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.