Class 6 Exam > Class 6 Notes > Mathematics (Maths) Class 6 > RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2
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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
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Document Description: RD Sharma Solutions: Basic Geometrical Tools Exercise 18.2 for Class 6 2024 is part of Mathematics (Maths) Class 6 preparation.
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