Class 6 Exam > Class 6 Notes > Mathematics (Maths) Class 6 > RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2
Basic Geometrical Tools Exercise 18.2 RD Sharma Solutions | Mathematics (Maths) Class 6 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
1. Mark two points, A and B on a piece of paper and join them. Measure this length. For each of the
following draw a line segment CD that is:
(i) equal to the segment AB
(ii) twice AB
(iii) three times AB
(iv) half AB
(v) collinear with AB and is equal to it.
Solution:
Mark points A and B on a piece of paper and join them as shown in figure.
Now place the ruler with its edge along AB to measure the length AB, such that the zero mark of the cm side of
the ruler coincides the point A. Read the mark on the ruler, which corresponds to the point B. The value obtained
on the ruler at point B is the length of AB i.e. AB = 5.6 cm
(i) In order to construct the line segment CD equal to AB, take a divider and open it, so that the end-point of one
of its arm is at A and the second arm end-point is at B. Now lift the divider and place the end-points of both hands
on the paper, where CD has to be drawn.
(ii) In order to construct a line segment twice AB, construct a line l and mark a point C on it. Take a divider and
open it so that end-points of both of its arms are at A and B. Then without disturbing its opening lift the divider,
place one end-point at point C and the other end-point on the line l an name it as point E. Now, lift the divider and
place one end-point at E and the other end-point on the line l, opposite to point C and name it as D.
Page 2
1. Mark two points, A and B on a piece of paper and join them. Measure this length. For each of the
following draw a line segment CD that is:
(i) equal to the segment AB
(ii) twice AB
(iii) three times AB
(iv) half AB
(v) collinear with AB and is equal to it.
Solution:
Mark points A and B on a piece of paper and join them as shown in figure.
Now place the ruler with its edge along AB to measure the length AB, such that the zero mark of the cm side of
the ruler coincides the point A. Read the mark on the ruler, which corresponds to the point B. The value obtained
on the ruler at point B is the length of AB i.e. AB = 5.6 cm
(i) In order to construct the line segment CD equal to AB, take a divider and open it, so that the end-point of one
of its arm is at A and the second arm end-point is at B. Now lift the divider and place the end-points of both hands
on the paper, where CD has to be drawn.
(ii) In order to construct a line segment twice AB, construct a line l and mark a point C on it. Take a divider and
open it so that end-points of both of its arms are at A and B. Then without disturbing its opening lift the divider,
place one end-point at point C and the other end-point on the line l an name it as point E. Now, lift the divider and
place one end-point at E and the other end-point on the line l, opposite to point C and name it as D.
(iii) In order to construct the line segment three times A, construct a line l and mark a point C on it. Take a divider
and open it such that the end-points of both of its arms are at A and B
Now lift the divider and place one end-point at point C and the other end-point on the line l and mark it as E.
Again lift the divider and place one end-point at E and the other end-point on the line l, opposite to point C and
mark it as F.
Again lift the divider and place one end-point at F and the other end-point on the line l, opposite to C and mark it
as D.
(iv) In order to construct the line segment that is half AB, construct a line l and mark a point C on it. With the help
of a ruler measure the line segment AB = 5.6 cm
So half of AB = 5.6/2 = 2.8 cm
Take a divider and open it such that one end arm is at 0 and the other end arm is at 2.8 cm.
Lift the divider and place one end at point C and the other end on the line l at point D.
Page 3
1. Mark two points, A and B on a piece of paper and join them. Measure this length. For each of the
following draw a line segment CD that is:
(i) equal to the segment AB
(ii) twice AB
(iii) three times AB
(iv) half AB
(v) collinear with AB and is equal to it.
Solution:
Mark points A and B on a piece of paper and join them as shown in figure.
Now place the ruler with its edge along AB to measure the length AB, such that the zero mark of the cm side of
the ruler coincides the point A. Read the mark on the ruler, which corresponds to the point B. The value obtained
on the ruler at point B is the length of AB i.e. AB = 5.6 cm
(i) In order to construct the line segment CD equal to AB, take a divider and open it, so that the end-point of one
of its arm is at A and the second arm end-point is at B. Now lift the divider and place the end-points of both hands
on the paper, where CD has to be drawn.
(ii) In order to construct a line segment twice AB, construct a line l and mark a point C on it. Take a divider and
open it so that end-points of both of its arms are at A and B. Then without disturbing its opening lift the divider,
place one end-point at point C and the other end-point on the line l an name it as point E. Now, lift the divider and
place one end-point at E and the other end-point on the line l, opposite to point C and name it as D.
(iii) In order to construct the line segment three times A, construct a line l and mark a point C on it. Take a divider
and open it such that the end-points of both of its arms are at A and B
Now lift the divider and place one end-point at point C and the other end-point on the line l and mark it as E.
Again lift the divider and place one end-point at E and the other end-point on the line l, opposite to point C and
mark it as F.
Again lift the divider and place one end-point at F and the other end-point on the line l, opposite to C and mark it
as D.
(iv) In order to construct the line segment that is half AB, construct a line l and mark a point C on it. With the help
of a ruler measure the line segment AB = 5.6 cm
So half of AB = 5.6/2 = 2.8 cm
Take a divider and open it such that one end arm is at 0 and the other end arm is at 2.8 cm.
Lift the divider and place one end at point C and the other end on the line l at point D.
(v) In order to construct a line segment CD collinear with AB and is equal to it, use a ruler along AB and draw the
line extended to AB. Take a divider and open it such that both of its end-points are on the points A and B. Lift the
divider and place its end-points of both of its arms on the extended line of AB and mark them as points C and D.
2. The end point P of a line segment PQ is against 4 cm mark and the end-point Q is against the mark
indicating 14.8 cm on a ruler. What is the length of the segment PQ?
Solution:
Now we have to extend the line segment QP towards point 0 of the ruler and take a point 0 on the line QP which
is extended corresponding to point 0 on the ruler.
From the diagram we know that
OP = 4 cm
OQ = 14.8 cm
It can be written as,
PQ = OQ – OP = (14.8 – 4) cm = 10.8 cm
3. Draw a line segment CD. Produce it to CE such that CE = 3CD.
Page 4
1. Mark two points, A and B on a piece of paper and join them. Measure this length. For each of the
following draw a line segment CD that is:
(i) equal to the segment AB
(ii) twice AB
(iii) three times AB
(iv) half AB
(v) collinear with AB and is equal to it.
Solution:
Mark points A and B on a piece of paper and join them as shown in figure.
Now place the ruler with its edge along AB to measure the length AB, such that the zero mark of the cm side of
the ruler coincides the point A. Read the mark on the ruler, which corresponds to the point B. The value obtained
on the ruler at point B is the length of AB i.e. AB = 5.6 cm
(i) In order to construct the line segment CD equal to AB, take a divider and open it, so that the end-point of one
of its arm is at A and the second arm end-point is at B. Now lift the divider and place the end-points of both hands
on the paper, where CD has to be drawn.
(ii) In order to construct a line segment twice AB, construct a line l and mark a point C on it. Take a divider and
open it so that end-points of both of its arms are at A and B. Then without disturbing its opening lift the divider,
place one end-point at point C and the other end-point on the line l an name it as point E. Now, lift the divider and
place one end-point at E and the other end-point on the line l, opposite to point C and name it as D.
(iii) In order to construct the line segment three times A, construct a line l and mark a point C on it. Take a divider
and open it such that the end-points of both of its arms are at A and B
Now lift the divider and place one end-point at point C and the other end-point on the line l and mark it as E.
Again lift the divider and place one end-point at E and the other end-point on the line l, opposite to point C and
mark it as F.
Again lift the divider and place one end-point at F and the other end-point on the line l, opposite to C and mark it
as D.
(iv) In order to construct the line segment that is half AB, construct a line l and mark a point C on it. With the help
of a ruler measure the line segment AB = 5.6 cm
So half of AB = 5.6/2 = 2.8 cm
Take a divider and open it such that one end arm is at 0 and the other end arm is at 2.8 cm.
Lift the divider and place one end at point C and the other end on the line l at point D.
(v) In order to construct a line segment CD collinear with AB and is equal to it, use a ruler along AB and draw the
line extended to AB. Take a divider and open it such that both of its end-points are on the points A and B. Lift the
divider and place its end-points of both of its arms on the extended line of AB and mark them as points C and D.
2. The end point P of a line segment PQ is against 4 cm mark and the end-point Q is against the mark
indicating 14.8 cm on a ruler. What is the length of the segment PQ?
Solution:
Now we have to extend the line segment QP towards point 0 of the ruler and take a point 0 on the line QP which
is extended corresponding to point 0 on the ruler.
From the diagram we know that
OP = 4 cm
OQ = 14.8 cm
It can be written as,
PQ = OQ – OP = (14.8 – 4) cm = 10.8 cm
3. Draw a line segment CD. Produce it to CE such that CE = 3CD.
Solution:
Draw a line l and take two points C and D on it.
Now take a divider and open it such that its end of both arms are at the points C and D.
Lift the divider and place its one end at D and the other end on the line l opposite to C and mark it as point A.
Again lift the divider and place its one end at A and the other end on the line l opposite to C and mark it as point
E.
We know that CD = DE = AE
It can be written as,
CE= CD + DE + AE
We know that CD = DE = AE
So we get
CE = = CD + CD + CD = 3CD
We get
CE = 3CD
4. If AB = 7.5 cm and CD = 2.5 cm, construct a segment whose length is equal to
(i) AB – CD
(ii) 2AB
(iii) 3CD
(iv) AB + CD
(v) 2AB + 3CD
Solution:
It is given that
AB = 7.5 cm and CD = 2.5 cm
Construct AB and CD
(i) Construct a line l and mark a point E on it.
Take a divider and open it such that both of its end arms are at points A and B. Now, lift the divider and place its
one end at E and other end at F on the line l. Reset the divider by making one end of its one hand is at C and the
end of the other hand is at D. Again lift the divider and place its one end at E and other end at G on the line l. FG
is required line segment having the length equal to (AB – CD).
Page 5
1. Mark two points, A and B on a piece of paper and join them. Measure this length. For each of the
following draw a line segment CD that is:
(i) equal to the segment AB
(ii) twice AB
(iii) three times AB
(iv) half AB
(v) collinear with AB and is equal to it.
Solution:
Mark points A and B on a piece of paper and join them as shown in figure.
Now place the ruler with its edge along AB to measure the length AB, such that the zero mark of the cm side of
the ruler coincides the point A. Read the mark on the ruler, which corresponds to the point B. The value obtained
on the ruler at point B is the length of AB i.e. AB = 5.6 cm
(i) In order to construct the line segment CD equal to AB, take a divider and open it, so that the end-point of one
of its arm is at A and the second arm end-point is at B. Now lift the divider and place the end-points of both hands
on the paper, where CD has to be drawn.
(ii) In order to construct a line segment twice AB, construct a line l and mark a point C on it. Take a divider and
open it so that end-points of both of its arms are at A and B. Then without disturbing its opening lift the divider,
place one end-point at point C and the other end-point on the line l an name it as point E. Now, lift the divider and
place one end-point at E and the other end-point on the line l, opposite to point C and name it as D.
(iii) In order to construct the line segment three times A, construct a line l and mark a point C on it. Take a divider
and open it such that the end-points of both of its arms are at A and B
Now lift the divider and place one end-point at point C and the other end-point on the line l and mark it as E.
Again lift the divider and place one end-point at E and the other end-point on the line l, opposite to point C and
mark it as F.
Again lift the divider and place one end-point at F and the other end-point on the line l, opposite to C and mark it
as D.
(iv) In order to construct the line segment that is half AB, construct a line l and mark a point C on it. With the help
of a ruler measure the line segment AB = 5.6 cm
So half of AB = 5.6/2 = 2.8 cm
Take a divider and open it such that one end arm is at 0 and the other end arm is at 2.8 cm.
Lift the divider and place one end at point C and the other end on the line l at point D.
(v) In order to construct a line segment CD collinear with AB and is equal to it, use a ruler along AB and draw the
line extended to AB. Take a divider and open it such that both of its end-points are on the points A and B. Lift the
divider and place its end-points of both of its arms on the extended line of AB and mark them as points C and D.
2. The end point P of a line segment PQ is against 4 cm mark and the end-point Q is against the mark
indicating 14.8 cm on a ruler. What is the length of the segment PQ?
Solution:
Now we have to extend the line segment QP towards point 0 of the ruler and take a point 0 on the line QP which
is extended corresponding to point 0 on the ruler.
From the diagram we know that
OP = 4 cm
OQ = 14.8 cm
It can be written as,
PQ = OQ – OP = (14.8 – 4) cm = 10.8 cm
3. Draw a line segment CD. Produce it to CE such that CE = 3CD.
Solution:
Draw a line l and take two points C and D on it.
Now take a divider and open it such that its end of both arms are at the points C and D.
Lift the divider and place its one end at D and the other end on the line l opposite to C and mark it as point A.
Again lift the divider and place its one end at A and the other end on the line l opposite to C and mark it as point
E.
We know that CD = DE = AE
It can be written as,
CE= CD + DE + AE
We know that CD = DE = AE
So we get
CE = = CD + CD + CD = 3CD
We get
CE = 3CD
4. If AB = 7.5 cm and CD = 2.5 cm, construct a segment whose length is equal to
(i) AB – CD
(ii) 2AB
(iii) 3CD
(iv) AB + CD
(v) 2AB + 3CD
Solution:
It is given that
AB = 7.5 cm and CD = 2.5 cm
Construct AB and CD
(i) Construct a line l and mark a point E on it.
Take a divider and open it such that both of its end arms are at points A and B. Now, lift the divider and place its
one end at E and other end at F on the line l. Reset the divider by making one end of its one hand is at C and the
end of the other hand is at D. Again lift the divider and place its one end at E and other end at G on the line l. FG
is required line segment having the length equal to (AB – CD).
(ii) Construct a line l and mark a point E on it. Take a divider and open such that both its end arm are at A and B
Lift the divider and place its one end at E and the other end as F on the line l
Again, lift the divider and place its one end at F and the other end on the line l, opposite to E and mark it as G
EG is the required line segment having the length equal to 2 AB.
(iii) Construct a line l and mark a point E on it. Take a divider and open it such that both of its ends are at C and D
respectively.
Lift the divider and place its end at point E on it and the other end F on the line l.
Again, lift the divider and mark the end as G on the line opposite to C.
Lift the divider and place its one end at G and another end H on the line l, opposite to E
EH is required line segment having the length equal to 3 CD
Read More
|
94 videos|347 docs|54 tests
|
Related Exams
About this Document
|
4.74/5 Rating |
|
Nov 23, 2024 Last updated |
Document Description: RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 for Class 6 2024 is part of Mathematics (Maths) Class 6 preparation.
The notes and questions for RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 have been prepared according to the Class 6 exam syllabus. Information about RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 covers topics
like and RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 Example, for Class 6 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2.
Introduction of RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 in English is available as part of our Mathematics (Maths) Class 6
for Class 6 & RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 in Hindi for Mathematics (Maths) Class 6 course.
Download more important topics related with notes, lectures and mock test series for Class 6
Exam by signing up for free. Class 6: Basic Geometrical Tools Exercise 18.2 RD Sharma Solutions | Mathematics (Maths) Class 6
Description
Full syllabus notes, lecture & questions for Basic Geometrical Tools Exercise 18.2 RD Sharma Solutions | Mathematics (Maths) Class 6 - Class 6 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 6 | Best notes, free PDF download
Information about RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2
In this doc you can find the meaning of RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 defined & explained in the simplest way possible. Besides explaining types of
RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 tests, examples and also practice Class 6
tests
|
Explore Courses for Class 6 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Exam
,Important questions
,Basic Geometrical Tools Exercise 18.2 RD Sharma Solutions | Mathematics (Maths) Class 6
,Semester Notes
,Free
,video lectures
,Sample Paper
,Basic Geometrical Tools Exercise 18.2 RD Sharma Solutions | Mathematics (Maths) Class 6
,ppt
,Basic Geometrical Tools Exercise 18.2 RD Sharma Solutions | Mathematics (Maths) Class 6
,MCQs
,Summary
,past year papers
,Extra Questions
,shortcuts and tricks
,study material
,Viva Questions
,Objective type Questions
,practice quizzes
,mock tests for examination
,Previous Year Questions with Solutions
;
Additional Information about RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 for Class 6 Preparation
RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 Free PDF Download
The RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 is an invaluable resource that delves deep into the core of the Class 6 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 now and kickstart your journey towards success in the Class 6 exam.
Importance of RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2
The importance of RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 cannot be overstated, especially for Class 6 aspirants.
This document holds the key to success in the Class 6 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 Notes
RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 Notes on EduRev are your ultimate resource for success.
RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 Class 6 Questions
The "RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 Class 6 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 6 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 on the App
Students of Class 6 can study RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 is prepared as per the latest Class 6 syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup