# MSBSHSE Solutions For Class 8 Maths Part 1 Chapter 2- Parallel Lines and Transversals

MSBSHSE Solutions For Class 8 Maths Part 1 Chapter 2- Parallel Lines and Transversals are available here. This chapter mainly deals with transversal, angles made by a transversal, corresponding angles, interior angles, alternate angles, properties of angles formed by two parallel lines and transversal, property of corresponding angles, property of alternate angles, property of interior angles. The subject experts at BYJUâ€™S outline the concepts in a clear and precise manner, based on the IQ level of students. Our solution module utilises numerous shortcut tips and practical examples to explain all the exercise questions in a simple and easily understandable language. If you wish to obtain an excellent score, solving Maharashtra Board Solutions for Class 8 is a must.

Here, the students will learn various techniques about the concepts. The solutions to all questions in Maharashtra State Board Class 8 Textbook Part 1 are given here in a detailed and step by step way to help the students understand more effectively. Students can download the solutions of Maharashtra Board from the provided links.

## Download the PDF of Maharashtra Board Solutions For Class 8 Maths Part 1 Chapter 2- Parallel Lines and Transversals

### Access answers to Maharashtra Board Solutions For Class 8 Maths Part 1 Chapter 2- Parallel Lines and Transversals

Practice set 2.1 PAGE NO: 08

1. In the adjoining figure, each angle is shown by a letter. Fill in the boxes with the help of the figure.

Solution:

From the above given figure we can say that,

For corresponding angles:

(1) ForÂ âˆ p

âˆ w is the angle which is on the same side and same direction of transversal.

Hence,Â âˆ wÂ is the corresponding angle toÂ âˆ p.

(2) ForÂ âˆ q

âˆ x is the angle which is on the same side and same direction of transversal.

Hence,Â âˆ xÂ is the corresponding angle toÂ âˆ q.

(3) ForÂ âˆ r

âˆ y is the angle which is on the same side and same direction of transversal.

Hence, âˆ r is the corresponding angle toÂ âˆ y.

(4) ForÂ âˆ s

âˆ z is the angle which is on the same side and same direction of transversal.

Hence,Â âˆ s is the corresponding angle toÂ âˆ z.

For Interior alternate angles:

(5) ForÂ âˆ s

The angel which is in the inner side as well as on the opposite side of transversal and its opposite angle isÂ âˆ x.

Hence,Â âˆ s andÂ âˆ x form pair of Interior Alternate angel.

(6) ForÂ âˆ w

The angel which is in the inner side as well as on the opposite side of transversal and its opposite angle isÂ âˆ r.

Hence,Â âˆ w andÂ âˆ r form pair of Interior Alternate angel.

2. Observe the angles shown in the figure and write the following pair of angles.

(1) Interior alternate angles
(2) Corresponding angles
(3) Interior angles

Solution:

(1) For Interior alternate angles:

When these angels are in the inner side they are called Interior alternateÂ angels.

• ForÂ âˆ b, âˆ b andÂ âˆ h form pair of Interior Alternate angel.
• ForÂ âˆ c, âˆ cÂ andÂ âˆ e form pair of Interior Alternate angel.

(2) For Corresponding angles

If the arms on the transversal of a pair of angles are in the same direction and the other arms are on the same side of the transversal, then it is called a pair of corresponding angles.

• ForÂ âˆ a, âˆ a is the corresponding angle toÂ âˆ e.
• ForÂ âˆ b, âˆ bÂ is the corresponding angle toÂ âˆ f.
• ForÂ âˆ d,Â âˆ d is the corresponding angle toÂ âˆ h.
• ForÂ âˆ c, âˆ c is the corresponding angle toÂ âˆ g.

(3) Interior angles

A pair of angles which are on the same side of the transversal and inside the given lines is called a pair of interior angles.

• For âˆ b, âˆ b andÂ âˆ e form pair of interior angels.
• For âˆ c, âˆ cÂ andÂ âˆ h form pair of interior angels.

Practice set 2.2 PAGE NO: 11

1. Choose the correct alternative.

(1) In the adjoining figure, if lineÂ mÂ || lineÂ nÂ and lineÂ pÂ is a transversal then findÂ x.

A. 135Â°
B. 90Â°
C. 45Â°
D. 40Â°

Solution:

From the given figure we have 3x and x. 3x and x form a pair of interior angle.

By using the property of interior angels. [We know that, each pair of interior angles formed by two parallel lines and their transversal is of supplementary angles i.e. 180Â°.]

x + 3x = 180

4x = 180

x = 180/4

= 45o

âˆ´Â The value of x is 45Â°.

(2) In the adjoining figure, if lineÂ aÂ || lineÂ bÂ and lineÂ lÂ is a transversal then findÂ x.

A. 90Â°
B. 60Â°
C. 45Â°
D. 30Â°

Solution:

From the figure we have 4x and 2x. 4x and 2x form a pair of interior angle.

By using the property of interior angels

4x + 2x = 180Â°

6x = 180Â°

x = 180/6

= 30o

âˆ´Â The value of x is 30o.

2. In the adjoining figure lineÂ pÂ || lineÂ q. LineÂ tÂ and lineÂ sÂ are transversals. Find the measure ofÂ âˆ x andÂ âˆ y using the measures of angles given in the figure.

Solution:

Given:

LineÂ pÂ || lineÂ q, lineÂ tÂ and lineÂ sÂ are transversals.

Let us find the measure ofÂ âˆ x andÂ âˆ y.

Firstly, Let us consider âˆ z as shown in figure.

Measure of âˆ z = 40Â° â€¦ (i) [Since, they are corresponding angles]

So,

mâˆ x + mâˆ z = 180Â° [Since, angles are in a linear pair]

mâˆ x + 40o = 180Â° [From equation (i)]

mâˆ x= 180Â° â€“ 40Â°

mâˆ x = 140Â°

Now, let us consider âˆ w as shown in the figure.

mâˆ w + 70Â° = 180Â° [Since, angles are in a linear pair]

mâˆ w = 180Â° â€“ 70Â°

mâˆ w = 110Â° â€¦(ii)

It is given that, line p || line q and line s is a transversal.

So, mâˆ y = mâˆ w [by using alternate angles]

mâˆ y =110Â° [From equation (ii)]

âˆ´ The measure of âˆ x is 140Â° and âˆ y is 110Â°.

3. In the adjoining figure. lineÂ pÂ || lineÂ q. lineÂ lÂ || lineÂ m. Find measures ofÂ âˆ a,Â âˆ b, andÂ âˆ c, using the measures of given angles. Justify your answers.

Solution:

Given:

Line p || line q and line l are transversal.

Line l || line m and line p is a transversal.

Line p || line q and line m is a transversal.

Firstly let us find the measure of âˆ a

Line p || line q and line l are transversal.

So,

mâˆ a + 80Â° = 180Â° [Since, they are interior angles]

mâˆ a= 180Â° â€“ 80Â°

mâˆ a= 100Â°

Now, line l || line m and line p is a transversal.

So, mâˆ c = 80Â° â€¦ (i) [By using exterior alternate angles]

Line p || line q and line m is a transversal.

So, mâˆ b = mâˆ c â€¦ [Since, they are corresponding angles]

mâˆ b = 80Â° â€¦ [From equation (i)]

âˆ´ The measure of âˆ a is 100Â°, mâˆ b is 80Â°, mâˆ c is 80Â°.

4. In the adjoining figure, lineÂ aÂ || lineÂ b. lineÂ lÂ is a transversal. Find the measures ofÂ âˆ x,Â âˆ y,Â âˆ zÂ using the given information.

Solution:

Given:

Line a || line b and line l is a transversal.

So,

mâˆ x = 105Â° â€¦ (i) [Since, it is a corresponding angle]

Now,

mâˆ y = mâˆ x [Since, they are vertically opposite angles]

So, mâˆ y = 105Â° â€¦[From equation (i)]

Now, mâˆ z + 105Â° = 180Â° [Since, angles are in a linear pair]

mâˆ z = 180Â°- 105Â°

mâˆ z = 75Â°

âˆ´ The measure of âˆ x is 105Â°, mâˆ y is 105Â°, mâˆ z is 75Â°.

5. In the adjoining figure, lineÂ pÂ || lineÂ lÂ || lineÂ q. FindÂ âˆ x with the help of the measures given in the figure.

Solution:

Let us mark few points as such,

It is given that,

Line p || line l and line IJ is a transversal.

So,

mâˆ IJN = mâˆ JIH [Since they are alternate angles]

mâˆ IJN = 40Â° â€¦ (i)

Now,

Line l || line q and line MJ is a transversal.

So, mâˆ MJN = mâˆ JMK [Since they are alternate angles]

mâˆ MJN = 30Â° â€¦ (ii)

Now, mâˆ x = mâˆ IJN + mâˆ MJN [By using angle addition property]

We get,

40Â° + 30Â° [From equation (i) and (ii)]

mâˆ x = 70Â°

âˆ´ The measure of âˆ x is 70o.

Practice set 2.3 PAGE NO: 13

1. Draw a lineÂ l. Take a point A outside the line. Through point A draw a line parallel to lineÂ l.

Solution:

Steps to construct:

a) Let us draw a line segment of any length. Mark it as CD.

b) Now from any point say P on that line segment draw a line perpendicular at any distance above and name that point A.

c) Now draw another perpendicular line say E of same length as of AP, and in same direction, name that point as F.

d) Draw a line through those points.

e) This line is parallel to given lineÂ l.

2. Draw a lineÂ l. Take a point T outside the line. Through point T draw a line parallel to lineÂ l.

Solution:

Steps to construct:

a) Let us draw a line segment of any length. Mark it as CD.

b) Now from any point say P on that line segment draw a line perpendicular at any distance above or and name that point T.

c) Now draw another perpendicular line say E of same length as of TP, and in same direction, name that point as F.

d) Draw a line through those points.

e) This line is parallel to given lineÂ l.

3. Draw a lineÂ m. DrawÂ aÂ lineÂ nÂ which is parallel to lineÂ mÂ at a distance ofÂ 4Â cm from it.

Solution:

Steps to construct:

a) Draw lineÂ l.

b) Take two points A and B on the lineÂ l.

c) Draw perpendicular lines above to the lineÂ lÂ from points A and B with a distance of 4cm, and mark that points as P and Q.

d) Join line PQ.

f) Line PQ is a line parallel to the lineÂ lÂ at a distance 4cm. Hence, the required line is obtained.

The lines in the same plane, which do not intersect each other are called parallel lines. If a line intersects two lines in two distinct points, then that line is called transversal of those two lines. To get to know more about the concepts covered in this chapter, students can refer to Parallel Lines and Transversals at BYJUâ€™S.

## Frequently Asked Questions on Maharashtra State Board Solutions for Class 8 Maths Chapter 2 Parallel Lines and Transversals

### Will I have to make any payment to access these Maharashtra State Board Solutions for Class 8 MathsÂ  Chapter 2 Parallel Lines and Transversals?

No, there is no need to pay any money. We have made these solutions available online free for download. Students can access them from our site by entering the login details. Or else, the scrollable PDF is also accessible.Â Those who wish to can also check out the questions and the solutions from our webpage.

### Are these Maharashtra State Board Solutions for Class 8 Maths Chapter 2 Parallel Lines and Transversals helpful?

Yes. These solutions are useful and they set the basis for questions that could get asked in the board exams. Students are also advised to practise these questions first and then refer back to the solutions to analyse their performance and then rectify the mistakes, so that they can avoid making any during the board exams.

### Is Maharashtra Board Class 8 Maths Solutions of Chapter 2 difficult?

No, Maharashtra Board Class 9 Maths Solutions of Chapter 2Â  is not half as difficult as it seems. You can easily understand concepts of the chapter if you refer to this chapterwise solutions . You can also go through your syllabus course structure and unit-wise weightage thoroughly to know how much weightage the chapter will carry for the final exam.