NCERT Solutions for Class 9 Maths Chapter 11 Constructions

 

NCERT solutions for class 9 maths chapter 11 Constructions: The chapter deals with the construction of certain angles and also triangles in certain conditions. In the previous chapters, you learnt to draw an approximate figure while writing the solutions. In this chapter, the scenario is different because the main purpose to introduce this chapter in the curriculum is to develop an approach to create an accurate geometric figure using some hints. CBSE NCERT solutions for class 9 maths chapter 11 Constructions has designed to provide you help in creating those accurate figures. Let's understand the kinds of questions which appear in this particular chapter. For example, you are asked to give a sketch of a roof of a triangular shape with a given perimeter and two base angles. The concepts learnt in this particular chapter can be used in the designing such cases.

CBSE NCERT solutions for class 9 maths chapter 11 Constructions has point by point solutions to every constructional problem. When there is any constructional project to start, then the first step to initiate that project is the planning and drawing. Geometrical constructions have their own line of study. If you want to pursue your career in this domain then the higher study related to this can be civil engineering and architecture. In this chapter, there are 2 exercises having total of 14 questions. Solutions of NCERT for class 9 maths chapter 11 Constructions have an in-depth solution to each and every question present in the practice exercises. For 360 degree help in your school examination preparation, you can use NCERT solutions for other chapters, classes, and subjects.

 

NCERT solutions for class 9 maths chapter 11 Constructions Excercise: 11.1

Q1 Construct an angle of 90o 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.

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.

construction

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.

construction

Hence, \angle POA = 90^{\circ }.

Justification:

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

So, join OR and OS and RQ. we obtain

construction 4

 

 

 

 

 

 

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:

constructino 5

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 45o 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.

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

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

step 3

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.

step 4

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.

step 5

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: 30o

Answer:

Steps to construction to follow:

Step 1: Draw a ray OY.

OY ray

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

step 2 aa

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.

step 3 a

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.

OY ray

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

step 2 aa

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

step 4 a

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.

construction 6

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.

construction 7

 

bisector

 


 

 

 

 

 

 

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

Answer:

Steps to construction to follow:

Step 1: Draw a ray OY.

OY ray

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

step 2 aa

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.

bisector xyzaa

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: 75o

Answer:

Steps to construction to follow:

construction 8

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: 105o

Answer:

The steps of construction to be followed:

question2

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: 135o

Answer:

The steps of construction to be followed:

construction abcd

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.

line segment AB

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

cut line segment AB

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.

triangle ABC

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 ]

 

NCERT solutions for class 9 maths chapter 11 Constructions Excercise: 11.2

Q1 Construct a triangle ABC in which BC = 7cm, \angle B = 75\degree and AB + AC = 13 cm.

Answer:

The steps of construction are as follows:

Step 1: Draw a line segment BC of 7cm length. Taking the help of protractor make an \angle XBC = 75^{\circ}.

Step 2: Now, cut a line segment BD having 13 cm on BX which is (AB+AC).

Triangle

Step 3: Now, join CD.

Step 4: Draw a perpendicular bisector of CD to intersect BD at a point A. Join AC. Then ABC is the required triangle

Hence, the required triangle is ABC.

Q2 Construct a triangle ABC in which BC = 8cm, \angle B = 45\degree° and AB - AC = 3.5cm

Answer:

The steps of construction to be followed:

Step 1: Draw a line segment BC = 8cm and make an angle of 45^{\circ} at point B i.e., \angle XBC.

XBC

Step 2: Now, cut the line segment BD = 3.5 cm on ray BX. i.e.(AB-AC).

Step 3: Join CD and draw a perpendicular bisector of CD i.e., PQ.

XCD 45

Step 4: Let the perpendicular bisector of CD intersects BX at point A. Then,

atep 3

Step 5: Join AC, to get the required triangle \triangle ABC.

 

Q3 Construct a triangle PQR in which QR = 6cm, \angle Q = 60 \degree and PR -PQ = 2cm.

Answer:

The steps of construction to be followed:

Step 1: Draw a ray QX and cut off a line segment QR which is equal to 6 cm in length.

Step 2: With an angle of 60^{\circ} with QR, construct a ray QY and extend it to form a line YQY'.

Step 3: Now, cut off a line segment QS equal to 2 cm from QY' and join RS.

                                             Y 

s2

   Y'

 

 

Step 4: Draw a perpendicular bisector of RS which intersects QY at a point P.

Step 5: Join PR to get the required triangle \triangle PQR.

 

Q4 Construct a triangle XYZ in which \angle Y = 30\degree, \angle Z = 90\degree° and XY + YZ + ZX = 11 cm.

Answer:

The steps of construction to be followed:

Step 1: For given XY+YZ+ZX = 11 cm, a line segment PQ =11 cm  is drawn.

Step 2: At points, P and Q angles of \angle RPQ = 30^{\circ} and \angle SQP =90^{\circ} are constructed respectively.

Q31

Step 3: Now, bisects the angle RPQ and SQP. The bisectors of these angles intersect each other at a point X.

Q32

Step 4: Construct the perpendicular bisector of PX and QX, name them as TU and WV respectively.

q33

Step 5: Let the bisector TU intersect PQ at Y and bisector WV intersect PQ at Z. Then XY and ZY are joined.

step aaa

Therefore, \triangle XYZ is the required triangle.

Q5 Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side is 18 cm.

Answer:

The steps of construction to follow:

Step 1: Draw a ray BX and Cut off a line segment BC=12cm from the ray.

Step 2: Now, construct an angle \angle XBY = 90^{\circ}.

Step 3: Cut off a line segment BD of length 18 cm on BY. Then join the CD.

step 3

Step 4: Now, construct a perpendicular bisector of CD which intersects BD at A and AC is joined.

abct

Thus, the constructed triangle is ABC.

NCERT solutions for class 9 maths chapter wise

Chapter No.

Chapter Name

Chapter 1

NCERT solutions for class 9 maths chapter 1 Number Systems

Chapter 2

CBSE NCERT solutions for class 9 maths chapter 2 Polynomials

Chapter 3

Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry

Chapter 4

NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables

Chapter 5

CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry

Chapter 6

Solutions of NCERT class 9 maths chapter 6 Lines And Angles

Chapter 7

NCERT solutions for class 9 maths chapter 7 Triangles

Chapter 8

CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals

Chapter 9

Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles

Chapter 10

NCERT solutions for class 9 maths chapter 10 Circles

Chapter 11

CBSE NCERT solutions for class 9 maths chapter 11 Constructions

Chapter 12

Solutions of NCERT class 9 maths chapter 12 Heron’s Formula

Chapter 13

NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes

Chapter 14

CBSE NCERT solutions for class 9 maths chapter 14 Statistics

Chapter 15

Solutions of NCERT class 9 maths chapter 15 Probability

NCERT solutions for class 9 subject wise

CBSE NCERT Solutions for Class 9 Maths

Solutions of NCERT Solutions for Class 9 Science

How to use NCERT solutions for class 9 maths chapter 11 Constructions

  • Understand the basic construction of angles using the previous class book.

  • Learn the constructional process of triangles and rectangles.

  • Go through some solved examples to understand the solutions pattern.

  • Apply the concepts learned in the practice exercises.

  • If you feel trouble in solving any problem then take the assistance of NCERT solutions for class 9 maths chapter 11 Constructions.

 

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