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.

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

## 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 = 120 ^{o}, find the measure of each of the remaining marked angles.**

**Solution:**

It is given that âˆ 1 = 120^{o}

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 = 180^{o}

By substituting the values

120^{o} + âˆ 2 = 180^{o}

On further calculation

âˆ 2 = 180^{o} – 120^{o}

By subtraction

âˆ 2 = 60^{o}

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

So we get

âˆ 1 = âˆ 3 = 120^{o}

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

So we get

âˆ 2 = âˆ 4 = 60^{o}

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 = 120^{o}

âˆ 2 = âˆ 6 = 60^{o}

âˆ 3 = âˆ 7 = 120^{o}

âˆ 4 = âˆ 8 = 60^{o}

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

**Solution:**

It is given that âˆ 7 = 80^{o}

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 = 180^{o}

By substituting the values

80^{o} + âˆ 8 = 180^{o}

On further calculation

âˆ 8 = 180^{o} – 80^{o}

By subtraction

âˆ 8 = 100^{o}

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

So we get

âˆ 7 = âˆ 5 = 80^{o}

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

So we get

âˆ 6 = âˆ 8 = 100^{o}

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 = 80^{o}

âˆ 2 = âˆ 6 = 100^{o}

âˆ 3 = âˆ 7 = 80^{o}

âˆ 4 = âˆ 8 = 100^{o}

**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 = 180^{o}

By substituting the values

2x + 3x = 180^{o}

On further calculation

5x = 180^{o}

By division

x = 180^{o}/5

x = 36^{o}

By substituting the value of x we get

âˆ 1 = 2x = 2 (36^{o}) = 72^{o}

âˆ 2 = 3x = 3 (36^{o}) = 108^{o}

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

So we get

âˆ 1 = âˆ 3 = 72^{o}

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

So we get

âˆ 2 = âˆ 4 = 108^{o}

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 = 72^{o}

âˆ 2 = âˆ 6 = 108^{o}

âˆ 3 = âˆ 7 = 72^{o}

âˆ 4 = âˆ 8 = 108^{o}

**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 = x^{o}

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 = 180^{o}

By substituting the values we get

âˆ BCD + 75^{o} = 180^{o}

On further calculation we get

âˆ BCD = 180^{o} – 75^{o}

By subtraction

âˆ BCD = 105^{o}

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

So we get

âˆ BCD = âˆ ABC = x = 105^{o}

âˆ ABC = x = 105^{o}

Therefore, the value of x is 105^{o}.

**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

x^{o} + âˆ ECD = 70^{o} 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 = 180^{o}

By substituting the values

âˆ ECD + 130^{o} = 180^{o}

On further calculation we get

âˆ ECD = 180^{o} – 130^{o}

By subtraction

âˆ ECD = 50^{o}

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

x^{o} + âˆ ECD = 70^{o}

x^{o} + 50^{o} = 70^{o}

On further calculation we get

x^{o }= 70^{o} – 50^{o}

By subtraction

x^{o }= 20^{o}

Therefore, the value of x is 20^{o}.

**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 = 75^{o}

âˆ EFG = y = 75^{o}

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

So we get

âˆ EFC + âˆ EFG = 180^{o}

It can also be written as

x + y = 180^{o}

By substituting the value of y we get

x + 75^{o} = 180^{o}

On further calculation we get

x = 180^{o} – 75^{o}

By subtraction

x = 105^{o}

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

125^{o} = y + z

125^{o} = 75^{o }+ z

On further calculation we get

z = 125^{o} – 75^{o}

By subtraction

z = 50^{o}

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

**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 = 65^{o}

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 = 35^{o}

âˆ DEB = x^{o}

From the figure we can write âˆ DEB as

âˆ DEB = âˆ BEG + âˆ GED

By substituting the values

x^{o} = 35^{o} + 65^{o}

By addition

x^{o} = 100^{o}

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 = 180^{o}

By substituting the values

25^{o} + âˆ FOD = 180^{o}

On further calculation

âˆ FOD = 180^{o} – 25^{o}

By subtraction

âˆ FOD = 155^{o}

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 = 180^{o}

By substituting the values

55^{o} + âˆ FOB = 180^{o}

On further calculation

âˆ FOB = 180^{o} – 55^{o}

By subtraction

âˆ FOB = 125^{o}

In order to find the value of x

x^{o} = âˆ FOB + âˆ FOD

By substituting the values

x^{o} = 125^{o} + 155^{o}

By addition

x^{o} = 280^{o}

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 = 180^{o}

By substituting the values

âˆ FEC + 124^{o} = 180^{o}

On further calculation

âˆ FEC = 180^{o} – 124^{o}

By subtraction

âˆ FEC = 56^{o}

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 = 180^{o}

By substituting the values

116^{o} + âˆ FEA = 180^{o}

On further calculation

âˆ FEA = 180^{o} – 116^{o}

By subtraction

âˆ FEA = 64^{o}

In order to find the value of x

x^{o} = âˆ FEA + âˆ FEC

By substituting the values

x^{o} = 64^{o} + 56^{o}

By addition

x^{o} = 120^{o}

Therefore, the value of x is 120^{o}

**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 = 20^{o}

It can also be written as

âˆ DCG = âˆ DCE – âˆ GCE

By substituting the values we get

âˆ DCG = 130^{o} – 20^{o}

By subtraction we get

âˆ DCG = 110^{o}

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 = 110^{o}

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 = 110^{o}

âˆ FAE = x = 110^{o}

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 = 75^{o}

It can be written as

âˆ EFQ = 75^{o}

Where

âˆ EFG + âˆ GFQ = 75^{o}

By substituting the values

25^{o} + y^{o} = 75^{o}

On further calculation

y^{o} = 75^{o} – 25^{o}

By subtraction

y^{o} = 50^{o}

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

So we get

âˆ BEF + âˆ EFQ = 180^{o}

By substituting the values

âˆ BEF + 75^{o} = 180^{o}

On further calculation

âˆ BEF = 180^{o} – 75^{o}

By subtraction

âˆ BEF = 105^{o}

We know that âˆ BEF can be written as

âˆ BEF = âˆ FEG + âˆ GEB

105^{o} = âˆ FEG + 20^{o}

On further calculation

âˆ FEG = 105^{o} – 20^{o}

By subtraction

âˆ FEG = 85^{o}

According to the â–³ EFG

We can write

x^{o} + 25^{o} + âˆ FEG = 180^{o}

By substituting the values

x^{o} + 25^{o} + 85^{o} = 180^{o}

On further calculation

x^{o} = 180^{o} – 25^{o} – 85^{o}

By subtraction

x^{o} = 180^{o} â€“ 110^{o}

x^{o} = 70^{o}

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 = 180^{o}

By substituting the values

75^{o} + âˆ ACD = 180^{o}

On further calculation

âˆ ACD = 180^{o} – 75^{o}

By subtraction

âˆ ACD = 105^{o}

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

So we get

âˆ ECF = âˆ ACD = 105^{o}

According to the â–³ CEF

We can write

âˆ ECF + âˆ CEF + âˆ EFC = 180^{o}

By substituting the values

105^{o} + x^{o} + 30^{o} = 180^{o}

On further calculation

x^{o} = 180^{o} – 105^{o} – 30^{o}

By subtraction

x^{o} = 180^{o} â€“ 135^{o}

x^{o} = 45^{o}

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 = 85^{o}

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 = 180^{o}

By substituting the values we get

85^{o} + âˆ QGH = 180^{o}

On further calculation

âˆ QGH = 180^{o} – 85^{o}

By subtraction

âˆ QGH = 95^{o}

We can also find the âˆ GHQ

âˆ GHQ + âˆ CHQ = 180^{o}

By substituting the values

âˆ GHQ + 115^{o} = 180^{o}

On further calculation

âˆ GHQ = 180^{o} – 115^{o}

By subtraction

âˆ GHQ = 65^{o}

According to the â–³ GHQ

We can write

âˆ GQH + âˆ GHQ + âˆ QGH = 180^{o}

By substituting the values

x^{o} + 65^{o} + 95^{o} = 180^{o}

On further calculation

x^{o} = 180^{o} – 65^{o} – 95^{o}

By subtraction

x^{o} = 180^{o} â€“ 160^{o}

x^{o} = 20^{o}

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

From the figure we know that âˆ BAD = âˆ ADC

So we get

z = 75

According to the â–³ ABO

We can write

âˆ ABO + âˆ BAO + âˆ BOA = 180^{o}

By substituting the values

x^{o} + 75^{o} + y^{o} = 180^{o}

35^{o} + 75^{o} + y^{o} = 180^{o}

On further calculation

y^{o} = 180^{o} – 35^{o} – 75^{o}

By subtraction

y^{o} = 180^{o} â€“ 110^{o}

y^{o} = 70^{o}

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 = 180^{o}â€¦.. (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 = 180^{o}

According to the diagram the above equation can be written as

âˆ DCE + (âˆ AEC +âˆ AEF) = 180^{o}

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

So we get

âˆ DCE + âˆ AEC + 180^{o} – âˆ BAE = 180^{o}

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 = r^{o} 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 = 180^{o}

By substituting the values we get

p^{o} + âˆ KFE = 180^{o}

So we get

âˆ KFE = 180^{o} – p^{o} â€¦â€¦. (2)

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

âˆ KFG + âˆ KFE = 180^{o} – p^{o} + r^{o}

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

âˆ EFG = 180^{o} – p^{o} + r^{o}

We know that âˆ EFG = q^{o}

q^{o} = 180^{o} – p^{o} + r^{o}

It can be written as

p + q â€“ r = 180^{o}

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

**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 = x^{o} = 60^{o} 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 = 60^{o}

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 = 110^{o}

60^{o} + âˆ QRD = 110^{o}

On further calculation

âˆ QRD = 110^{o} – 60^{o}

By subtraction

âˆ QRD = 50^{o}

According to the â–³ QRS

We can write

âˆ QRD + âˆ QSR + âˆ RQS = 180^{o}

By substituting the values

âˆ QRD + t^{o} + y^{o} = 180^{o}

50^{o} + t^{o} + 60^{o} = 180^{o}

On further calculation

t^{o} = 180^{o} – 50^{o} – 60^{o}

By subtraction

t^{o} = 180^{o} â€“ 110^{o}

t^{o} = 70^{o}

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

From the figure we know that z^{o} and t^{o} are alternate angles

So we get

z^{o} = t^{o} = 70^{o}

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

**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 = 90 ^{o}.**

**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 = 180^{o}

Dividing the entire equation by 2 we get

(1/2)âˆ BEF + (1/2) âˆ DFE = 90^{o}

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

âˆ GEF + âˆ GFE = 90^{o} â€¦â€¦. (1)

According to the â–³ GEF

We can write

âˆ GEF + âˆ GFE + âˆ EGF = 180^{o}

Based on equation (1) we get

90^{o} + âˆ EGF = 180^{o}

On further calculation

âˆ EGF = 180^{o} – 90^{o}

By subtraction

âˆ EGF = 90^{o}

Therefore, it is proved that âˆ EGF = 90^{o}

**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 = 180 ^{o}.**

**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 = 180^{o}

âˆ ABZ can also be written as âˆ ABC

âˆ ABC + âˆ EZB = 180^{o} 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 = 180^{o}

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

**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 = 90^{o}

So we can write it as

âˆ APB = âˆ PAB + âˆ PBA

By substituting the values

90^{o} = 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 = 90^{o}

We can write it as

âˆ 1 + âˆ 2 + âˆ 3 + âˆ 4 = 180^{o}

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

âˆ CAB + âˆ ABD = 180^{o}

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 = 110^{o}

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 = 90^{o}

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 = 90^{o}

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.

### Access other exercise solutions of Class 9 Maths Chapter 7: Lines and Angles

Exercise 7A Solutions 12 Questions

Exercise 7B Solutions 16 Questions

### 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.