NCERT Solutions For Class 9 Maths Chapter 11 Constructions

NCERT Solutions for Class 9 Maths Chapter 11 Constructions- The team of subject matter experts at GFG have curated NCERT Solutions for Class 9 Maths Chapter 11 Constructions to make sure that every student can understand how to solve Construction problems in a stepwise manner. This article provides solutions to all the problems asked in Class 9 Maths Chapter 11 Constructions of your NCERT textbook in a step-by-step manner as per the latest CBSE Syllabus 2023-24 and guidelines.

NCERT Class 9 Maths Chapter 11 Constructions is a foundational chapter in the learning of constructions. Understanding the methods for creating various types of triangles for various criteria has been made easier by the detailed NCERT Solutions For Class 9 Maths Chapter 11 Constructions provided in this article.

NCERT Class 9 Maths Chapter 11 Constructions covers topics such as:

• Basic Constructions
• Construction of Triangles

Solutions to all the exercises in the NCERT Class 9 Maths Chapter 11 Constructions have been collectively covered in NCERT Solutions for Class 9 Maths. They are regularly revised to check errors and updated according to the latest CBSE Syllabus 2023-24 and guidelines.

Class 9 Maths NCERT Solutions Chapter 11 Constructions Exercises
Class 9 NCERT Solutions- Chapter 11 Constructions – Exercise 11.1 – 5 Questions  (2 Short Answers, 2 Long Answer, 1 Very Long Answer)
Class 9 NCERT Solutions- Chapter 11 Constructions – Exercise 11.2 – 5 Questions (5 Very Long Answers)

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

Question 1. Construct an angle of 90° at the initial point of a given ray and justify the construction.

Solution:

Steps of construction

1. Take a ray with initial point A.
2. Taking care center and same radius draw an Arc of a circle which intersect AB at C.
3. With C as Centre and the same radius, draw an arc intersecting the previous arc at E.
4. With E as Centre and the same radius draw an arc which intersects the arc drawn in step 2 at F.
5. With E as Centre and the same radius, draw another arc, intersecting the previous arc at G.
6. Draw the ray AG.
7. Then ∠BAG is the required angle 90°

Justification:

Join AE, CE, EF and AE, AF

AC = CE = AE      [ by construction]

∴ ACE is an equilateral Triangle

⇒ ∠CAE = 60°    —————–1

Similarly, AE = EF = AF

∴Triangle AEF is an equilateral Triangle

⇒ ∠EAF = 60°

Because AG bisects ⇒ ∠EAF

∴∠GAE = 1\2 = 30° = 30°————2

1+2

∴∠CAE + ∠GAE = 60°+30°

∠GAB=30°

Question 2. Construct an angle of 45° at the initial point of a given ray and justify the construction.

Solution:

Step of Construction:

1. Take a ray AB with initial point A
2. Draw ∠BAF=90°
3. Taking C as Centre and radius more than   draw an arc.
4. Taking G as Centre and the same radius as before, draw another arc.
5. Taking G as Centre and the same radius as before, draw another arc. Intersecting previous arc at H.
6. Draw the ray AH.
7. Then ∠BAH is the required angle of 45°

Justification:

Join GH and HC    (construct)

In ∆ AHG and ∆ AHC

HG=HC…………….[arc of equal radii]

AG=AC……………..[radii of same arc]

AH=AH………………[common]

AHG≅AHC             [S.S.S]

∠HAG=∠HAC                [C.P.C.T]

But ∠HAG+∠HAC=90

∠HAG=∠HAC=90\2=45

∴∠BAH=45

Question 3. Draw the angles of the following measurement

i) 30°

Solution:

Step of construction

1. Draw a ray AB with initial point A.
2. With A as centre, draw an arc intersecting AB at c.
3. With c as centre and the same radius, draw another arc, intersecting the previously drawn arc at D.
4. Taking C and D as centre  and with the radius more than 1\2 DC draws arcs to intersect each other at E.
5. Draw ray AE. ∠EAB is the required angle of 30.

ii) 22 ½°

Solution:

Steps of construction

1. Take a ray AB
2. Draw an angle ∠AB=90° on point A.
3. Bisect ∠CAB and draw ∠DAB=45°
4. Bisect ∠DAB and draw ∠EAB
5. ∠EAB is required angle of 22 ½°

iii) 15°

Solution:

Steps of construction

1. Take a ray AB.
2. Draw an arc on AB, by taking A a center, which intersect AB at c.
3. From C with the same radius draw another re which intersect the previous  arc at D.
4. Join DA.
5. ∠DAB =60°
6. Bisect  ∠DAB and draw angle  EAB=30°
7. Bisect ∠EAB and draw ∠FAB
8. ∠FAB is the required angle.

Question 4. Construct the following angles and verify by measuring them by a protractor

(i) 75°

Solution:

Steps of construction

1. Draw a ray AB with initial point  A.
2. At point A draw an angle ∠CAB=90°
3. At point A draw ∠DAB=60°
4. Bisect ∠CAD, now ∠EAD=15°
5. ∠EAB=75°        {∠EAB=∠EAD+∠DAB=15°+60°=75°}

(ii) 105°

Solution:

Steps of construction

1. Draw a ray AB with initial point  A.
2. At point A draw an angle ∠CAB=90°
3. At point A draw ∠DAB=120°
4. Bisect ∠CAD, now ∠EAD=15°
5. ∠EAB=75°        {∠EAB=∠EAC+∠CAB=15°+90°=105°}

(iii) 135°

Solution:

Steps of construction

1. Draw a ray AB with initial point A.
2. At point A draw an angle ∠CAB=120°
3. At point A draw ∠DAB=150°
4. Bisect ∠CAD, now ∠EAC=15°
5. ∠EAB=135°        {∠EAB=∠EAC+∠CAB=15°+120°=135°}

Question 5. Construct an equilateral triangle, given its side and justify the construction.

Solution:

Steps of construction

1. Draw a line segment of AB of a given length.
2. With A and B as centre and radius equal to AB draw arcs to intersect each other at c.
3. Join AC and BC.

Then ABC is the required equilateral triangle.

Justification:

AB=AC    ……………. [by construction]

AB=BC    ……………..[by construction]

AB=AC=BC

Hence, ∆ABC is required equilateral triangle.

NCERT Solutions for Class 9 Maths Chapter 11 Constructions: Exercise 11.2

Question 1. Construct a ∆ ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm.

Solution:

Steps of construction:

1. Draw a line segment BC base of cm is drawn.
2. At point B draw an angle of 75°.
3. Cut BD =13cm from BY.
4. Join ∠D which intersect BD at A.
5. Join AC. Now triangle ABC is the required triangle

Question 2. Construct a ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 35 cm.

Solution:

Steps of construction:

1. Draw  a line segment BC=8cm.
2. At point B, draw angle 45°.
3. Cut BD=3.5 from BY.
4. Join CD.
5. Draw perpendicular bisector of CD, which construct BY at A.
6. Join AC. NOW, ABC is the required triangle.

Question 3. Construct a ∆ ABC in which QR = 6 cm, ∠Q = 60° and PR – PQ = 2 cm.

Solution:

Steps of construction:

1. Draw a line segment QR=6cm.
2. At point Q draw angle 60°.
3. Extend PQ to Y’.
4. Cut QS =2cm from QY’.
5. Join RS.
6. Draw perpendicular bisector of RS which intersect QY at P.
7. Join PR. Now, PQR is the required triangle.

Question 4. Construct a ∆ XYZ in which ∠Y = 30°, ∠Y = 90° and XY + YZ + ZX = 11 cm.

Solution:

Steps of construction:

1. Draw a line segment AB=11cm.
2. At point A draw ∠BAP=30°.
3. At point B draw angle 90°.
4. Draw the bisector of ∠BAP and ∠ABR which intersect each other at X.
5. Join AX and BX.
6. Draw perpendicular bisector of AX and BX which intersect AB on Y and Z respectively.
7. Join XY and XZ. Then XYZ is the required triangle.

Question 5. Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.

Solution:

Steps of construction:

1. Draw a line segment of BC=12cm.
2. At point B draw angle b=90°
3. Cut BD =18cm.
4. Join CD.
5. Draw perpendicular bisector of CD which intersect BD at point A.
6. Join AC. Now ABC is the required triangle.

Key Features of NCERT Solutions for Class 9 Maths Chapter 11 Constructions

• NCERT Solutions provides accurate and complete solutions for problems given in NCERT textbooks.
• NCERT solutions are developed by the GfG team, with a focus on students’ benefit and for all the chapters of class 9 including Construction.
• These solutions are very accurate and comprehensive, which can help students prepare for any academic as well as competitive exam.

FAQs on NCERT Solutions for Class 9 Maths Chapter 11 Constructions

1. Why is it important to learn constructions?

Constructions involve using a compass and straightedge to create accurate geometric figures, such as perpendicular lines, angles, triangles, and bisectors. By learning constructions, students learn to visualize and manipulate geometric shapes, improving their problem-solving abilities. Additionally, constructions serve as a foundation for more advanced geometry topics and practical applications in fields like architecture, engineering, and design.

2. What topics are covered in NCERT Solutions for Class 9 Maths Chapter 11 Constructions?

NCERT Solutions for Class 9 Maths Chapter 11-Constructions includes such as constructions of angle bisectors and triangles under given criteria.

3. How can NCERT Solutions for Class 9 Maths Chapter 11 Constructions help me?

NCERT Solutions for Class 9 Maths Chapter 11-Constructions can help you solve the NCERT exercise without any limitations. If you are stuck on a problem, you can find its solution in these solutions and free yourself from the frustration of being stuck on some question.

4. How many exercises are there in NCERT Solutions for Class 9 Maths Chapter 11 Constructions?

There are 2 exercises in the Class 9 Maths Chapter 11-Constructions which covers all the important topics and sub-topics.

5. Where can I find NCERT Solutions for Class 9 Maths Chapter 11 Constructions?

You can find these NCERT Solutions in this article created by our team of experts at GeeksforGeeks.