NCERT Solutions for Exercise 11.1 Class 9 Maths Chapter 11

NCERT Solutions for Exercise 11.1 Class 9 Maths Chapter 11 - Constructions

Edited By Vishal kumar | Updated on Oct 10, 2023 01:36 PM IST

NCERT Solutions for Class 9 Maths Chapter 11: Constructions Exercise 11.1- Download Free PDF

NCERT Solutions for Class 9 Maths Chapter 11: Constructions Exercise 11.1- In Chapter 9th class maths exercise 11.1 answers Maths, titled "Constructions," you'll learn how to design perfect geometric shapes with only a compass and a straightedge. This amazing journey begins with ex 11.1 class 9 It assigns you tasks that need you to precisely produce various types of angles.

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NCERT Solutions for Class 9 Maths Chapter 11: Constructions Exercise 11.1- Download Free PDF
Download PDF of NCERT Solutions for Class 9 Maths Chapter 11 – Constructions Exercise 11.1
Access Constructions Class 9 Maths Chapter 11 Excercise: 11.1
Q5 Construct an equilateral triangle, given its side and justify the construction. Answer: The following steps to make an equilateral triangle: Step 1: Draw a line segment AB = 4 cm. Step 2: With A and B as centres, make two arcs in the line segment AB. Mark it as D and E respectively. Step 3: Now, with D and E as centres, make the two arcs cutting the previous arcs respectively, and forming an angle of each. Step 4: Extend the lines of A and B until they intersect each other at point C. Hence, triangle constructed is ABC which is equilateral. Now, Justification ; Since the angles constructed are of each, so the third angle will also be .
More About NCERT Solutions for Class 9 Maths Exercise 11.1
Benefits of NCERT Solutions for Class 9 Maths Exercise 11.1
key Features of Class 9 Maths Chapter 11 Exercise 11.1
NCERT Solutions of Class 10 Subject Wise
NCERT Solutions for Class 9 Maths

NCERT Solutions for Exercise 11.1 Class 9 Maths Chapter 11 - Constructions

But don't worry, because the NCERT Solutions for this class 9 maths chapter 11 exercise 11.1 serve as helpful guides. They are created by professionals and include detailed, step-by-step instructions for each construction. Plus, you can get them for free in PDF format, so you can use them even if you don't have access to the internet. So, let us begin confidently mastering the art of geometry creations!

Along with Class 9 Maths chapter 11 exercise 11.6 the following exercise is also present.

Constructions Exercise 11.2

** According to the revised NCERT Syllabus 2023-24 update, this chapter has been removed.

Download PDF of NCERT Solutions for Class 9 Maths Chapter 11 – Constructions Exercise 11.1

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Access Constructions Class 9 Maths Chapter 11 Excercise: 11.1

Q1 Construct an angle of 90 ^o at the initial point of a given ray and justify the construction.

Answer:

The steps of construction to follow:

Step 1: Draw a ray OP.

Then, take O as the centre and any radius draw an arc cutting OP at Q.

1640600746099

Step 2: Now, taking Q as the centre and with the same radius as before draw an arc cutting the previous arc at R. Repeat the process with R to cut the previous arc at S.

1640600957166

Step 3: Take R and S as centre draw the arc of radius more than the half of RS and draw two arcs intersecting at A. Then, join OA.

1640600972011

Hence, $\angle POA = 90^{\circ }$ .

Justification:

We need to justify, $\angle POA = 90^{\circ }$

So, join OR and OS and RQ. we obtain

1640600984472

By construction OQ = OS = QR.

So, $\triangle ROQ$ is an equilateral triangle. Similarly $\triangle SOR$ is an equilateral triangle.

So, $\angle SOR = 60^{\circ}$

Now, $\angle ROQ = 60^{\circ}$ that means $\angle ROP = 60^{\circ}$ .

Then, join AS and AR:

1640601001877

Now, in triangles OSA and ORA:

$SR = SR$ (common)

$AS = AR$ (Radii of same arcs)

$OS = OR$ (radii of the same arcs)

So, $\angle SOA = \angle ROA = \frac{1}{2}\angle SOR$

Therefore, $\angle ROA = 30^{\circ}$

and $\angle POA = \angle ROA+\angle POR = 30^{\circ} +60^{\circ} =90^{\circ}$

Hence, justified.

Q2 Construct an angle of 45 ^o at the initial point of a given ray and justify the construction.

Answer:

The steps of construction to follow:

Step 1: Draw a ray OY.

Then, take O as the centre and any radius, mark a point A on the arc ABC.

1640601062706

Step 2: Now, taking A as the centre and the same radius, mark a point B on the arc ABC.

1640601050776

Step 3: Take B as a centre and the same radius, mark a point C on the arc ABC.

1640601072959

Step 4: Now, taking C and B as centre one by one, draw an arc from each centre intersecting each other at a point X.

1640601086817

Step 5: X and O are joined and a ray making an angle $90^{\circ}$ with OY is formed.

Let the arc AC touches OX at E

Step 6: With A and E as centres, 2 arcs are marked intersecting each other at D and the bisector of angle XOY is drawn.

1640601100159

Justification:

By construction we have,

$\angle XOY = 90^{\circ}$

We constructed the bisector of $\angle XOY$ as $\angle DOY$

Thus,

$\angle DOY = \frac{1}{2}\angle XOY = \frac{1}{2}\times90^{\circ} = 45^{\circ}$

Q3 (i) Construct the angles of the following measurements: 30 ^o

Answer:

Steps to construction to follow:

Step 1: Draw a ray OY.

1640601126922

Step 2: Now, take A as a center and take any radius, then draw an arc AB cutting OY at A.

1640601139457

Step 3: Take A and B as centres, draw 2 arcs are marked intersecting each other at X and hence, the bisector of $30^{\circ}$ is constructed.

1640601151573

Thus, $\angle XOY$ is the required angle.

Q3.Construct the angles of the following measurements: (i) 30o
Edit Q

Q3 (ii) Construct the angles of the following measurements: $22\frac{1}{2}\degree$

Answer:

Steps to construction to follow:

Step 1: Draw a ray OY.

1640601174811

Step 2: Now, take A as a centre and take any radius, then mark a point B on the arc

1640601184856

Step 3: Take B as a centre with the same radius, mark a point C on the arc ABC.

1640601195448

Step 4: Now, taking B and C as centres simultaneously, then draw an arc from each centre intersecting each other at a point X. Then join X and O and a ray making an angle with OY is formed.

1640601206833

Let the arc AC touches OX at E

Step 5: Now, with A and E as centres, mark 2 arcs which intersect each other at D and we obtain the bisector of the angle $XOY$ .

1640601218774

1640601229663

Q3 (iii) Construct the angles of the following measurements: 15 ^o

Answer:

Steps to construction to follow:

Step 1: Draw a ray OY.

1640601261458

Step 2: Now, take A as a centre and take any radius, draw an arc AB which cuts OY at A.

1640601272131

Step 3: Take A and B as a centre, then mark 2 arcs which intersect each other at X and Hence, the bisector is constructed of $30^{\circ}$ .

Step 4: Now, with A and E as centres, mark 2 arcs which intersect each other at D and we obtain the bisector of the angle $XOY$ .

1640601288755

1640601299418

Thus, the angle of $15^{\circ}$ is obtained which is $\angle EOY.$

Q4 (i) Construct the following angles and verify by measuring them by a protractor: 75 ^o

Answer:

Steps to construction to follow:

1640601336495

Step 1: Draw a ray OY.

Step 2: Now, taking O as the centre draw an arc ABC.

Step 3: On taking A as a centre, draw two arcs B and C on the arc ABC.

Step 4: Now, taking B and C as centres, arcs are made to intersect at point E and the $\angle EOY = 90^{\circ}$ is constructed.

Step 5: Taking A and C as centres, arcs are made to intersect at D.

Step 6: Now, join OD and hence, $\angle DOY = 75^{\circ}$ is constructed.

Hence, $\angle DOY$ is the required angle.

Q4 (ii) Construct the following angles and verify by measuring them by a protractor: 105 ^o

Answer:

The steps of construction to be followed:

1640601361664

Step 1: Draw a ray OY.

Step 2: Then, taking O as a centre, draw an arc ABC.

Step 3: Now, with A as a centre, draw two arcs B and C which are made on the arc ABC.

Step 4: Taking B and C as centres simultaneously, arcs are made to intersect at E and $\angle EOY =90^{\circ}$ is constructed.

Step 5: With B and C as centres, arcs are made to intersect at X.

Step 6: Join the OX and we get $\angle XOY =105^{\circ}$ is constructed.

Thus, the angle $XOY$ is $105^{\circ}.$

Q4 (iii) Construct the following angles and verify by measuring them by a protractor: 135 ^o

Answer:

The steps of construction to be followed:

1640601385098

Step 1: Draw a ray DY.

Step 2: Draw an arc ACD with O as a center.

Step 3: Now, with A as a centre, draw two arcs B and C on the arc ACD.

Step 4: Taking B and C as centres, arcs are made to intersect at E and the angle formed is $\angle EOY = 90^{\circ}$ .

Step 5: Take F and D as centres, draw arcs to intersect at point X or the bisector of angle EOD is made.

Step 6: Join OX and the $\angle XOY = 135^{\circ}$ is made.

Hence, the angle required $\angle XOY$ is $135^{\circ}$ .

Q5 Construct an equilateral triangle, given its side and justify the construction.
Answer:
The following steps to make an equilateral triangle:

Step 1: Draw a line segment AB = 4 cm.

Step 2: With A and B as centres, make two arcs in the line segment AB. Mark it as D and E respectively.

Step 3: Now, with D and E as centres, make the two arcs cutting the previous arcs respectively, and forming an angle of $60^{\circ }$ each.
Step 4: Extend the lines of A and B until they intersect each other at point C.

Hence, triangle constructed is ABC which is equilateral.

Now, Justification ;
Since the angles constructed are of $60^{\circ }$ each, so the third angle will also be $60^{\circ }$ .
$\left [ \because The\ sum\ of\ all\ angles\ of\ triangle = 180^{\circ} \right ]$

This chapter and NCERT Solutions for Class 9 Maths exercise 11.1 are majorly based on the construction like construction of angle, side bisectors and angle bisector.

NCERT solutions for exercise 11.1 Class 9 Maths chapter 11 provides us the basic core concepts which includes the construction of triangles parallelograms circles and many different complex figures which are majorly from standard 10 perspectives.

In order to drawn an angle bisector as described in NCERT solutions for Class 9 Maths chapter 11 exercise 11.1, we first draw an arc that cuts both the sides of the angle then we draw two arcs of keeping their radius larger. Thus, then we join the point where these two new arcs meet to the point of the centre of the angle this line formed is called as the angle bisector.

Figure below shows the formation of an angle bisector by following the steps from exercise 11.1 Class 9 Maths

1640605777434

We follow simply this type of procedure in order to get the side bisector also

Benefits of NCERT Solutions for Class 9 Maths Exercise 11.1

Class 9 Maths chapter 11 exercise 11.1, is based construction and most of the crucial implication of construction.
From Class 9 Maths chapter 11 exercise 11.1 we get to learn the construction of general angles.
Understanding the concepts from NCERT solutions for Class 9 Maths exercise 11.1 will make the ideas and question from better expectations (like Class 10) simpler for us.

key Features of Class 9 Maths Chapter 11 Exercise 11.1

Step-by-Step Solutions: NCERT Solutions for class 9 maths ex 11.1 step-by-step instructions for each construction problem. This helps students follow the process with ease.
Clear and Concise: 9th class maths exercise 11.1 answers are written in clear and concise language, making it accessible for students of all levels.
Accessible PDF Format: The ex 11.1 class 9 solutions are available for free download in PDF format, allowing students to access them offline and at their convenience.
Concept Reinforcement: This class 9 ex 11.1 reinforces students' understanding of geometrical concepts and helps develop their construction skills, fostering attention to detail.
Preparation for Advanced Topics: The constructions taught in this chapter are foundational for more advanced topics in geometry that students will encounter in higher classes.

Also, See

NCERT Solutions of Class 10 Subject Wise

Frequently Asked Questions (FAQs)

1. What is the core concept for NCERT solutions for Class 9 maths exercise 11.1?

This Exercise deals with the general construction of side bisector and angle bisector along with the making of some general angles.

2. How can you define an angle bisector?

An angle bisector is a line or beam that separates an angle into two harmonious or equivalent points.

3. What is a side bisector?

A side bisector is a line that divides the line segment into congruent parts. It makes an angle of 90° with the line that is being bisected.

4. How many side bisectors are possible for a triangle?

A triangle has three side bisectors, each one for its sides.

5. Can we get 45-degree angle if we are given an angle of 90 degrees?

Yes, we can get 45 degrees if we just bisect the right angle (90 degree) into two equal halves.

6. For a given angle how many bisectors are possible?

Two bisectors are possible for a given angle. Interior and exterior angle bisector.

7. For a given line segment how many bisectors are possible?

A line segment has a maximum of only one side bisector unlike that of an angle.

8. Can we get an angle of 15 degrees from a given angle of 60 degree ?

Yes, we can get an angle of 15 degrees from the given angle of 60 Degree. First, we have to bisect the 60 degree angle into two 30 degrees then further bisect the 30 degree angle into two parts, thus we will end up with a 15 degree angle.

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