NCERT Solutions for Class 10 Maths Chapter 11 Constructions

NCERT Solutions for Class 10 Maths Chapter 11 Constructions - Free PDF

NCERT Solutions for Class 10 Chapter 11, Constructions is well crafted by subject-matter experts in Vedantu. They have developed the NCERT Solutions as per the latest syllabus set by CBSE board. Vedantu also provides relevant notes for the Maths NCERT Solutions Class 10 to give a better understanding of the concept. 

You can download the free pdf format of the NCERT Solutions Chapter 11 from Vedantu’s official website. NCERT Solutions for other subjects for other classes are also available on Vedantu. If you have any queries relating to the concepts, you can reach out to our experienced teachers. 

In this chapter, you will learn the concept of determining a point dividing a line segment internally given a ratio, construction of similar triangles, construction of a tangent to a circle, construction of a pair of tangents and construction of a pair of tangents which are inclined to each other at an angle. Below are some basic reference notes that will help you solve the NCERT Solutions of Chapter 11. 

Vedantu also provides free CBSE Solutions for all the classes. You can download NCERT Solution for Class 10 Science PDF on vedantu.com for free to score more marks in the examinations.

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NCERT Solutions for Class 10 Maths Chapter 11 Constructions part-1

NCERT Solutions for Class 10 Maths Chapter 11 Constructions - PDF Download

Introduction

Construction of basic figures like a triangle, bisecting a line segment or drawing a perpendicular at a point on a line requires a ruler with a bevelled edge, a sharply pointed pencil preferably a type of set squares and a pair of compasses for justification of the method used. Some basic knowledge of geometry is required like proportionality theorem, concept of similar triangles, etc. have a look at to cover all the questions from exercise 11.1 and exercise 11.2

You can Find the Solutions of All the Maths Chapters below.

NCERT Solutions for Class 10 Maths


Division of a Line Segment

To divide a line segment internally in a given ratio m:n, we take the following steps: 

Steps of Construction:

Step 1: Draw a line segment of a given length and name it AB.

Step 2: Draw a ray making an acute angle with AB and let the ray be AX.

Step 3: Along AX, mark off (m + n) points A1,  A2, ……., Am, Am+1,....., Am+n (for ex: if the ratio is to 

be 2:3, then we mark of 5 (= 2+ 3) points)

Step 4: Join BAAm+n

Step 5: Draw a line through point Am parallel to Am+n B and make an angle equal to ∠AAm+n B. 

Let this line meet Ab at a point P. Then this point is the required point which divides AB 

internally in the ratio m:n.

(Image to be added soon)


Constructing a Triangle Similar to a Given Triangle

Here we construct a triangle similar to a given triangle. The constructed triangle may be smaller or larger than the given triangle. So we define the following term:

Scale factor: Scale factor is the ratio of the sides of any figure to be constructed with the corresponding measurements of the given figure.

Let ABC be the given triangle by using the given & suppose we want to construct a triangle similar to ABC, such that each of its sides is (m/n)th of the corresponding sides of ABC.


The Following Are the Steps to Be Taken for Construction a Triangle When M<n:

Step 1: Construct the given ABC by using the given data.

Step 2: Take AB as the base of the given ABC.

Step 3: At one end, say A of AB construct an acute angle ∠BAX below the base AB.

Step 4: Along AX mark off n points A1, A2,A3, ……., Am, Am+1,....., An such that AA1= A1A2 = A2A3 

= A3 = Am-1Am = …….. = An-1An.

Step 5: Join AnB.

Step 6: Start from A & reach to the point AnB which meets AB at B’

Step 7: From B’ draw B’C’ || CB meeting AC at C’.

Then AB’C’ is the required triangle each of whose sides is (m/n)th of the corresponding sides of ABC.


Construction of Tangents to a Circle

We know that when a point lies inside a circle no tangent can be drawn to the circle from this point. If a point lies on the circle at that point but if the point lies outside the circle, two agents can be drawn to the circle from the point. 


Constructing a Tangent to a Circle at a Ive Point, We Consider Two Cases:

Case A: When the Centre of the Circle Is Known, the Steps of Construction Are:

(Image to be added soon)

Step 1: We take a point O on the plane of the paper and using a compass & ruler, we 

  draw a circle of given radius. 

Step 2: Let there be a point P on the circle.

Step 3: Join OP                                                                                                             

Step 4: Construct ∠OPT = 90°.

Step 5: Produce TP to T’ to obtain the line TPT’ as the required tangent.


Case B: When the Centre of the Circle Is Not Known, Then the Steps of Construction Are:

(Image to be added soon)

Step 1: A chord PQ is drawn to the circle through the given point P on the circle.

Step 2: P & Q are joined to a point R in the major arc of the circle.

Step 3: Construct ∠QPT equal to ∠QRP on the opposite side of the chord PQ.

Step 4: Produce TP to T’ to obtain  TPT’ as the required tangent. 


Construction of Tangents to a Circle from an External Point

We know that two tangents can be drawn to a circle from an external point. The cases may arise---

Case A: When the Centre of the Circle Is Known, the Steps of Construction Are:

(Image to be added soon)

Step 1: Given external point P is joined to centre O of the circle.

Step 2: Draw a perpendicular bisector of OP, intersecting OP at Q.

Step 3: Draw a circle with Q as center and OQ = QP as radius , intersecting the given circle at T & T’.

Step 4: Join PT & PT’.

Then PT & PT’ are the two tangents to the circle drawn from the external point P.


Case B: When Centre of the Circle Is Not Known, Then the Steps of Construction Are:

P is the external point & a circle is given with diameter AB.

(Image to be added soon)

Step 1: From P draw a secant PAB intersecting the given circle at A & B.

Step 2: Produce AP to C, such that AP = PC.

Step 3: Locate midpoint of BC as M, & draw a semicircle.

Step 4: Draw a perpendicular PD on BC intersecting the semicircle at D.

Step 5: With P as centre & radius PD draw arcs intersecting the given circle at T & T’.

Step 6: Join PT & PT’.

Then PT & PT’ are the two required tangents drawn on the given circle from the external point P

Strategy for Preparing for Exams

For any exam, enough practice and revision are required. Even time management and how to tackle tricky questions are very significant during the exams. NCERT Solutions provided by Vedantu will definitely give you a good practice on the variety of questions. The notes along with the solution will give you clarity about the concept. Constructions is an important geometry chapter in Maths and pre knowledge of constructing an angle bisector and drawing a perpendicular line will be helpful. The NCERT Solutions design by Vedantu will give you a clear idea about the question papers that you will get in exams. The entire solution in Vedantu is designed in a concise manner and in step wise method. You can adopt the same method for exams. You will also learn how to manage time and deal with difficult questions in exams with the help of Vedantu’s NCERT Solutions. 


Advantage of using Vedantu’s NCERT Solution

There are many NCERT Solutions on the net but Vedantu provides 100% accurate and the latest solutions as per the syllabus under the strict guidelines set by the CBSE Board. Vedantu’s NCERT solutions have benefitted countless students till today. Experienced teachers have designed the solutions in a very simple way that makes it self explanatory and you will be thorough with the concepts at your fingertips. Vedantu also provides live sessions which is the highlight of Vedantu. This will boost the confidence of the students and will help master the subject. You can take the live video on any device and from anywhere.  For strengthening your concept understanding, you can completely rely on the NCERT Solutions provided by Vedantu. This is the right platform for you to revise your subjects and score more marks in exams. You can download the pdf format from Vedantu’s website which is available for free. You can download on any device and practice as per your convenient time. You can carry the pdf anywhere and anytime. 

FAQs (Frequently Asked Questions)

1. What do you understand by the scale factor of any geometrical figure?

Scale factor is the ratio of the sides of any figure to be constructed with the corresponding measurements of the given figure.

2. Why should you refer to Vedantu for NCERT Solutions for Class 10 Chapter 11?

You should refer to Vedantu for NCERT Solutions for Class 10 Chapter 11 because Vedantu provides the latest and very comprehensive explanation of all NCERT Solution of Chapter 11. The experienced faculties in Vedantu have created the solutions just for you after intensive research. The solutions for Chapter 11 have many illustrated examples that will give you a complete revision of the chapter and you can prepare well for the exams. The NCERT Solutions provided by Vedantu are designed in a unique way starting from easy questions and as you proceed the level of the questions will become complicated. In this way you can solve any type of question whether it is easy or difficult.

3. What are the topics covered in the Chapter 11 for Class 10?

The topics that are covered in the Chapter 11 for Class 10 are how to determine a point dividing a line segment internally given a ratio, construction of similar triangles, constructing a tangent to a circle, and constructing a pair of tangents which are inclined to each other at an angle.

4. How can I improve my score through Vedantu?

Vedantu helps students to strengthen their foundation in all subjects and develop the ability to tackle different kinds of questions given in the textbooks. Free pdf of each subject topic wise  is available on the website. NCERT Maths book for class 10 guide will definitely help the students significantly to improve their performance in academics.

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