# NCERT Solutions for Class 9 Maths Chapter 11 Constructions

**NCERT solutions for class 9 maths chapter 11 Constructions: **The chapter construction class 9 deals with the construction of certain angles and also triangles in certain conditions. In the previous NCERT book chapters, you learnt to draw an approximate figure while writing the solutions. In this NCERT book construction class 9 chapter, the scenario is different because the main purpose to introduce this chapter in the curriculum is to develop an approach to creating an accurate geometric figure using some hints. NCERT solutions for class 9 maths chapter 11 Constructions have been designed to provide you with help in creating those accurate figures. Let's understand the kinds of questions which appear in this particular chapter 11 maths class 9. 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 11 maths class 9 can be used in designing such cases.

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NCERT solutions for class 9 maths chapter 11 Constructions have point-by-point solutions to every constructional problem. When there is any construction 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 construction class 9 NCERT book chapter, there are 2 exercises having a total of 14 questions. NCERT solutions 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 NCERt book chapters, classes, and subjects. Here you will get NCERT solutions for class 9 also.

**Also read:**

## ** Constructions Class 9 NCERT 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.

** 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.

** 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.

Hence, .

** Justification: **

We need to justify,

So, join OR and OS and RQ. we obtain

By construction OQ = OS = QR.

So, is an equilateral triangle. Similarly is an equilateral triangle.

So,

Now, that means .

Then, join AS and AR:

Now, in triangles OSA and ORA:

(common)

(Radii of same arcs)

(radii of the same arcs)

So,

Therefore,

and

** 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.

** 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 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 5: ** X and O are joined and a ray making an angle 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.

** Justification: **

By construction we have,

We constructed the bisector of as

Thus,

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

** Answer: **

Steps to construction to follow:

** Step 1: ** Draw a ray OY.

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

** Step 3: ** Take A and B as centres, draw 2 arcs are marked intersecting each other at X and hence, the bisector of is constructed.

Thus, 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:

** Answer: **

Steps to construction to follow:

** Step 1: ** Draw a ray OY.

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

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

** 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.

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 .

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

** Answer: **

Steps to construction to follow:

** Step 1: ** Draw a ray OY.

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

** 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 .

** 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 .

** Thus, the angle of is obtained which is **

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

** Answer: **

Steps to construction to follow:

** 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 is constructed.

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

Step 6: Now, join OD and hence, is constructed.

** Hence, 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:

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 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 is constructed.

** Thus, the angle is **

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

** Answer: **

The steps of construction to be followed:

** 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 .

** 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 is made.

** Hence, the angle required is . **

** 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 .

**Constructions Class 9 NCERT Excercise:**** 11.2 **

** Q1 ** Construct a triangle ABC in which , and .

** 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 .

** Step 2: ** Now, cut a line segment BD having 13 cm on BX which is

** 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 , ° and

** Answer: **

The steps of construction to be followed:

** Step 1: ** Draw a line segment BC = 8cm and make an angle of at point B i.e.,

** Step 2: ** Now, cut the line segment BD = 3.5 cm on ray BX. i.e. .

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

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

** Step 5: ** Join AC, to get the required triangle

** Q3 ** Construct a triangle PQR in which , and .

** 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 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

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

** Q4 ** Construct a triangle XYZ in which , ° and .

** Answer: **

The steps of construction to be followed:

** Step 1: ** For given , a line segment is drawn.

** Step 2: ** At points, P and Q angles of and are constructed respectively.

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

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

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

** Therefore, 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 from the ray.

** Step 2: ** Now, construct an angle .

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

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

** Thus, the constructed triangle is ABC. **

**Here students can find construction class 9 exercise NCERT solutions**

**NCERT solutions for class 9 maths chapter wise **

Chapter No. | Chapter Name |

Chapter 1 | Number Systems |

Chapter 2 | Polynomials |

Chapter 3 | Coordinate Geometry |

Chapter 4 | Linear Equations In Two Variables |

Chapter 5 | Introduction to Euclid's Geometry |

Chapter 6 | Lines And Angles |

Chapter 7 | Triangles |

Chapter 8 | Quadrilaterals |

Chapter 9 | Areas of Parallelograms and Triangles |

Chapter 10 | Circles |

Chapter 11 | Constructions |

Chapter 12 | Heron’s Formula |

Chapter 13 | Surface Area and Volumes |

Chapter 14 | Statistics |

Chapter 15 | Probability |

** NCERT solutions for class 9 subject wise **

** 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.

**Also Check NCERT Books and NCERT Syllabus here:**

## Frequently Asked Question (FAQs) - NCERT Solutions for Class 9 Maths Chapter 11 Constructions

**Question: **What are the important topics in chapter Constructions ?

**Answer: **

Basic Construction and Construction of triangles are two important topics in this chapter. Students can prioritise the important chapters according to the NCERT syllabus and practice them to get good hold and score well in the exam.

**Question: **Does CBSE provides the solutions of NCERT for class 9 ?

**Answer: **

No, CBSE doesn’t provide NCERT solutions for any class or subject but interested students can find NCERT book solutions above in this article that are listed according to chapter and subject wise.

**Question: **Where can I find the complete solutions of NCERT for class 9 ?

**Answer: **

Here you will get the detailed NCERT solutions for class 9 by clicking on the link.

**Question: **Where can I find the complete solutions of NCERT for class 9 maths ?

**Answer: **

Here you will get the detailed NCERT solutions for class 9 maths by clicking on the link.

**Question: **Which is the official website of NCERT ?

**Answer: **

NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.

**Question: **Does CBSE class 9 maths is tough ?

**Answer: **

In CBSE class 9 maths there are some new topics, and maths in previous classes were basic and very simple, so some students may find it tough. Overall CBSE class 9 maths is not tough at all.

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