sum of Co-interior angles of parallel lines is 180Â°

EF || CD

âˆ DCE + âˆ CEF = 35 + 145 = 180Â° â€¦(2)

From (1) and (2)

AB || EF

[Since, Lines parallel to the same line are parallel to each other]Hence Proved.

**Question 5 : In figure, if AB || CD and CD || EF, find âˆ ACE.**

**Solution: **

Given: CD || EF

âˆ FEC + âˆ ECD = 180Â°

[Sum of co-interior angles is supplementary to each other]=> âˆ ECD = 180Â° â€“ 130Â° = 50Â°

Also, BA || CD

=> âˆ BAC = âˆ ACD = 70Â°

[Alternative angles of parallel lines are equal]But, âˆ ACE + âˆ ECD =70Â°

=> âˆ ACE = 70Â° â€” 50Â° = 20Â°

** Question 6: In figure, PQ || AB and PR || BC. If âˆ QPR = 102Â°, determine âˆ ABC. Give reasons.**

**Solution:** Extend line AB to meet line PR at point G.

Given: PQ || AB,

âˆ QPR = âˆ BGR =102Â°

[Corresponding angles of parallel lines are equal]And PR || BC,

âˆ RGB+ âˆ CBG =180Â°

[Corresponding angles are supplementary]âˆ CBG = 180Â° â€“ 102Â° = 78Â°

Since, âˆ CBG = âˆ ABC

=>âˆ ABC = 78Â°

**Question 7 : In figure, state which lines are parallel and why?**

**Solution: **

We know, If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel

From figure:

=> âˆ EDC = âˆ DCA = 100Â°

Lines DE and AC are intersected by a transversal DC such that the pair of alternate angles are equal.

So, DE || AC

**Question 8: In figure, if l||m, n || p and âˆ 1 = 85Â°, find âˆ 2.**

**Solution:**

Given: âˆ 1 = 85Â°

As we know, when a line cuts the parallel lines, the pair of alternate interior angles are equal.

=> âˆ 1 = âˆ 3 = 85Â°

Again, co-interior angles are supplementary, so

âˆ 2 + âˆ 3 = 180Â°

âˆ 2 + 55Â° =180Â°

âˆ 2 = 180Â° â€“ 85Â°

âˆ 2 = 95Â°

**Question 9 : If two straight lines are perpendicular to the same line, prove that they are parallel to each other.**

**Solution: **

Let lines l and m are perpendicular to n, then

âˆ 1= âˆ 2=90Â°

Since, lines l and m cut by a transversal line n and the corresponding angles are equal, which shows that, line l is parallel to line m.

**Question 10: Prove that if the two arms of an angle are perpendicular to the two arms of another angle, then the angles are either equal or supplementary.**

**Solution:** Let the angles be âˆ ACB and âˆ ABD

Let AC perpendicular to AB, and CD is perpendicular to BD.

To Prove : âˆ ACD = âˆ ABD OR âˆ ACD + âˆ ABD =180Â°

Proof :

In a quadrilateral,

âˆ A+ âˆ C+ âˆ D+ âˆ B = 360Â°

[ Sum of angles of quadrilateral is 360Â° ]=> 180Â° + âˆ C + âˆ B = 360Â°

=> âˆ C + âˆ B = 360Â° â€“180Â°

Therefore, âˆ ACD + âˆ ABD = 180Â°

And âˆ ABD = âˆ ACD = 90Â°

Hence, angles are equal as well as supplementary.

## RD Sharma Solutions for Class 9 Maths Chapter 8 Exercise 8.4

RD Sharma Solutions Class 9 Maths Chapter 8 Lines and Angles Exercise 8.4 is based on following topics and subtopics:

- Angles made by a transversal with two lines
- Alternate interior angles
- Consecutive interior angles
- Corresponding angles axiom