# RS Aggarwal Solutions for Class 9 Chapter 7: Lines and Angles Exercise 7C

Class 9 Maths introduces students to various new topics which are important from exam point of view. The students without knowledge about the concepts very often memorize the answers by heart. This leads to forgetting the steps while appearing for the exams. The objective of preparing solutions is to help students understand the methods of solving. While solving the problems from RS Aggarwal textbook, the students can make use of the PDF as a reference guide to perform well in exams. RS Aggarwal Solutions for Class 9 Maths Chapter 7 Lines and Angles Exercise 7C are provided here.

## Access RS Aggarwal Solutions for Class 9 Chapter 7: Lines and Angles Exercise 7C

Exercise 7(C)

1. In the given figure, l || m and a transversal t cuts them. If âˆ 1 = 120o, find the measure of each of the remaining marked angles.

Solution:

It is given that âˆ 1 = 120o

From the figure we know that âˆ 1 and âˆ 2 form a linear pair of angles

So it can be written as

âˆ 1 + âˆ 2 = 180o

By substituting the values

120o + âˆ 2 = 180o

On further calculation

âˆ 2 = 180o – 120o

By subtraction

âˆ 2 = 60o

From the figure we know that âˆ 1 and âˆ 3 are vertically opposite angles

So we get

âˆ 1 = âˆ 3 = 120o

From the figure we know that âˆ 2 and âˆ 4 are vertically opposite angles

So we get

âˆ 2 = âˆ 4 = 60o

It is given that, l || m and t is a transversal

So the corresponding angles according to the figure is written as

âˆ 1 = âˆ 5 = 120o

âˆ 2 = âˆ 6 = 60o

âˆ 3 = âˆ 7 = 120o

âˆ 4 = âˆ 8 = 60o

2. In the figure, l || m and a transversal t cuts them. If âˆ 7 = 80o, find the measure of each of the remaining marked angles.

Solution:

It is given that âˆ 7 = 80o

From the figure we know that âˆ 7 and âˆ 8 form a linear pair of angles

So it can be written as

âˆ 7 + âˆ 8 = 180o

By substituting the values

80o + âˆ 8 = 180o

On further calculation

âˆ 8 = 180o – 80o

By subtraction

âˆ 8 = 100o

From the figure we know that âˆ 7 and âˆ 5 are vertically opposite angles

So we get

âˆ 7 = âˆ 5 = 80o

From the figure we know that âˆ 6 and âˆ 8 are vertically opposite angles

So we get

âˆ 6 = âˆ 8 = 100o

It is given that, l || m and t is a transversal

So the corresponding angles according to the figure is written as

âˆ 1 = âˆ 5 = 80o

âˆ 2 = âˆ 6 = 100o

âˆ 3 = âˆ 7 = 80o

âˆ 4 = âˆ 8 = 100o

3. In the figure, l || m and a transversal t cuts them. If âˆ 1: âˆ 2 = 2: 3, find the measure of each of the marked angles.

Solution:

It is given that âˆ 1: âˆ 2 = 2: 3

From the figure we know that âˆ 1 and âˆ 2 form a linear pair of angles

So it can be written as

âˆ 1 + âˆ 2 = 180o

By substituting the values

2x + 3x = 180o

On further calculation

5x = 180o

By division

x = 180o/5

x = 36o

By substituting the value of x we get

âˆ 1 = 2x = 2 (36o) = 72o

âˆ 2 = 3x = 3 (36o) = 108o

From the figure we know that âˆ 1 and âˆ 3 are vertically opposite angles

So we get

âˆ 1 = âˆ 3 = 72o

From the figure we know that âˆ 2 and âˆ 4 are vertically opposite angles

So we get

âˆ 2 = âˆ 4 = 108o

It is given that, l || m and t is a transversal

So the corresponding angles according to the figure is written as

âˆ 1 = âˆ 5 = 72o

âˆ 2 = âˆ 6 = 108o

âˆ 3 = âˆ 7 = 72o

âˆ 4 = âˆ 8 = 108o

4. For what value of x will the lines l and m be parallel to each other?

Solution:

If the lines l and m are parallel it can be written as

3x â€“ 20 = 2x + 10

We know that the two lines are parallel if the corresponding angles are equal

3x â€“ 2x = 10 + 20

On further calculation

x = 30

Therefore, the value of x is 30.

5. For what value of x will the lines l and m be parallel to each other?

Solution:

We know that both the angles are consecutive interior angles

So it can be written as

3x + 5 + 4x = 180

On further calculation we get

7x = 180 â€“ 5

By subtraction

7x = 175

By division we get

x = 175/ 7

x = 25

Therefore, the value of x is 25.

6. In the figure, AB || CD and BC || ED. Find the value of x.

Solution:

From the given figure we know that

AB and CD are parallel line and BC is a transversal

We know that âˆ BCD and âˆ ABC are alternate angles

So we can write it as

âˆ BCD + âˆ ABC = xo

We also know that BC || ED and CD is a transversal

From the figure we know that âˆ BCD and âˆ EDC form a linear pair of angles

So it can be written as

âˆ BCD + âˆ EDC = 180o

By substituting the values we get

âˆ BCD + 75o = 180o

On further calculation we get

âˆ BCD = 180o – 75o

By subtraction

âˆ BCD = 105o

From the figure we know that âˆ BCD and âˆ ABC are vertically opposite angles

So we get

âˆ BCD = âˆ ABC = x = 105o

âˆ ABC = x = 105o

Therefore, the value of x is 105o.

7. In the figure, AB || CD || EF. Find the value of x.

Solution:

It is given that AB || CD and BC is a transversal

From the figure we know that âˆ BCD and âˆ ABC are alternate interior angles

So we get

âˆ ABC = âˆ BCD

In order to find the value of x we can write it as

xo + âˆ ECD = 70o â€¦â€¦. (1)

It is given that CD || EF and CE is a transversal

From the figure we know that âˆ ECD and âˆ CEF are consecutive interior angles

So we get

âˆ ECD + âˆ CEF = 180o

By substituting the values

âˆ ECD + 130o = 180o

On further calculation we get

âˆ ECD = 180o – 130o

By subtraction

âˆ ECD = 50o

Now by substituting âˆ ECD in equation (1) we get

xo + âˆ ECD = 70o

xo + 50o = 70o

On further calculation we get

xo = 70o – 50o

By subtraction

xo = 20o

Therefore, the value of x is 20o.

8. In the given figure, AB || CD. Find the values of x, y and z.

Solution:

It is given that AB || CD and EF is a transversal

From the figure we know that âˆ AEF and âˆ EFG are alternate angles

So we get

âˆ AEF = âˆ EFG = 75o

âˆ EFG = y = 75o

From the figure we know that âˆ EFC and âˆ EFG form a linear pair of angles

So we get

âˆ EFC + âˆ EFG = 180o

It can also be written as

x + y = 180o

By substituting the value of y we get

x + 75o = 180o

On further calculation we get

x = 180o – 75o

By subtraction

x = 105o

From the figure based on the exterior angle property it can be written as

âˆ EGD = âˆ EFG + âˆ FEG

By substituting the values in the above equation we get

125o = y + z

125o = 75o + z

On further calculation we get

z = 125o – 75o

By subtraction

z = 50o

Therefore, the values of x, y and z are 105o, 75o and 50o.

9. In each of the figures given below, AB || CD. Find the value of x in each case.

(i)

(ii)

(iii)

Solution:

(i) Draw a line at point E which is parallel to CD and name it as EG

It is given that EG || CD and ED is a transversal

From the figure we know that âˆ GED and âˆ EDC are alternate interior angles

So we get

âˆ GED = âˆ EDC = 65o

EG || CD and AB || CD

So we get EG || AB and EB is a transversal

From the figure we know that âˆ BEG and âˆ ABE are alternate interior angles

So we get

âˆ BEG = âˆ ABE = 35o

âˆ DEB = xo

From the figure we can write âˆ DEB as

âˆ DEB = âˆ BEG + âˆ GED

By substituting the values

xo = 35o + 65o

xo = 100o

Therefore, the value of x is 100.

(ii) Draw a line OF which is parallel to CD

So we get

OF || CD and OD is a transversal

From the figure we know that âˆ CDO and âˆ FOD are consecutive angles

So we get

âˆ CDO + âˆ FOD = 180o

By substituting the values

25o + âˆ FOD = 180o

On further calculation

âˆ FOD = 180o – 25o

By subtraction

âˆ FOD = 155o

We also know that OF || CD and AB || CD

So we get OF || AB and OB is a transversal

From the figure we know that âˆ ABO and âˆ FOB are consecutive angles

So we get

âˆ ABO + âˆ FOB = 180o

By substituting the values

55o + âˆ FOB = 180o

On further calculation

âˆ FOB = 180o – 55o

By subtraction

âˆ FOB = 125o

In order to find the value of x

xo = âˆ FOB + âˆ FOD

By substituting the values

xo = 125o + 155o

xo = 280o

Therefore, the value of x is 280.

(iii) Draw a line EF through point O which is parallel to CD

So we get EF || CD and EC is a transversal

From the figure we know that âˆ FEC and âˆ ECD are consecutive interior angles

So we get

âˆ FEC + âˆ ECD = 180o

By substituting the values

âˆ FEC + 124o = 180o

On further calculation

âˆ FEC = 180o – 124o

By subtraction

âˆ FEC = 56o

We know that EF || CD and AB || CD

So we get EF || AB and AE is a transversal

From the figure we know that âˆ BAE and âˆ FEA are consecutive interior angles

So we get

âˆ BAE + âˆ FEA = 180o

By substituting the values

116o + âˆ FEA = 180o

On further calculation

âˆ FEA = 180o – 116o

By subtraction

âˆ FEA = 64o

In order to find the value of x

xo = âˆ FEA + âˆ FEC

By substituting the values

xo = 64o + 56o

xo = 120o

Therefore, the value of x is 120o

10. In the given figure, AB || CD. Find the value of x.

Solution:

Draw a line through point C and name it as FG where FG || AE

We know that CG || BE and CE is a transversal

From the figure we know that âˆ GCE and âˆ CEA are alternate angles

So we get

âˆ GCE = âˆ CEA = 20o

It can also be written as

âˆ DCG = âˆ DCE – âˆ GCE

By substituting the values we get

âˆ DCG = 130o – 20o

By subtraction we get

âˆ DCG = 110o

We also know that AB || CD and FG is a transversal

From the figure we know that âˆ BFC and âˆ DCG are corresponding angles

So we get

âˆ BFC = âˆ DCG = 110o

We know that FG || AE and AF is a transversal

From the figure we know that âˆ BFG and âˆ FAE are corresponding angles

So we get

âˆ BFG = âˆ FAE = 110o

âˆ FAE = x = 110o

Therefore, the value of x is 110.

11. In the given figure, AB || PQ. Find the values of x and y.

Solution:

It is given that AB || PQ and EF is a transversal

From the figure we know that âˆ CEB and âˆ EFQ are corresponding angles

So we get

âˆ CEB = âˆ EFQ = 75o

It can be written as

âˆ EFQ = 75o

Where

âˆ EFG + âˆ GFQ = 75o

By substituting the values

25o + yo = 75o

On further calculation

yo = 75o – 25o

By subtraction

yo = 50o

From the figure we know that âˆ BEF and âˆ EFQ are consecutive interior angles

So we get

âˆ BEF + âˆ EFQ = 180o

By substituting the values

âˆ BEF + 75o = 180o

On further calculation

âˆ BEF = 180o – 75o

By subtraction

âˆ BEF = 105o

We know that âˆ BEF can be written as

âˆ BEF = âˆ FEG + âˆ GEB

105o = âˆ FEG + 20o

On further calculation

âˆ FEG = 105o – 20o

By subtraction

âˆ FEG = 85o

According to the â–³ EFG

We can write

xo + 25o + âˆ FEG = 180o

By substituting the values

xo + 25o + 85o = 180o

On further calculation

xo = 180o – 25o – 85o

By subtraction

xo = 180o â€“ 110o

xo = 70o

Therefore, the value of x is 70.

12. In the given figure, AB || CD. Find the value of x.

Solution:

It is given that AB || CD and AC is a transversal.

From the figure we know that âˆ BAC and âˆ ACD are consecutive interior angles

So we get

âˆ BAC + âˆ ACD = 180o

By substituting the values

75o + âˆ ACD = 180o

On further calculation

âˆ ACD = 180o – 75o

By subtraction

âˆ ACD = 105o

From the figure we know that âˆ ECF and âˆ ACD are vertically opposite angles

So we get

âˆ ECF = âˆ ACD = 105o

According to the â–³ CEF

We can write

âˆ ECF + âˆ CEF + âˆ EFC = 180o

By substituting the values

105o + xo + 30o = 180o

On further calculation

xo = 180o – 105o – 30o

By subtraction

xo = 180o â€“ 135o

xo = 45o

Therefore, the value of x is 45.

13. In the given figure, AB || CD. Find the value of x.

Solution:

It is given that AB || CD and PQ is a transversal

From the figure we know that âˆ PEF and âˆ EGH are corresponding angles

So we get

âˆ PEF = âˆ EGH = 85o

From the figure we also know that âˆ EGH and âˆ QGH form a linear pair of angles

So we get

âˆ EGH + âˆ QGH = 180o

By substituting the values we get

85o + âˆ QGH = 180o

On further calculation

âˆ QGH = 180o – 85o

By subtraction

âˆ QGH = 95o

We can also find the âˆ GHQ

âˆ GHQ + âˆ CHQ = 180o

By substituting the values

âˆ GHQ + 115o = 180o

On further calculation

âˆ GHQ = 180o – 115o

By subtraction

âˆ GHQ = 65o

According to the â–³ GHQ

We can write

âˆ GQH + âˆ GHQ + âˆ QGH = 180o

By substituting the values

xo + 65o + 95o = 180o

On further calculation

xo = 180o – 65o – 95o

By subtraction

xo = 180o â€“ 160o

xo = 20o

Therefore, the value of x is 20.

14. In the given figure, AB || CD. Find the value of x, y and z.

Solution:

It is given that AB || CD and BC is a transversal

From the figure we know that âˆ ABC = âˆ BCD

So we get

x = 35

It is also given that AB || CD and AD is a transversal

So we get

z = 75

According to the â–³ ABO

We can write

âˆ ABO + âˆ BAO + âˆ BOA = 180o

By substituting the values

xo + 75o + yo = 180o

35o + 75o + yo = 180o

On further calculation

yo = 180o – 35o – 75o

By subtraction

yo = 180o â€“ 110o

yo = 70o

Therefore, the value of x, y and z is 35, 70 and 75.

15. In the given figure, AB || CD. Prove that âˆ BAE – âˆ ECD = âˆ AEC.

Solution:

Construction a line EF which is parallel to AB and CD through the point E

We know that EF || AB and AE is a transversal

From the figure we know that âˆ BAE and âˆ AEF are supplementary

So it can be written as

âˆ BAE + âˆ AEF = 180oâ€¦.. (1)

We also know that EF || CD and CE is a transversal

From the figure we know that âˆ DCE and âˆ CEF are supplementary

So it can be written as

âˆ DCE + âˆ CEF = 180o

According to the diagram the above equation can be written as

âˆ DCE + (âˆ AEC +âˆ AEF) = 180o

From equation (1) we know that âˆ AEF can be written as 180o – âˆ BAE

So we get

âˆ DCE + âˆ AEC + 180o – âˆ BAE = 180o

So we get

âˆ BAE – âˆ DCE = âˆ AEC

Therefore, it is proved that âˆ BAE – âˆ DCE = âˆ AEC

16. In the given figure, AB || CD. Prove that p + q â€“ r = 180.

Solution:

Draw a line KH passing through the point F which is parallel to both AB and CD

We know that KF || CD and FG is a transversal

From the figure we know that âˆ KFG and âˆ FGD are alternate angles

So we get

âˆ KFG = âˆ FGD = ro â€¦â€¦. (1)

We also know that AE || KF and EF is a transversal

From the figure we know that âˆ AEF and âˆ KFE are alternate angles

So we get

âˆ AEF + âˆ KFE = 180o

By substituting the values we get

po + âˆ KFE = 180o

So we get

âˆ KFE = 180o – po â€¦â€¦. (2)

By adding both the equations (1) and (2) we get

âˆ KFG + âˆ KFE = 180o – po + ro

From the figure âˆ KFG + âˆ KFE can be written as âˆ EFG

âˆ EFG = 180o – po + ro

We know that âˆ EFG = qo

qo = 180o – po + ro

It can be written as

p + q â€“ r = 180o

Therefore, it is proved that p + q â€“ r = 180o

17. In the given figure, AB || CD and EF || GH. Find the values of x, y, z and t.

Solution:

From the figure we know that âˆ PRQ = xo = 60o as the vertically opposite angles are equal

We know that EF || GH and RQ is a transversal

From the figure we also know that âˆ PRQ and âˆ RQS are alternate angles

So we get

âˆ PRQ = âˆ RQS

âˆ x = âˆ y = 60o

We know that AB || CD and PR is a transversal

From the figure we know that âˆ PRD and âˆ APR are alternate angles

So we get

âˆ PRD = âˆ APR

It can be written as

âˆ PRQ + âˆ QRD = âˆ APR

By substituting the values we get

x + âˆ QRD = 110o

60o + âˆ QRD = 110o

On further calculation

âˆ QRD = 110o – 60o

By subtraction

âˆ QRD = 50o

According to the â–³ QRS

We can write

âˆ QRD + âˆ QSR + âˆ RQS = 180o

By substituting the values

âˆ QRD + to + yo = 180o

50o + to + 60o = 180o

On further calculation

to = 180o – 50o – 60o

By subtraction

to = 180o â€“ 110o

to = 70o

We know that AB || CD and GH is a transversal

From the figure we know that zo and to are alternate angles

So we get

zo = to = 70o

Therefore, the values of x, y, z and t are 60o, 60o, 70o and 70o.

18. In the given figure, AB || CD and a transversal t cuts them at E and F respectively. If EG and FG are the bisectors of âˆ BEF and âˆ EFD respectively, prove that âˆ EGF = 90o.

Solution:

We know that AB || CD and t is a transversal cutting at points E and F

From the figure we know that âˆ BEF and âˆ DFE are interior angles

So we get

âˆ BEF + âˆ DFE = 180o

Dividing the entire equation by 2 we get

(1/2)âˆ BEF + (1/2) âˆ DFE = 90o

According to the figure the above equation can further be written as

âˆ GEF + âˆ GFE = 90o â€¦â€¦. (1)

According to the â–³ GEF

We can write

âˆ GEF + âˆ GFE + âˆ EGF = 180o

Based on equation (1) we get

90o + âˆ EGF = 180o

On further calculation

âˆ EGF = 180o – 90o

By subtraction

âˆ EGF = 90o

Therefore, it is proved that âˆ EGF = 90o

19. In the given figure, AB || CD and a transversal t cuts them at E and F respectively. If EP and FQ are the bisectors of âˆ AEF and âˆ EFD respectively, prove that EP || FQ.

Solution:

We know that AB || CD and t is a transversal

From the figure we know that âˆ AEF and âˆ EFD are alternate angles

So we get

âˆ AEF = âˆ EFD

Dividing both the sides by 2 we get

(1/2) âˆ AEF = (1/2) âˆ EFD

So we get

âˆ PEF = âˆ EFQ

The alternate interior angles are formed only when the transversal EF cuts both FQ and EP.

Therefore, it is proved that EP || FQ.

20. In the given figure, BA || ED and BC || EF. Show that âˆ ABC = âˆ DEF.

Solution:

Extend the line DE to meet the line BE at the point Z.

We know that AB || DZ and BC is a transversal

From the figure we know that âˆ ABC and âˆ DZC are corresponding angles

So we get

âˆ ABC = âˆ DZC â€¦.. (1)

We also know that EF || BC and DZ is a transversal

From the figure we know that âˆ DZC and âˆ DEF are corresponding angles

So we get

âˆ DZC = âˆ DEF â€¦â€¦ (2)

Considering both the equation (1) and (2) we get

âˆ ABC = âˆ DEF

Therefore, it is proved that âˆ ABC = âˆ DEF

21. In the given figure, BA || ED and BC || EF. Show that âˆ ABC + âˆ DEF = 180o.

Solution:

Extend the line ED to meet the line BC at the point Z

We know that AB || EZ and BC is a transversal

From the figure we know that âˆ ABZ and âˆ EZB are interior angles

So we get

âˆ ABZ + âˆ EZB = 180o

âˆ ABZ can also be written as âˆ ABC

âˆ ABC + âˆ EZB = 180o â€¦â€¦ (1)

We know that EF || BC and EZ is a transversal

From the figure we know that âˆ BZE and âˆ ZEF are alternate angles

So we get

âˆ BZE = âˆ ZEF

âˆ ZEF can also be written as âˆ DEF

âˆ BZE = âˆ DEF â€¦.. (2)

By substituting equation (1) in (2) we get

âˆ ABC + âˆ DEF = 180o

Therefore, it is proved that âˆ ABC + âˆ DEF = 180o

22. In the given figure, m and n are two plane mirrors perpendicular to each other. Show that the incident ray CA is parallel to the reflected ray BD.

Solution:

Construct a line m and n from A and B intersect at P

So we get

OB âŠ¥ m and OC âŠ¥ n

So m âŠ¥ n

We can also write it as

OB âŠ¥ OC

Since APB is a right angle triangle

We know that âˆ APB = 90o

So we can write it as

âˆ APB = âˆ PAB + âˆ PBA

By substituting the values

90o = âˆ 2 + âˆ 3

We know that angle of incidence is equal to the angle of reflection

So we get

âˆ 1 = âˆ 2 and âˆ 4 = âˆ 3

It can be written as

âˆ 1 + âˆ 4 = âˆ 2 + âˆ 3 = 90o

We can write it as

âˆ 1 + âˆ 2 + âˆ 3 + âˆ 4 = 180o

We know that âˆ 1 + âˆ 2 = âˆ CAB and âˆ 3 + âˆ 4 = âˆ ABD

âˆ CAB + âˆ ABD = 180o

According to the diagram âˆ CAB and âˆ ABD are consecutive interior angles when the transversal AB cuts BD and CA.

Therefore, it is proved that CA || BD.

23. In the figure given below, state which lines are parallel and why?

Solution:

According to the figure we know that âˆ BAC and âˆ ACD are alternate angles

So we get

âˆ BAC = âˆ ACD = 110o

We know that âˆ BAC and âˆ ACD are alternate angles when the transversal AC cuts CD and AB.

Therefore, AB || CD.

24. Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.

Solution:

According to the figure consider the two parallel lines as m and n

We know that p âŠ¥ m and q âŠ¥ n

So we get

âˆ 1 = âˆ 2 = 90o

We know that m || n and p is a transversal

From the figure we know that âˆ 1 and âˆ 3 are corresponding angles

So we get

âˆ 1 = âˆ 3

We also know that

âˆ 2 = âˆ 3 = 90o

We know that âˆ 2 and âˆ 3 are corresponding angles when the transversal n cuts p and q.

So we get p || q.

Therefore, it is shown that the two lines which are perpendicular to two parallel lines are parallel to each other.

### RS Aggarwal Solutions Class 9 Maths Chapter 7 – Lines and Angles Exercise 7C

RS Aggarwal Solutions Class 9 Maths Chapter 7 Lines and Angles Exercise 7C explains in brief about the results on parallel lines, angles formed when a transversal cuts two lines and corresponding angles axiom.

### Key features of RS Aggarwal Solutions for Class 9 Maths Chapter 7: Lines and Angles Exercise 7C

• The solutions are prepared according to the RS Aggarwal textbook of latest CBSE syllabus.
• The faculty explains the problems in a simple language based on the understanding level of the students.
• PDF of the solutions can be used while solving problems to get their doubts cleared and have clear cut idea about the concepts.
• Helping students to score well in the board exams is the main vision of preparing the solutions.