NCERT Solutions for Class 10 Maths Chapter 11 Constructions  Free PDF
CBSE Class 10 Maths Chapter 11 is based on the construction of different geometric figures. This chapter focuses on teaching how to draw dividing points for line segments and closed figures. This fundamental chapter needs the complete study material for proper preparation. This material should contain the solutions for Class 10 Maths Constructions exercises.
The questions in the exercises have been solved by the subject experts of Vedantu by following the latest CBSE syllabus and guidelines. It will help students to figure out the ideal methods of approaching these fundamental problems and use the geometric concepts explained in this chapter perfectly.
NCERT Solutions for Class 10 Chapter 11, Constructions is well crafted by subjectmatter 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.
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Significance of CBSE Class 10 Maths Constructions Exercise Solutions
This chapter holds immense importance in the Class 10 Maths syllabus as it teaches how to draw different points and figures. The stepwise presentation of the process of drawing such geometric figures and points in the book will become easier to follow when students have the right study material.
This material must contain the NCERT solutions for all the exercises included in this chapter. The exercise problems have been compiled by the CBSE board to check the preparation and comprehension levels of the students. This is where the solutions framed by the experts for all the questions in this chapter will be very useful.
Once a section is done, students will proceed to solve the exercise questions. The solutions to these questions can be referred to find accurate answers. They can check their answers and find out where they need to focus more. Hence, the NCERT solutions for Class 10 Maths Construction can be used as an assessment tool for the preparation of this subject.
How to use Class 10 Maths Construction Solutions?
Focus on the theory part of the chapter. Use the handson method to copy what the textbook is instructing you to do. Follow the steps and achieve what this chapter explains in a section. Once done, practice the process repeatedly.
When you are done practising, proceed to the respective exercise to solve the given problems. Refer to the solutions here and resolve your doubts. Focus on how the experts have skilfully used the chapter theorems and fundamental geometric concepts to draw such figures and answered the exercise questions.
Now, practice following the methods and steps mentioned in the solutions to make the exercise problems easily comprehensible. This is how you can brilliantly use the solutions for all the exercises in this chapter.
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.
Overview of the Exercises Covered in the NCERT Solutions Class 10 Maths Ch11 Constructions
NCERT Solutions for Chapter 11 of Class 10 Maths are prepared by the extremely talented team of subject experts at Vedantu. Students will get a thorough idea of constructing geometric figures when they go through these NCERT Solutions for Ch11 for the CBSE Term 2 exam. There are
Exercise 11.1: There are a total of 7 sums in this exercise. These sums familiarize students with the construction of line segments and dividing them into parts as per the given ratio, and with the construction of various triangles. For example, some of the sums of this exercise require students to construct isosceles triangles. Students should solve and practice all sums given in this exercise to achieve better accuracy in constructions of geometric figures in the exam.
Exercise 11.2: Like the previous exercise, the second exercise of Class 10 Maths Chapter 11 comprises a total of 7 sums. These sums cover the constructions of circles, tangents, lines, rightangled triangles, etc. Students will also need to justify some of the constructions by using the concepts and theorems of triangles, circles, etc.
Exercises under NCERT Solutions for Class 10 Maths Chapter 11 Constructions
NCERT Solutions for Class 10 Maths Chapter 11, "Constructions", is a chapter that focuses on the construction of various geometric shapes using a ruler and compass. The chapter contains three exercises that cover different aspects of construction, including constructing triangles when given their side lengths and angles, constructing similar triangles, and verifying the construction of different shapes.
Exercise 11.1: This exercise involves constructing a triangle using a ruler and compass when the lengths of its three sides are given. The exercise contains six questions that test the student's understanding of constructing triangles and their ability to use ruler and compass accurately. The solutions provide stepbystep instructions and diagrams to help students visualize the construction process.
Exercise 11.2: This exercise involves constructing a triangle using a ruler and compass when two sides and an angle are given. The exercise contains six questions that test the student's understanding of constructing triangles and their ability to use ruler and compass accurately. The solutions provide stepbystep instructions and diagrams to help students visualize the construction process.
Exercise 11.3: This exercise involves constructing a similar triangle using a ruler and compass when a triangle is given. The exercise contains four questions that test the student's understanding of the concept of similarity of triangles and their ability to use ruler and compass accurately. The solutions provide stepbystep instructions and diagrams to help students visualize the construction process.
Access NCERT Solutions for Class 10 Mathematics Chapter 11 – Constructions
NCERT Solutions for Class 10 Math Chapter 11 Constructions are provided on this page in a comprehensive manner, where one can find a stepbystep solution to all the questions covered in the Chapter 11. NCERT Solutions for Class 10 Chapter 11 Construction are prepared by subject experts under the guidelines of NCERT to assist students in their academic as well as competitive examination. Get free NCERT Solutions for Class 10 Math, Chapter 11 Constructions at Vedantu to accelerate the exam preparation. All the questions of NCERT exercises are solved using diagrams with a stepbystep procedure for construction. NCERT Solutions given here help students boost their concepts and clear doubts.
Topics Covered In Class 10 Math 11 Construction
11.1; Introduction to Construction
11.2: Division of Line Segment
11.3: Construction of Tangent To A Circle
Class 10 Chapter 11 Construction Summary
In this chapter, students will learn how to divide a line segment in a given ratio. Students can divide a line segment by measuring the length and then marking a point on the line that divides it in a given ratio. Here, students will learn how to divide a line segment in a given ratio using the compass.
Here, students will also learn how to construct a tangent to a circle. You should know if a point lies inside a circle that no tangent to a circle can be formed whereas if a point lies outside the circle then there is only one tangent to a circle at this point and is perpendicular to the radius to this point.
1. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts. Give a proper justification for the construction.
Answer
Given: Draw a line segment of length 7.6 cm and divide it in the ratio 5:8.
To find: Measure the two parts. Give a proper justification of the construction.
A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows.
Step 1: Draw line segment ${\text{AB}}$ of $7.6\;{\text{cm}}$ and draw a ray AX making an acute angle with line segment ${\text{AB}}$
Step 2: Locate $13( = 5 + 8)$ points, ${A_1},{A_2},{A_3},{A_4} \ldots \ldots \ldots {A_{13}}$, on $A X$ such that $A{A_1} = {A_1}{A_2} = {A_2}{A_3}$ and so on.
Step 3: Join ${\text{B}}{{\text{A}}_{13}}$.
Step 4: Through the point ${{\text{A}}_5}$, draw a line parallel to ${\text{B}}{{\text{A}}_{13}}$ (by making an angle equal to $\angle {\text{A}}{{\text{A}}_{13}}\;{\text{B}}$ ) at ${{\text{A}}_5}$ intersecting ${\text{AB}}$ at point ${\text{C}}$.
${\text{C}}$ is the point dividing line segment ${\text{AB}}$ of $7.6\;{\text{cm}}$ in the required ratio of $5: 8$. The lengths of ${\text{AC}}$ and ${\text{CB}}$ can be measured. It comes out to $2.9\;{\text{cm}}$ and $4.7\;{\text{cm}}$ respectively.
Justification
The construction can be justified by proving that $\dfrac{{{\text{AC}}}}{{{\text{CB}}}} = \dfrac{5}{8}$
By construction, we have . By applying Basic proportionality theorem for the triangle ${\text{A}}{{\text{A}}_{13}}\;{\text{B}}$, we obtain
$\dfrac{{{\text{AC}}}}{{{\text{CB}}}} = \dfrac{{{\text{A}}{{\text{A}}_5}}}{{\;{{\text{A}}_5}\;{{\text{A}}_{13}}}} \ldots (1)$
From the figure, it can be observed that ${\text{A}}{{\text{A}}_5}$ and ${{\text{A}}_5}\;{{\text{A}}_{13}}$ contain 5 and 8 equal divisions of line segments respectively. $\dfrac{{{\text{A}}{{\text{A}}_5}}}{{\;{{\text{A}}_5}\;{{\text{A}}_{13}}}} = \dfrac{5}{8} \ldots $ (2)
On comparing equations (1) and (2), we obtain $\dfrac{{{\text{AC}}}}{{{\text{CB}}}} = \dfrac{5}{8}$
This justifies the construction.
2. Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar to it whose sides are \[\dfrac{2}{3}\] of the corresponding sides of the first triangle. Give a proper justification of the construction.
Answer
Given: Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar to it whose sides are \[\dfrac{2}{3}\] of the corresponding sides of the first triangle.
To find: Give a proper justification of the construction.
Step 2: Draw a ray AX making an acute angle with line ${\text{AB}}$ on the opposite side of vertex ${\text{C}}$.
Step 3: Locate 3 points ${{\text{A}}_1},\;{{\text{A}}_2},\;{{\text{A}}_3}($ as 3 is greater between 2 and 3) on line ${\text{AX}}$ such that ${\text{A}}{{\text{A}}_1} = {{\text{A}}_1}\;{{\text{A}}_2} = $ ${{\text{A}}_2}\;{{\text{A}}_3}$
Step 4: Join ${\text{B}}{{\text{A}}_3}$ and draw a line through ${{\text{A}}_2}$ parallel to ${\text{B}}{{\text{A}}_3}$ to intersect ${\text{AB}}$ at point ${\text{B}}$ '. Step 5. Draw a line through B' parallel to the line ${\text{BC}}$ to intersect ${\text{AC}}$ at ${\text{C}}$ '. $\vartriangle {\text{A}}{{\text{B}}^\prime }{{\text{C}}^\prime }$ is the required triangle.
The construction can be justified by proving that ${\text{A}}{{\text{B}}^\prime } = \dfrac{2}{3}{\text{AB}},{{\text{B}}^\prime }{{\text{C}}^\prime } = \dfrac{2}{3}{\text{BC}},{\text{A}}{{\text{C}}^\prime } = \dfrac{2}{3}{\text{AC}}$
By construction, we have
$\therefore \angle {\text{A}}{{\text{B}}^\prime }{{\text{C}}^\prime } = \angle {\text{ABC}}$ (Corresponding angles)
In $\vartriangle {\text{A}}{{\text{B}}^\prime }{{\text{C}}^\prime }$ and $\vartriangle {\text{ABC}}$,
$\angle {\text{ABC}} = \angle {\text{A}}{{\text{B}}^\prime }{\text{C}}$ (Proved above)
$\angle {\text{BAC}} = \angle {{\text{B}}^\prime }{\text{A}}{{\text{C}}^\prime }$ (Common)
$\therefore \vartriangle {\text{A}}{{\text{B}}^\prime }{{\text{C}}^\prime } \sim \vartriangle {\text{ABC}}$ (AA similarity criterion)
$ \Rightarrow \dfrac{{{\text{A}}{{\text{B}}^\prime }}}{{{\text{AB}}}} = \dfrac{{{{\text{B}}^\prime }{{\text{C}}^\prime }}}{{{\text{BC}}}} = \dfrac{{{\text{A}}{{\text{C}}^\prime }}}{{{\text{AC}}}} \ldots \ldots $ (1)
In $\vartriangle {\text{A}}{{\text{A}}_2}\;{{\text{B}}^\prime }$ and $\vartriangle {\text{A}}{{\text{A}}_3}\;{\text{B}}$,
$\angle {{\text{A}}_2}{\text{A}}{{\text{B}}^\prime } = \angle {{\text{A}}_3}{\text{AB}}$ (Common)
$\angle {\text{A}}{{\text{A}}_2}\;{{\text{B}}^\prime } = \angle {\text{A}}{{\text{A}}_3}\;{\text{B}}$ (Corresponding angles)
$\therefore \vartriangle {\text{A}}{{\text{A}}_2}\;{{\text{B}}^\prime }$ and $\vartriangle {\text{A}}{{\text{A}}_3}\;{\text{B}}$ (AA similarity criterion)
$ \Rightarrow \dfrac{{{\text{A}}{{\text{B}}^\prime }}}{{{\text{AB}}}} = \dfrac{{{\text{A}}{{\text{A}}_2}}}{{{\text{A}}{{\text{A}}_3}}}$
$ \Rightarrow \dfrac{{{\text{A}}{{\text{B}}^\prime }}}{{{\text{AB}}}} = \dfrac{2}{3}\quad \ldots \ldots $
From equations (1) and (2),
$\dfrac{{{\text{A}}{{\text{B}}^\prime }}}{{{\text{AB}}}} = \dfrac{{{{\text{B}}^\prime }{{\text{C}}^\prime }}}{{{\text{BC}}}} = \dfrac{{{\text{A}}{{\text{C}}^\prime }}}{{{\text{AC}}}} = \dfrac{2}{3}$
$ \Rightarrow {\text{A}}{{\text{B}}^\prime } = \dfrac{2}{3}{\text{AB}},{{\text{B}}^\prime }{{\text{C}}^\prime } = \dfrac{2}{3}{\text{BC}},{\text{A}}{{\text{C}}^\prime } = \dfrac{2}{3}{\text{AC}}$
This justifies the construction.
3. Construct a triangle with sides $5\;{\text{cm}},6\;{\text{cm}}$ and $7\;{\text{cm}}$ and then another triangle whose sides are $\dfrac{7}{5}$ of the corresponding sides of the first triangle. Give a proper justification of the construction.
Answer
Given: Construct a triangle with sides $5\;{\text{cm}},6\;{\text{cm}}$ and $7\;{\text{cm}}$ and then another triangle whose sides are $\dfrac{7}{5}$ of the corresponding sides of the first triangle.
To find: Give a proper justification of the construction
Step 1: Draw a line segment ${\text{AB}}$ of $5\;{\text{cm}}$. Taking ${\text{A}}$ and ${\text{B}}$ as centre, draw arcs of $6\;{\text{cm}}$ and $5\;{\text{cm}}$ radius respectively. Let these arcs intersect each other at point C. $\vartriangle {\text{ABC}}$ is the required triangle having length of sides as $5\;{\text{cm}},6\;{\text{cm}}$, and $7\;{\text{cm}}$ respectively.
Step 2: Draw a ray AX making acute angle with line ${\text{AB}}$ on the opposite side of vertex ${\text{C}}$. Step 3. Locate 7 points, ${A_1},{A_2},{A_3},{A_4}{A_5},{A_6},{A_7}$ (as 7 is greater between 5 and 7 ), on line AX such that ${\text{A}}{{\text{A}}_1} = {{\text{A}}_1}\;{{\text{A}}_2} = {{\text{A}}_2}\;{{\text{A}}_3} = {{\text{A}}_3}\;{{\text{A}}_4} = {{\text{A}}_4}\;{{\text{A}}_5} = {{\text{A}}_5}\;{{\text{A}}_6} = {{\text{A}}_6}\;{{\text{A}}_7}$
Step 4: Join ${\text{B}}{{\text{A}}_5}$ and draw a line through ${{\text{A}}_7}$ parallel to ${\text{B}}{{\text{A}}_5}$ to intersect extended line segment ${\text{AB}}$ at point B'.
Step 5: Draw a line through B' parallel to ${\text{BC}}$ intersecting the extended line segment ${\text{AC}}$ at ${{\text{C}}^\prime }.\vartriangle {\text{A}}{{\text{B}}^\prime }{{\text{C}}^\prime }$ is the required triangle.
Justification
The construction can be justified by proving that ${\text{A}}{{\text{B}}^\prime } = \dfrac{7}{5}{\text{AB}},{{\text{B}}^\prime }{{\text{C}}^\prime } = \dfrac{7}{5}{\text{BC}},{\text{A}}{{\text{C}}^\prime } = \dfrac{7}{5}{\text{AC}}$
In $\vartriangle {\text{ABC}}$ and $\vartriangle {\text{A}}{{\text{B}}^\prime }{{\text{C}}^\prime }$,
$\angle {\text{ABC}} = \angle {\text{A}}{{\text{B}}^\prime }{{\text{C}}^\prime }$ (Corresponding angles)
$\angle {\text{BAC}} = \angle {{\text{B}}^\prime }{\text{A}}{{\text{C}}^\prime }$ (Common)
$\therefore \vartriangle {\text{ABC}} \sim \Delta {\text{A}}{{\text{B}}^\prime }{{\text{C}}^\prime }$ (AA similarity criterion)
$ \Rightarrow \dfrac{{{\text{AB}}}}{{{\text{A}}{{\text{B}}^\prime }}} = \dfrac{{{\text{BC}}}}{{{{\text{B}}^\prime }{{\text{C}}^\prime }}} = \dfrac{{{\text{AC}}}}{{{\text{A}}{{\text{C}}^\prime }}} \ldots (1)$
In $\vartriangle {\text{A}}{{\text{A}}_5}\;{\text{B}}$ and $\vartriangle {\text{A}}{{\text{A}}_7}\;{{\text{B}}^\prime }$,
$\angle {{\text{A}}_5}{\text{AB}} = \angle {{\text{A}}_7}{\text{A}}{{\text{B}}^\prime }$ (Common)
$\angle {\text{A}}{{\text{A}}_5}\;{\text{B}} = \angle {\text{A}}{{\text{A}}_7}\;{{\text{B}}^\prime }$ (Corresponding angles)
$\therefore \Delta {\text{A}}{{\text{A}}_5}\;{\text{B}} \sim \;\Delta {\text{A}}{{\text{A}}_7}\;{{\text{B}}^\prime }$ (AA similarity criterion)
$ \Rightarrow \dfrac{{{\text{A}}{{\text{B}}^\prime }}}{{{\text{AB}}}} = \dfrac{{{\text{A}}{{\text{A}}_5}}}{{{\text{A}}{{\text{A}}_7}}}$
$ \Rightarrow \dfrac{{{\text{AB}}}}{{{\text{A}}{{\text{B}}^\prime }}} = \dfrac{5}{7}\quad \ldots \ldots $
On comparing equations (1) and (2), we obtain $\dfrac{{{\text{AB}}}}{{{\text{A}}{{\text{B}}^\prime }}} = \dfrac{{{\text{BC}}}}{{{{\text{B}}^\prime }{{\text{C}}^\prime }}} = \dfrac{{{\text{AC}}}}{{{\text{A}}{{\text{C}}^\prime }}} = \dfrac{5}{7}$
$ \Rightarrow {\text{A}}{{\text{B}}^\prime } = \dfrac{7}{5}{\text{AB}},{{\text{B}}^\prime }{{\text{C}}^\prime } = \dfrac{7}{5}{\text{BC}},{\text{A}}{{\text{C}}^\prime } = \dfrac{7}{5}{\text{AC}}$
This iustifies the construction.
4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose side are $1\dfrac{1}{2}$ times the corresponding sides of the isosceles triangle. Give a proper justification of the construction.
Answer
Given: Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose side are $1\dfrac{1}{2}$ times the corresponding sides of the isosceles triangle.
To find: Give a proper justification of the construction.
A $\vartriangle {\text{A}}{{\text{B}}^\prime }{{\text{C}}^\prime }$ whose sides are $\dfrac{3}{2}$ times of $\Delta {\text{ABC}}$ can be drawn as follows.
Step 1: Draw a line segment ${\text{AB}}$ of $8\;{\text{cm}}$. Draw arcs of same radius on both sides of the line segment while taking point ${\text{A}}$ and ${\text{B}}$ as its centre. Let these arcs intersect each other at ${\text{O}}$ and ${{\text{O}}^\prime }$. Join ${\text{O}}{{\text{O}}^\prime }$. Let OO' intersect ${\text{AB}}$ at ${\text{D}}$.
Step 2: Taking D as centre, draw an arc of $4\;{\text{cm}}$ radius which cuts the extended line segment ${\text{O}}{{\text{O}}^\prime }$ at point C. An isosceles $\vartriangle {\text{ABC}}$ is formed, having ${\text{CD}}$ (altitude) as $4\;{\text{cm}}$ and ${\text{AB}}$ (base) as $8\;{\text{cm}}$.
Step 3: Draw a ray AX making an acute angle with line segment ${\text{AB}}$ on the opposite side of vertex ${\text{C}}$. Step 4. Locate 3 points (as 3 is greater between 3 and 2) ${{\text{A}}_1},\;{{\text{A}}_2}$, and ${{\text{A}}_3}$ on ${\text{AX}}$ such that ${\text{A}}{{\text{A}}_1} = {{\text{A}}_1}\;{{\text{A}}_2} = $ ${{\text{A}}_2}\;{{\text{A}}_3}$
Step 5: Join ${\text{B}}{{\text{A}}_2}$ and draw a line through ${{\text{A}}_3}$ parallel to ${\text{B}}{{\text{A}}_2}$ to intersect extended line segment ${\text{AB}}$ at point B'.
Step 6: Draw a line through B' parallel to BC intersecting the extended line segment ${\text{AC}}$ at ${{\text{C}}^\prime }.\vartriangle {\text{A}}{{\text{B}}^\prime }{{\text{C}}^\prime }$ is the required triangle.
Justification
The construction can be justified by proving that ${\text{A}}{{\text{B}}^\prime } = \dfrac{2}{3}{\text{AB}},{{\text{B}}^\prime }{{\text{C}}^\prime } = \dfrac{3}{2}{\text{BC}},{\text{A}}{{\text{C}}^\prime } = \dfrac{3}{2}{\text{AC}}$
In $\vartriangle {\text{ABC}}$ and $\Delta {\text{A}}{{\text{B}}^\prime }{{\text{C}}^\prime }$,
$\angle {\text{ABC}} = \angle {\text{A}}{{\text{B}}^\prime }{{\text{C}}^\prime }$ (Corresponding angles)
$\angle {\text{BAC}} = \angle {{\text{B}}^\prime }{\text{A}}{{\text{C}}^\prime }$ (Common)
$\therefore \vartriangle {\text{ABC}} \sim \Delta {\text{A}}{{\text{B}}^\prime }{{\text{C}}^\prime }$ (AA similarity criterion)
$ \Rightarrow \dfrac{{{\text{AB}}}}{{{\text{A}}{{\text{B}}^\prime }}} = \dfrac{{{\text{BC}}}}{{{{\text{B}}^\prime }{{\text{C}}^\prime }}} = \dfrac{{{\text{AC}}}}{{{\text{A}}{{\text{C}}^\prime }}} \ldots (1)$
In $\Delta {\text{A}}{{\text{A}}_2}\;{\text{B}}$ and $\vartriangle {\text{A}}{{\text{A}}_3}\;{{\text{B}}^\prime }$
$\angle {{\text{A}}_2}{\text{AB}} = \angle {{\text{A}}_3}{\text{A}}{{\text{B}}^\prime }$ (Common)
$\angle {\text{A}}{{\text{A}}_2}\;{\text{B}} = \angle {\text{A}}{{\text{A}}_3}\;{{\text{B}}^\prime }$ (Corresponding angles)
$\therefore \Delta {\text{A}}{{\text{A}}_2}\;{\text{B}} \sim \vartriangle {\text{A}}{{\text{A}}_3}\;{{\text{B}}^\prime }$ (AA similarity criterion)
$\dfrac{{AB}}{{A{B^\prime }}} = \dfrac{{A{A_2}}}{{A{A_3}}} = \dfrac{2}{3}$
On comparing equations (1) and (2), we obtain $\dfrac{{{\text{AB}}}}{{{\text{A}}{{\text{B}}^\prime }}} = \dfrac{{{\text{BC}}}}{{{{\text{B}}^\prime }{{\text{C}}^\prime }}} = \dfrac{{{\text{AC}}}}{{{\text{A}}{{\text{C}}^\prime }}} = \dfrac{2}{3}$
$ \Rightarrow {\text{A}}{{\text{B}}^\prime } = \dfrac{3}{2}{\text{AB}},{{\text{B}}^\prime }{{\text{C}}^\prime } = \dfrac{3}{2}{\text{BC}},{\text{A}}{{\text{C}}^\prime } = \dfrac{3}{2}{\text{AC}}$
This justifies the construction.
5. Draw a triangle ${\text{ABC}}$ with side ${\text{BC}} = 6\;{\text{cm}},{\text{AB}} = 5\;{\text{cm}}$ and $\angle {\text{ABC}} = {60^\circ }.$ Then construct a triangle whose sides are $\dfrac{3}{4}$ of the corresponding sides of the triangle ${\text{ABC}}$. Give a proper justification of the construction.
Answer
Given: Draw a triangle ${\text{ABC}}$ with side ${\text{BC}} = 6\;{\text{cm}},{\text{AB}} = 5\;{\text{cm}}$ and $\angle {\text{ABC}} = {60^\circ }.$ Then construct a triangle whose sides are $\dfrac{3}{4}$ of the corresponding sides of the triangle ${\text{ABC}}$.
To find: Give a proper justification of the construction.
A $\Delta {{\text{A}}^\prime }{\text{B}}{{\text{C}}^\prime }$ whose sides are $\dfrac{3}{4}$ th of the corresponding sides of $\vartriangle {\text{ABC}}$ can be drawn as follows.
Step1: Draw a $\Delta {\text{ABC}}$ with side ${\text{BC}} = 6\;{\text{cm}},{\text{AB}} = 5\;{\text{cm}}$ and $\angle {\text{ABC}} = {60^\circ }$.
Step 2: Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
Step 3: Locate 4 points (as 4 is greater in 3 and 4), ${{\text{B}}_1},\;{{\text{B}}_2},\;{{\text{B}}_3},\;{{\text{B}}_4}$, on line segment BX. Step 4. Join ${{\text{B}}_4}{\text{C}}$ and draw a line through ${{\text{B}}_3}$, parallel to ${{\text{B}}_4}{\text{C}}$ intersecting ${\text{BC}}$ at ${\text{C}}$ ' Step 5. Draw a line through C' parallel to AC intersecting ${\text{AB}}$ at ${{\text{A}}^\prime }.\Delta {{\text{A}}^\prime }{\text{B}}{{\text{C}}^\prime }$ is the required triangle.
Justification
The construction can be justified by proving ${\text{A}}{{\text{B}}^\prime } = \dfrac{3}{4}{\text{AB}},{\text{B}}{{\text{C}}^\prime } = \dfrac{3}{4}{\text{BC}},{{\text{A}}^\prime }{{\text{C}}^\prime } = \dfrac{3}{4}{\text{AC}}$
In $\vartriangle {{\text{A}}^\prime }{\text{B}}{{\text{C}}^\prime }$ and $\vartriangle {\text{ABC}}$,
$\angle {{\text{A}}^\prime }{{\text{C}}^\prime }{\text{B}} = \angle {\text{ACB}}$ (Corresponding angles)
$\angle {{\text{A}}^\prime }{\text{B}}{{\text{C}}^\prime } = \angle {\text{ABC}}$ (Common)
$\therefore \Delta {{\text{A}}^\prime }{\text{B}}{{\text{C}}^\prime } \sim \vartriangle {\text{ABC}}$ (AA similarity criterion)
$ \Rightarrow \dfrac{{{{\text{A}}^\prime }{\text{B}}}}{{{\text{AB}}}} = \dfrac{{{\text{B}}{{\text{C}}^\prime }}}{{{\text{BC}}}} = \dfrac{{{{\text{A}}^\prime }{{\text{C}}^\prime }}}{{{\text{AC}}}} \ldots $ (1)
In $\Delta {\text{B}}{{\text{B}}_3}{{\text{C}}^\prime }$ and $\Delta {\text{B}}{{\text{B}}_4}{\text{C}}$,
$\angle {{\text{B}}_3}{\text{B}}{{\text{C}}^\prime } = \angle {{\text{B}}_4}{\text{BC}}$ (Common)
$\angle {\text{B}}{{\text{B}}_3}{{\text{C}}^\prime } = \angle {\text{B}}{{\text{B}}_4}{\text{C}}$ (Corresponding angles)
$\therefore \Delta {\text{B}}{{\text{B}}_3}{{\text{C}}^\prime } \sim \Delta {\text{B}}{{\text{B}}_4}{\text{C}}$ (AA similarity criterion)
$ \Rightarrow \dfrac{{{\text{B}}{{\text{C}}^\prime }}}{{{\text{BC}}}} = \dfrac{{{\text{B}}{{\text{B}}_3}}}{{{\text{B}}{{\text{B}}_4}}}$
$ \Rightarrow \dfrac{{{\text{B}}{{\text{C}}^\prime }}}{{{\text{B}}{{\text{C}}^\prime }}} = \dfrac{3}{4}\quad \ldots \ldots .(2)$
From equations (1) and (2), we obtain $\dfrac{{{{\text{A}}^\prime }{\text{B}}}}{{{\text{AB}}}} = \dfrac{{{\text{B}}{{\text{C}}^\prime }}}{{{\text{BC}}}} = \dfrac{{{{\text{A}}^\prime }{{\text{C}}^\prime }}}{{{\text{AC}}}} = \dfrac{3}{4}$
$ \Rightarrow {{\text{A}}^\prime }{\text{B}} = \dfrac{3}{4}{\text{AB}},{\text{B}}{{\text{C}}^\prime } = \dfrac{3}{4}{\text{BC}},{\text{A}}{{\text{C}}^\prime } = \dfrac{3}{4}{\text{AC}}$
This justifies the construction.
6. Draw a triangle ${\text{ABC}}$ with side ${\text{BC}} = 7\;{\text{cm}},\angle {\text{B}} = {45^\circ },\angle {\text{A}} = {105^\circ }$. Then, construct a triangle whose sides are $4/3$ times the corresponding side of $\Delta {\text{ABC}}$. Give a proper justification of the construction.
Answer
Given: Draw a triangle ${\text{ABC}}$ with side ${\text{BC}} = 7\;{\text{cm}},\angle {\text{B}} = {45^\circ },\angle {\text{A}} = {105^\circ }$. Then, construct a triangle whose sides are $4/3$ times the corresponding side of $\Delta {\text{ABC}}$.
To find: Give a proper justification of the construction.
$\angle {\text{B}} = {45^\circ },\angle {\text{A}} = {105^\circ }$
Sum of all interior angles in a triangle is ${180^\circ }$. $\angle {\text{A}} + \angle {\text{B}} + \angle {\text{C}} = {180^\circ }$
${105^\circ } + {45^\circ } + \angle {\text{C}} = {180^\circ }$
$\angle {\text{C}} = {180^\circ }  {150^\circ }$
$\angle {\text{C}} = {30^\circ }$
The required triangle can be drawn as follows.
Step 1: Draw a $\vartriangle {\text{ABC}}$ with side ${\text{BC}} = 7\;{\text{cm}},\angle {\text{B}} = {45^\circ },\angle {\text{C}} = {30^\circ }$.
Step 2: Draw a ray BX making an acute angle with ${\text{BC}}$ on the opposite side of vertex ${\text{A}}$. Step 3. Locate 4 points (as 4 is greater in 4 and 3), ${{\text{B}}_1},\;{{\text{B}}_2},\;{{\text{B}}_3},\;{{\text{B}}_4}$, on BX. Step 4. Join ${{\text{B}}_3}{\text{C}}$. Draw a line through ${{\text{B}}_4}$ parallel to ${{\text{B}}_3}{\text{C}}$ intersecting extended ${\text{BC}}$ at ${\text{C}}$ '. Step 5. Through ${{\text{C}}^\prime }$, draw a line parallel to ${\text{AC}}$ intersecting extended line segment at ${\text{C}}$. $\vartriangle {{\text{A}}^\prime }{\text{B}}{{\text{C}}^\prime }$ is the required triangle.
Justification
The construction can be justified by proving that ${\text{A}}{{\text{B}}^\prime } = \dfrac{4}{3}{\text{AB}},{\text{B}}{{\text{C}}^\prime } = \dfrac{4}{3}{\text{BC}},{{\text{A}}^\prime }{{\text{C}}^\prime } = \dfrac{4}{3}{\text{AC}}$
In $\vartriangle {\text{ABC}}$ and $\vartriangle {{\text{A}}^\prime }{\text{B}}{{\text{C}}^\prime }$,
$\angle {\text{ABC}} = \angle {{\text{A}}^\prime }{\text{B}}{{\text{C}}^\prime }$ (Common)
$\angle {\text{ACB}} = \angle {{\text{A}}^\prime }{{\text{C}}^\prime }{\text{B}}$ (Corresponding angles)
$\therefore \Delta {\text{ABC}} \sim \Delta {{\text{A}}^\prime }{\text{B}}{{\text{C}}^\prime }$ (AA similarity criterion)
$ \Rightarrow \dfrac{{{\text{AB}}}}{{{{\text{A}}^\prime }{\text{B}}}} = \dfrac{{{\text{BC}}}}{{{\text{B}}{{\text{C}}^\prime }}} = \dfrac{{{\text{AC}}}}{{{{\text{A}}^\prime }{{\text{C}}^\prime }}} \ldots $ (1)
In $\vartriangle {\text{B}}{{\text{B}}_3}{\text{C}}$ and $\vartriangle {\text{B}}{{\text{B}}_4}{{\text{C}}^\prime }$,
$\angle {{\text{B}}_3}{\text{BC}} = \angle {{\text{B}}_4}{\text{B}}{{\text{C}}^\prime }$ (Common)
$\angle {\text{B}}{{\text{B}}_3}{\text{C}} = \angle {\text{B}}{{\text{B}}_4}{{\text{C}}^\prime }$ (Corresponding angles)
$\therefore \angle {\text{B}}{{\text{B}}_3}{\text{C}} \sim \angle {\text{B}}{{\text{B}}_4}{{\text{C}}^\prime }$ (AA similarity criterion)
$ \Rightarrow \dfrac{{{\text{BC}}}}{{{\text{B}}{{\text{C}}^\prime }}} = \dfrac{{{\text{B}}{{\text{B}}_3}}}{{{\text{B}}{{\text{B}}_4}}}$
$ \Rightarrow \dfrac{{{\text{BC}}}}{{{\text{B}}{{\text{C}}^\prime }}} = \dfrac{3}{4}\quad \cdots \cdots $
On comparing equations (1) and (2), we obtain $\dfrac{{{\text{AB}}}}{{{{\text{A}}^\prime }{\text{B}}}} = \dfrac{{{\text{BC}}}}{{{\text{B}}{{\text{C}}^\prime }}} = \dfrac{{{\text{AC}}}}{{{{\text{A}}^\prime }{{\text{C}}^\prime }}} = \dfrac{3}{4}$
$ \Rightarrow {{\text{A}}^\prime }{\text{B}} = \dfrac{4}{3}{\text{AB}},{\text{B}}{{\text{C}}^\prime } = \dfrac{4}{3}{\text{BC}},{{\text{A}}^\prime }{{\text{C}}^\prime } = \dfrac{4}{3}{\text{AC}}$
7. This justifies the construction.Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. then construct another triangle whose sides are $\dfrac{5}{3}$ times the corresponding sides of the given triangle. Give the justification of the construction.
Answer
Given: Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. then construct another triangle whose sides are $\dfrac{5}{3}$ times the corresponding sides of the given triangle.
To find: Give the justification of the construction,
It is given that sides other than hypotenuse are of lengths $4\;{\text{cm}}$ and $3\;{\text{cm}}.$ Clearly, these will be perpendicular to each other. The required triangle can be drawn as follows.
Step 1: Draw a line segment ${\text{AB}} = 4\;{\text{cm}}$. Draw a ray SA making ${90^\circ }$ with it.
Step 2: Draw an arc of $3\;{\text{cm}}$ radius while taking A as its centre to intersect ${\text{SA}}$ at ${\text{C}}$. Join ${\text{BC}}$. $\Delta {\text{ABC}}$ is the required triangle.
Step 3: Draw a ray AX making an acute angle with ${\text{AB}}$, opposite to vertex ${\text{C}}$.
Step 4: Locate 5 points (as 5 is greater in 5 and 3 ), ${{\text{A}}_1},\;{{\text{A}}_2},\;{{\text{A}}_3},\;{{\text{A}}_4},\;{{\text{A}}_5}$, online segment ${\text{AX}}$ such that ${\text{A}}{{\text{A}}_1}$ $ = {{\text{A}}_1}\;{{\text{A}}_2} = {{\text{A}}_2}\;{{\text{A}}_3} = {{\text{A}}_3}\;{{\text{A}}_4} = {{\text{A}}_4}\;{{\text{A}}_5}$
Step 5: Join ${{\text{A}}_3}\;{\text{B}}$. Draw a line through ${{\text{A}}_5}$ parallel to ${{\text{A}}_3}\;{\text{B}}$ intersecting extended line segment ${\text{AB}}$ at ${\text{B}}$ '. Step 6. Through B', draw a line parallel to BC intersecting extended line segment ${\text{AC}}$ at ${\text{C}}$ '. $\vartriangle {\text{A}}{{\text{B}}^\prime }{{\text{C}}^\prime }$ is the required triangle.
Justification
The construction can be justified by proving that ${\text{A}}{{\text{B}}^\prime } = \dfrac{5}{3}{\text{AB}},{{\text{B}}^\prime }{{\text{C}}^\prime } = \dfrac{5}{3}{\text{BC}},{\text{A}}{{\text{C}}^\prime } = \dfrac{5}{3}{\text{AC}}$In $\vartriangle {\text{ABC}}$ and $\vartriangle {\text{A}}{{\text{B}}^\prime }{\text{C}}$,
$\angle {\text{ABC}} = \angle {\text{A}}{{\text{B}}^\prime }{{\text{C}}^\prime }$ (Corresponding angles)
$\angle {\text{BAC}} = \angle {{\text{B}}^\prime }{\text{A}}{{\text{C}}^\prime }$ (Common)
$\therefore \vartriangle {\text{ABC}} \sim \vartriangle {\text{A}}{{\text{B}}^\prime }{{\text{C}}^\prime }$ (AA similarity criterion)
$ \Rightarrow \dfrac{{{\text{AB}}}}{{{\text{A}}{{\text{B}}^\prime }}} = \dfrac{{{\text{BC}}}}{{{{\text{B}}^\prime }{{\text{C}}^\prime }}} = \dfrac{{{\text{AC}}}}{{{\text{A}}{{\text{C}}^\prime }}}\quad \cdots $
In $\vartriangle {\text{A}}{{\text{A}}_3}\;{\text{B}}$ and $\vartriangle {\text{A}}{{\text{A}}_5}\;{{\text{B}}^\prime }$,
$\angle {{\text{A}}_3}{\text{AB}} = \angle {{\text{A}}_5}{\text{A}}{{\text{B}}^\prime }$ (Common)
$\angle {\text{A}}{{\text{A}}_3}\;{\text{B}} = \angle {\text{A}}{{\text{A}}_5}\;{{\text{B}}^\prime }$ (Corresponding angles)
$\therefore \Delta {\text{A}}{{\text{A}}_3}\;{\text{B}} \sim \Delta {\text{A}}{{\text{A}}_5}\;{{\text{B}}^\prime }$ (AA similarity criterion)
$ \Rightarrow \dfrac{{{\text{AB}}}}{{{\text{A}}{{\text{B}}^\prime }}} = \dfrac{{{\text{A}}{{\text{A}}_3}}}{{{\text{A}}{{\text{A}}_5}}}$
$ \Rightarrow \dfrac{{{\text{AB}}}}{{{\text{A}}{{\text{B}}^\prime }}} = \dfrac{3}{5}$
On comparing equations (1) and (2), we obtain $\dfrac{{{\text{AB}}}}{{{\text{A}}{{\text{B}}^\prime }}} = \dfrac{{{\text{BC}}}}{{{{\text{B}}^\prime }{{\text{C}}^\prime }}} = \dfrac{{{\text{AC}}}}{{{\text{A}}{{\text{C}}^\prime }}} = \dfrac{3}{5}$
$ \Rightarrow {\text{A}}{{\text{B}}^\prime } = \dfrac{5}{3}{\text{AB}},{{\text{B}}^\prime }{{\text{C}}^\prime } = \dfrac{5}{3}{\text{BC}},{\text{A}}{{\text{C}}^\prime } = \dfrac{5}{3}{\text{AC}}$
This justifies the construction.
EXERCISE NO: 11.2
1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Give a proper justification of the construction.
Answer
Given: Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
To prove: Give a proper justification of the construction.
Justification
The construction can be justified by proving that PQ and PR are the tangents to the circle (whose centre is O and radius is 6 cm). For this, join OQ and OR.
∠PQO is an angle in the semicircle. We know that the angle in a semicircle is a right angle.
\[\therefore \angle PQO{\text{ }} = {\text{ }}90^\circ \]
\[ \Rightarrow OQ \bot PQ\]
Since OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly, PR is a tangent of the circle.
2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation. Give the justification of the construction.
Answer
Given: Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
To prove: Give the justification of the construction.
It can be observed that PQ and PR are of length 4.47 cm each.
In $\Delta {\text{PQO}}$,
Since ${\text{PQ}}$ is a tangent, $\angle {\text{PQO}} = {90^\circ }$
${\text{PO}} = 6\;{\text{cm}}$
${\text{QO}} = 4\;{\text{cm}}$
Applying Pythagoras theorem in $\Delta {\text{PQO}}$, we obtain ${\text{P}}{{\text{Q}}^2} + {\text{Q}}{{\text{O}}^2} = {\text{P}}{{\text{Q}}^2}$
$P{Q^2} + {(4)^2} = {(6)^2}$
${\text{P}}{{\text{Q}}^2} + 16 = 36$
${\text{P}}{{\text{Q}}^2} = 36  16$
${\text{P}}{{\text{Q}}^2} = 20$
${\text{PQ}} = 2\sqrt 5 $
${\text{PQ}} = 4.47\;{\text{cm}}$
Justification
The construction can be justified by proving that PQ and PR are the tangents to the circle (whose centre is O and radius is 4 cm). For this, let us join OQ and OR.
\[\angle PQO\] is an angle in the semicircle. We know that the angle in a semicircle is a right angle.
\[ \therefore \angle PQO{\text{ }} = {\text{ }}90^\circ \]
\[\Rightarrow OQ \bot PQ \]
Since OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly, PR is a tangent of the circle
3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q. Give the justification of the construction.
Answer
Given: Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
To prove: Give the justification of the construction.
Justification
The construction can be justified by proving that RV, RW, SY, and SX are the tangents to the circle (whose centre is O and radius is 3 cm). For this, join OV, OW, OX, and OY.
\[\angle RVO\]is an angle in the semicircle. We know that the angle in a semicircle is a right angle.
\[ \therefore \angle RVO{\text{ }} = {\text{ }}90^\circ \]
\[ \Rightarrow OV \bot RV \]
Since OV is the radius of the circle, RV has to be a tangent of the circle. Similarly, OW, OX, and OY are the tangents of the circle
4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of \[60^\circ \]. Give a proper justification of the construction.
Answer
Given: Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of \[60^\circ \].
To prove: Give a proper justification of the construction.
Justification
The construction can be justified by proving that $\angle {\text{APB}} = {60^\circ }$ By our construction $\angle {\text{OAP}} = {90^\circ }$
$\angle {\text{OBP}} = {90^\circ }$
And $\angle {\text{AOB}} = {120^\circ }$
We know that the sum of all interior angles of a quadrilateral $ = {360^\circ }$ $\angle {\text{OAP}} + \angle {\text{AOB}} + \angle {\text{OBP}} + \angle {\text{APB}} = {360^\circ }$
${90^\circ } + {120^\circ } + {90^\circ } + \angle {\text{APB}} = {360^\circ }$
$\angle {\text{APB}} = {60^\circ }$
This justifies the construction.
5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle. Give a proper justification for the construction.
Answer
Given: Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
To prove: Give a proper justification of the construction.
Justification
The construction can be justified by proving that AS and AR are the tangents of the circle (whose centre is B and radius is 3 cm) and BP and BQ are the tangents of the circle (whose centre is A and radius is 4 cm). For this, join AP, AQ, BS, and BR.
\[ \therefore \angle ASB{\text{ }} = {\text{ }}90^\circ \]
\[ \Rightarrow BS \bot AS \]
Since BS is the radius of the circle, AS has to be a tangent of the circle. Similarly, AR, BP, and BQ are the tangents.
6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and \[\angle B{\text{ }} = {\text{ }}90^\circ \]. BD is the perpendicular from B on AC. The circle through B, C, and D is drawn. Construct the tangents from A to this circle. Give a proper justification of the construction.
Answer
Given: Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and \[\angle B{\text{ }} = {\text{ }}90^\circ \]. BD is the perpendicular from B on AC. The circle through B, C, and D is drawn. Construct the tangents from A to this circle.
To prove: Give a proper justification of the construction.
Justification
The construction can be justified by proving that AG and AB are the tangents to the circle. For this, join EG.
\[\angle AGE\] is an angle in the semicircle. We know that an angle in a semicircle is a right angle.
\[ \therefore \angle AGE{\text{ }} = {\text{ }}90^\circ \]
\[ \Rightarrow EG \bot AG \]
Since EG is the radius of the circle, AG has to be a tangent of the circle.
Already,
\[ \angle B{\text{ }} = {\text{ }}90^\circ \]
\[ \Rightarrow AB \bot BE \]
Since BE is the radius of the circle, AB has to be a tangent of the circle.
7. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circles. Give a proper justification of the construction.
Answer
Given: Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circles.
To prove: Give a proper justification of the construction.
Justification
The construction can be justified by proving that PV and PW are the tangents to the circle. For this, first of all, it has to be proved that O is the centre of the circle. Let us join OV and OW.
We know that the perpendicular bisector of a chord passes through the centre.
\[ \Rightarrow OV \bot PV \]
Since OV is the radius of the circle, PV has to be a tangent of the circle. Similarly, PW is a tangent of the circle.
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 = Am1Am = …….. = An1An.
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
Key Features for NCERT Solutions for Class 10 Maths Chapter 11 Constructions
The key features of the NCERT Class 10 Maths Chapter 11 Solutions as given below will help you to understand how useful these solutions are for your exam preparation.
Stepbystep instructions for all constructions
Prepared by subject experts
Detailed explanations and reasoning for every sum
Easy to access online and offline
Available in PDF format
Free to download
Absolutely in accordance with the rules of construction for Class 10 Term 2 Maths
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.
Key Features for NCERT Solutions for Class 10 Maths Chapter 11 Constructions
The key features of the NCERT Class 10 Maths Chapter 11 Solutions as given below will help you to understand how useful these solutions are for your exam preparation.
Stepbystep instructions for all constructions
Prepared by subject experts
Detailed explanations and reasoning for every sum
Easy to access online and offline
Available in PDF format
Free to download
Absolutely in accordance with the rules of construction for Class 10 Term 2 Maths
Advantage of Using Vedantu’s NCERT Solution
Vedantu stands out among various platforms for providing 100% accurate and uptodate NCERT Solutions, strictly adhering to CBSE Board guidelines.
These solutions, meticulously crafted by experienced teachers, are designed to be selfexplanatory, ensuring a thorough understanding of concepts.
Vedantu's live sessions, a notable highlight, boost student confidence and aid in mastering subjects. Accessible on any device from anywhere, these sessions enhance concept understanding.
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The free availability of PDF formats on Vedantu's website allows convenient download and practice on any device at one's preferred time, providing flexibility in study routines.
Specifically for Class 10 Maths Chapter 11 Constructions, Vedantu offers free PDF solutions on its mobile app and official website. These solutions cover exercises 11.1 and 11.2, enhancing students' comprehension of geometry concepts in Class 10 CBSE Maths Term 2.
The chapter includes constructions related to triangles, line segments, circles, tangents, etc. The stepbystep explanations in Vedantu's NCERT Solutions for Class 10 Maths Chapter 11 serve as valuable tools for effective exam preparation.
Download NCERT Solutions for Class 10 Maths Construction Exercises
Get the free PDF version of the solutions for this chapter. Learn how simply the experts have solved the geometry problems for your benefit. Focus on how to use the theories and construction methods explained in this chapter to score more in the exams.
NCERT Solutions for Class 10 Maths
Conclusion
Vedantu's NCERT Solutions for Class 10 Maths Chapter 11 on Constructions offer a reliable and userfriendly resource for students. Crafted by experienced teachers, these NCERT solutions are accurate, uptodate, and designed to enhance understanding. The platform's live sessions boost confidence and aid in mastering the subject, available on any device, anywhere. Conveniently accessible in free PDF format, Vedantu's NCERT solutions empower students to revise and practice at their own pace. Whether constructing triangles, line segments, circles, or tangents, these solutions serve as invaluable tools for effective exam preparation, reinforcing concepts for better performance.
FAQs on NCERT Solutions for Class 10 Maths Chapter 11  Constructions
1. What do you understand by the scale factor of any geometrical 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.
5. What is the basic concept of construction in Chapter 11 Constructions of Class 10 Maths?
A ruler with a bevelled edge, a highly pointed pencil, ideally a kind of set squares, and a pair of compasses for justification of the technique employed are required for the construction of fundamental figures such as a triangle, bisecting a line segment, or drawing a perpendicular at a point on a line. It is the basic concept of construction. Refer to Vedantu’s NCERT Solutions for Class 10 Chapter 11 to understand the concepts of this chapter.
6. What all constructions can be learnt in Chapter 11 Constructions of Class 10 Maths?
In this chapter, you'll learn how to determine a point by internally dividing a line segment given a ratio, how to construct similar triangles, how to construct a tangent to a circle, how to construct a pair of tangents, and how to construct a pair of tangents that are inclined to each other at an angle. You will learn more new definitions and interesting techniques in this chapter.
7. How much time do students need to complete Chapter 11 Constructions of Class 10 Maths?
There are only two exercises in Chapter 11 of Class 10 Maths. It will not take much time to do these exercises if your concepts are clear. Construction needs correct skills and patience. A lot of practice is required to do the construction in the right way. Vedantu offers NCERT Solutions for Class 10 Chapter 11 to help students understand the chapter better and score good marks in exams.
8. How many questions and examples are there in Chapter 11 Constructions of Class 10 Maths?
There are only two exercises in Chapter 11 of Class 10 Maths. Completing the exercises won’t take much time if your concepts are clear. So, there are 14 questions in and just two examples along with it to practise. Practicing all the questions honestly will be enough. If you have doubts, you can refer to NCERT Solutions for Class 10 Chapter 11 offered by Vedantu at free of cost available on the official website and on the Vedantu app. These solutions are prepared by experts in an easy to understand language and students can download them for free.
9. What should be the strategy for preparing Chapter 11 Constructions of Class 10 Maths?
A certain amount of practise and review is necessary for every exam. During the examinations, time management and how to approach difficult problems are also crucial. Vedantu's NCERT Solutions will offer you plenty of practise with a wide range of questions. The annotations, as well as the answer, will help you understand the idea. Construction is a crucial geometry subject in Math, and prior knowledge of how to create an angle bisector and draw a perpendicular line will come in handy.