RS Aggarwal Solutions for Class 7 Maths Chapter 14 – Properties of Parallel Lines is the best study material for those students who are finding difficulties in solving problems. Our specialist teachers formulate these exercises to help you with your exam preparation to achieve good marks in Maths. These solutions can help students clear doubts quickly and help in learning the topic effectively. Students who wish to score good marks in Maths practise RS Aggarwal Class 7 Solutions.

These Properties of Parallel Lines solutions are available for download in pdf format and provide solutions to all the questions provided in Class 7 Maths Chapter 14 wherein problems are solved step by step with detailed explanations. Download pdf of Class 7 Chapter 14 in their respective links.

## Download PDF of RS Aggarwal Solutions for Class 7 Maths Chapter 14 – Properties of Parallel lines

### Access answer to chapter 14 – Properties of Parallel Lines

__Exercise 14__

**1. In the given figure, l âˆ¥ m and t is a transversal. If âˆ 5 = 70 ^{o}, find the measure of each of the angles âˆ 1, âˆ 3, âˆ 4 and âˆ 8.**

**Solution:- **

Two parallel lines l and m be cut by a transversal t, forming angles.

It is given that âˆ 5 = 70^{o}.

Now,

âˆ 5 = âˆ 3 = 70^{o} â€¦ [âˆµ alternate interior angles]

âˆ 5 + âˆ 8 = 180^{o} â€¦ [âˆµ linear pair]

70^{o} + âˆ 8 = 180^{o}

âˆ 8 = 180^{o} â€“ 70

âˆ 8 = 110^{o}

âˆ 1 = âˆ 3= 70^{o} â€¦ [âˆµ vertically opposite angles]

âˆ 3 + âˆ 4 = 180^{o} â€¦ [âˆµ linear pair]

70 + âˆ 4 = 180^{o}

âˆ 4 = 180^{o} â€“ 70^{o}

âˆ 4 = 110^{o}

âˆ´ âˆ 1 = 70^{o}, âˆ 3 = 70^{o}, âˆ 4 = 110^{o} and âˆ 8 = 110^{o}.

**2. In the given figure, l âˆ¥ m and t is a transversal. If âˆ 1 and âˆ 2 are in the ratio 5: 7, find the measure of each of the angles âˆ 1, âˆ 2, âˆ 3 and âˆ 8.**

**Solution:- **

Two parallel lines l and m be cut by a transversal t, forming angles.

It is given that âˆ 1: âˆ 2 = 5: 7

Let the measures of angel be 5x and 7x,

Then,

5x + 7x = 180

12x = 180

X = 180/12

X = 15

âˆ´ âˆ 1 = 5x = 5 Ã— 15 = 75^{o}

âˆ 2 = 7x = 7 Ã— 15 = 105^{o}

We know that,

âˆ 2 + âˆ 3 = 180^{o} â€¦ [âˆµ linear pair]

105^{o }+ âˆ 3 = 180^{o}

= âˆ 3 = 180 â€“ 105

= âˆ 3 = 75^{o}

âˆ 3 + âˆ 6 = 180^{o} â€¦ [âˆµ the sum of the consecutive interior angle is 180^{o}]

75^{o} + âˆ 6 = 180^{o}

âˆ 6 = 180 â€“ 75

âˆ 6 = 105^{o}

Now âˆ 6 = âˆ 8 = 105^{o} â€¦ [âˆµ vertically opposite angles are equal]

âˆ´ âˆ 1= 75^{o}, âˆ 2 = 105^{o}, âˆ 3 = 75^{o} and âˆ 8 = 105^{o}.

**3. Two parallel lines l and m are cut by a transversal t. if the interior angles of the same side of t be (2x â€“ 8) ^{o} and (3x – 7)^{o}, find the measures of each of these angles.**

**Solution:-**

Two parallel lines l and m be cut by a transversal t, forming angles.

It is given that the interior angles of the same side of t be (2x â€“ 8)^{o} and (3x – 7)^{o}.

Let the interior angles of the same side of t be,

âˆ 3 = (2x â€“ 8)^{o} and âˆ 4 = (3x â€“ 7)^{o}

We know that the sum of the consecutive interior angle is 180^{o}

Then,

âˆ 3 + âˆ 4 = 180^{o}

= (2x â€“ 8) + (3x â€“ 7) = 180

= 2x â€“ 8 + 3x â€“ 7 = 180

= 5x â€“ 15 = 180

= 5x = 180 + 15

= 5x = 195

= x = 195/5

= x = 39

Now substitute the value x in the given equation to get the âˆ 3 and âˆ 4

âˆ´ âˆ 3 = (2x â€“ 8) = ((2 Ã— 39) â€“ 8) = (78 – 8) = 70^{o}

âˆ 4 = (3x â€“ 7) = ((3 Ã— 39) â€“ 7) = (117 â€“ 7) = 110^{o}

**4. In the given figure, l âˆ¥ m, if s and t be transversals such that s is not parallel to t, find he values of x and y.**

**Solution:-**

Two parallel lines l and m be cut by a transversal lines s and t, forming angles.

From the figure, we have,

âˆ 1 = âˆ 3 = 50^{O} â€¦ [âˆµCorresponding angles]

And,

âˆ 1 + x = 180^{o} â€¦ [âˆµlinear pair]

= x^{o }=180^{o} â€“ 50

= x^{o} = 130^{o}

Now,

âˆ 2 = âˆ 4 = 65^{o} â€¦ [âˆµCorresponding angles]

And,

âˆ 2 + y = 180^{o} â€¦ [âˆµlinear pair]

= 65 + y = 180^{o}

= y = 180 â€“ 65

= y = 115^{o}

âˆ´The value of x =130^{o} and y = 115^{o}

**5. In the figure, âˆ B = 65 ^{o} and âˆ c = 45^{o} in Î” ABC and DAE âˆ¥ BC. If âˆ DAB = x^{o} and âˆ EAC = y^{o}, find the values of x and y.**

**Solution:-**

From the question, we have,

âˆ B = 65^{o}, âˆ c = 45^{o}, DAE âˆ¥ BC

The given lines are parallel,

âˆ B = x^{o} = 65^{o}

âˆ C = y^{o} = 45^{o}

**6. In the adjoining figure, it is given that CE âˆ¥ BA, âˆ BAC = 80 ^{o} and âˆ ECD = 35^{o}.**

**Find (i) âˆ ACE, (ii) âˆ ACB, (iii) âˆ ABC.**

**Solution:-**

From the question,

CEâˆ¥BA, âˆ BAC = 80^{o}, âˆ ECD = 35^{o}.

Now,

(i) âˆ BAC = âˆ ACE = 80^{o} â€¦ [âˆµ Alternate angles]

(ii) âˆ ACB,

= âˆ ACB + âˆ ACD = 180^{o} â€¦ [âˆµ Linear pair]

= âˆ ACB + âˆ ACE + âˆ ECD = 180^{o}

= âˆ ACB + 80 + 35 = 180

= âˆ ACB + 125 = 180

= âˆ ACB = 180 â€“ 115

= âˆ ACB = 65^{o}

(iii) âˆ ABC

Let us consider Î” ABC,

âˆ ABC + âˆ ACB + âˆ BAC = 180^{o}

= âˆ ABC + 65 + 80 = 180

= âˆ ABC + 145 = 180

= âˆ ABC = 180 â€“ 145

= âˆ ABC = 35^{o}

**7. In the adjoining figure, it is being given that AO âˆ¥ CD, OB âˆ¥ CE and âˆ AOB = 50 ^{o}. Find the measure of âˆ ECD.**

**Solution:-**

From the question,

Given that, AO âˆ¥ CD, OB âˆ¥ CE and âˆ AOB = 50^{o}

âˆ AOD = âˆ CDB = 50^{o}

Then,

âˆ ECD + âˆ CDB = 180^{o} â€¦ [âˆµ Consecutive interior angles are supplementary]

âˆ ECD + 50 = 180

âˆ ECD = 180 â€“ 50

âˆ ECD = 130^{o}

## RS Aggarwal Solutions for Class 7 Maths Chapter 14 – Properties of Parallel Lines

Chapter 14 – Properties of Parallel Lines contains 1 exercise and the RS Aggarwal Solutions available on this page provide solutions to the questions present in the exercises. Now, let us have a look at some of the concepts discussed in this chapter.

- Parallel Lines
- Distance Between Parallel Lines
- Transversal
- Angles Formed When a Transversal CutsTwo Parallel Lines

### Chapter Brief of RS Aggarwal Solutions for Class 7 Maths Chapter 14 – Properties of Parallel Lines

RS Aggarwal Solutions for Class 7 Maths Chapter 14 – Properties of Parallel Lines. The chapter covers types of angles forming when a transversal cuts two parallel lines and also properties of angles.The solutions are solved in such a way that students will understand clearly.