In the previous section we saw that, equally spaced red parallel lines will cut equal distances on the cyan lines. In this section, we will see a practical application of this property.
Consider a line 7 cm in length. We can easily divide this line into two equal parts:
Consider a line 7 cm in length. We can easily divide this line into two equal parts:
• One method is to draw a
perpendicular bisector of the line
• Another method is to mark a
point, 3.5 cm from one end of the line
■ But what if we want to divide
the 7 cm long line into 3 equal parts?
We cannot take exact 7⁄3
cm on a scale. So we must think of an alternate
method
• We have seen that a set of
equally spaced parallel lines will cut any line into equal parts.
• In our case, we want 3 equal
parts. So we will need 4 parallel lines. These parallel lines should
be equally spaced. The easiest way to achieve this is as follows:
1. Draw a horizontal cyan line 6 cm
long
2. Mark 2 cm points on it
3. Draw red vertical lines
through the markings and also through the ends. This is shown in the fig.18.9(a) below:
 |
| Fig.18.9 |
• The single cyan line, and the four red parallel lines together is our
‘tool’. We are going to use this tool to divide a 7 cm long line
into 3 equal parts.
[Note that any set of 'equally spaced parallel lines' can be used as a tool. They can be vertical, horizontal or slanting. Also the spacing must be appropriate for the problem under consideration]
4. Mark any convenient point A on
the first vertical line. With A as centre, draw an arc of radius 7
cm, cutting the last vertical line at B. Join AB. This is shown in
fig.b
5. AB is a line of length 7 cm,
and it is divided into 3 equal parts. The job is done.
A different situation:
In the above example, the cyan
line and the 4 red parallel lines is our ‘tool’. The 7 cm long
line is our ‘subject’. We brought the subject to the tool. But
this may not be always possible.
Consider the plan of a building. A 7
cm line AB in the plan may represent the wall of a room. We want to
divide that wall into 3 equal parts. Obviously, we cannot take out
the line PQ and bring it to the tool. So we have to bring the tool to
the subject. How can that be done?
1. In the fig.18.10(a) below, a 6 cm
long cyan line AC is drawn in any convenient direction from the point A.
 |
| Fig.18.10 |
2. 2 cm points are marked on it
3. B is joined to C
4. Through the 2 cm points on the
cyan line, draw red lines parallel to BC
5. These red parallel lines are
equally spaced, and so, they will divide PQ into 3 equal parts. Note that we do not need to draw a red line through A. This is because it coincides with one end of the 'subject', which is the 7 cm long line AB.
Division using circle
In our note books, we have
ruled lines. Those lines are parallel, and are equally spaced as
shown in fig.18.11(a). Can we use them to divide a line of any given length,
into equal parts?
We can. Suppose we want to
divide a line of length ‘x’ into equal parts. The procedure is as
follows:
 |
| Fig.18.11 |
1. Mark a convenient point O on a
line, near the lower left corner of the page
2. With O as centre, draw an arc
(shown in green colour in fig.18.11.b) of radius x cm, cutting the ruled
lines
3. Mark the points of
intersection of the arc with the ruled lines as A, B, C etc.,
4. So OA, OB, OC etc., will all have
the same length x cm
5. Now consider OA. Three equally
spaced parallel lines pass through OA. They are:
♦ One line through O
♦ One line through A
♦ One intermediate line
• So the line OA is divided into
2 equal parts
6. Consider OB. Four equally
spaced parallel lines pass through OB. They are:
♦ One line through O
♦ One line through A
♦ Two intermediate lines
• So the line OB is divided into
3 equal parts
7. Consider OC. Five equally
spaced parallel lines pass through OC. They are:
♦ One line through O
♦ One line through A
♦ Three intermediate lines
• So the line OC is divided into
4 equal parts
■ Thus we see that the equally
spaced ruled lines are a convenient tool for division of lines into
equal parts.
Now we will see a solved
example which shows the application of both theorems 18.1 and 18.2
Solved example 18.1
Draw a horizontal line 9 cm
long and divide it into 7 equal parts. Draw another 9 cm long
horizontal line and divide it in the ratio 3:4
Solution:
Part (i): When a 9 cm long
line is divided into 7 equal parts, each part will be 9⁄7
= 12⁄7
cm long. We cannot measure this length
accurately. So we use an alternate method as shown in fig.18.12(a):
 |
| Fig.18.12 |
1. AB is the 9 cm long line. Draw
another cyan line AC, 7 cm long in any convenient direction.
[We can draw AC at any convenient direction. But for this particular problem, we will draw AC at a convenient angle, say 30o or 35o with AB. We must note down this angle so that it can be used in part (ii) of the problem]
2. Mark 1 cm points on AC. Thus
AC is divided into 7 equal parts
3. Join BC. Draw red lines
parallel to BC through the 1 cm points on AC.
4. These red parallel lines will
divide AB into 7 equal parts. This is a direct application of theorem
18.2
Part (ii): A 9 cm long line PQ
is to be divided in the ratio 3:4. Let the point of division be T
• If the line is divided into 7
equal parts, each part will be equal to 9⁄7
cm.
• PT will take up 3 of those
parts. So PT = 3 × 9⁄7
= 27⁄7
= 36⁄7
cm.
• QT will take up 4 of those
parts. So QT = 4 × 9⁄7
= 36⁄7
= 51⁄7
cm.
• We cannot measure these
lengths accurately. So we use the alternate method as shown in fig.b
1. PQ is the 9 cm long line. Draw
another cyan line PR, 7 cm long at any convenient angle to PQ. We will use the same ∠CAB used in fig.a. This will help to make a
comparison.
2. Mark a point S at a distance
of 3 cm from P. So PS = 3 cm and SR = 4 cm. Thus PR is divided in the
ratio 3:4. But we want such a division on PQ
3. Join QR. Draw red line
parallel to QR through S.
4. This red parallel line will
meet PQ at T
5. Then PT : TQ = PS: SR =3 : 4.
This is a direct application of theorem 18.1. Because, the red
parallel lines cut through two lines PR and PQ
■ Note that 3 × 12⁄7
= 36⁄7
and 4 × 12⁄7
= 51⁄7
. Also note that the line ST in fig.b is the same 'thick red line' in
fig.a
Solved example 18.2
Draw a rectangle whose sides
are in the ratio 3:5, and perimeter is 17 cm
Solution:
• We have: perimeter p = 2 ×
(l+b) = 17 cm. So (l+b) = 17⁄2
= 8.5 cm
• That is., sum of the sides =
8.5 cm.
■ So we have two conditions to
satisfy:
♦ Sum of sides must be equal
to 8.5 cm (∵ perimeter is to be 17 cm)
♦ Ratio of sides must be
equal to 3:5
The steps are given below:
1. Draw a line AB of length 8.5
cm. divide it at a point E in the ratio 3:5. This is shown in the
fig.18.13(a):
 |
| Fig.18.13 |
The procedure for such a
division is clear from the fig.a. However, we will write the steps:
• First we draw a 8 cm long line
AC at any convenient angle with AB
• Then we mark D at a distance
of 5 cm from A. So AD:DC = 5:3
• Next we join CB, and then draw
DE parallel to CB. Thus AE:EB = 5:3
• Now we have the 8.5 cm long
line AB divided in the ratio 5:3
2. The longer segment AE is the
length, and the shorter segment EB is the width of the rectangle.
• So we must make EB
perpendicular to AE
3. For that, draw EF’
perpendicular to AB. With E as centre, and EB as radius, draw an arc
cutting EF’ at F.
4. So EF = EB, and EF is the
width of the rectangle.
5. Draw FG parallel to AE and AG
parallel to EF
■ AEFG is the required
rectangle.
The same rectangle AEFG can be drawn in a different position. It is shown in fig. 18.14(b) below:
 |
| Fig.18.14 |
The steps are the same upto (2). In (3), there is a change. We will rewrite the steps from (3):
3. Draw EF’ perpendicular to AB towards the bottom side of line AB. With E as centre, and EB as radius, draw an arc cutting EF’ at F.
4. So EF = EB, and EF is the width of the rectangle.
5. Draw FG parallel to AE and AG parallel to EF
■ AEFG is the required rectangle. The advantage of using this method is that, the final rectangle will not over lap with the construction lines.
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