In the previous section we proved that the ratio is constant for 3 parallel lines. In this section, we will see 4 Parallel lines. Also later in this section, we will see equally spaced parallel lines.
In fig.18.5 below, 4 red parallel lines
Line A, Line B, Line C and Line D are drawn in red colour. Across these
parallel lines, Line 0 is drawn in a perpendicular direction. Lines
(1), (2) and (3) are drawn in random directions. These transverse
lines are drawn in cyan colour.
These cyan lines intersect the red
parallel lines at A0 ,
B0 ,
A1 ,
A2 etc.,
For our present discussion, we want the distances between these
points of intersection. Those distances are also marked in the fig.18.5.
 |
| Fig.18.5 |
From the fig, we can see that
the distance A0B0= 1.50 cm, B0C0
= 4.50 cm, A2B2
= 1.82 cm etc.,
Once we mark all the distances, we can begin the calculations:
Once we mark all the distances, we can begin the calculations:
1. Take the ratio A0B0
: B0C0
: C0D0
• It will be equal to 1.50 :
4.50 : 3.00
• Reducing to the lowest
term, 1.50 : 4.50 : 3.00 is equal to 1:3:2 [∵ 4.50 ÷ 1.50 = 3 and
3.00 ÷ 1.5 = 2]
• That is., A0B0
: B0C0
: C0D0
= 1:3:2
2. Now take the corresponding
ratio along Line 1.
• We get A1B1
: B1C1
: C1D1
= 1.55 : 4.65 : 3.10
• Reducing to the lowest
term, 1.55 : 4.65 : 3.10 is equal to 1:3:2 [∵ 4.65 ÷ 1.55 = 3 and
3.10 ÷ 1.55 = 2]
• That is., A1B1
: B1C1
: C1D1
= 1:3:2
3. Now take the corresponding
ratio along Line 2.
• We get A2B2
: B2C2
: C2D2
= 1.82 : 5.46 : 3.64
• Reducing to the lowest
term, 1.82 : 5.46 : 3.64 is equal to 1:3:2 [∵ 5.46 ÷ 1.82 = 3 and
3.64 ÷ 1.82 = 2]
• That is., A2B2
: B2C2
: C2D2
= 1:3:2
4. Finally, take the
corresponding ratio along the last Line 3
• We get A2B2
: B2C2
: C2D2
= 1.82 : 5.46 : 3.64
• Reducing to the lowest
term, 1.82 : 5.46 : 3.64 is equal to 1:3:2 [∵ 5.46 ÷ 1.82 = 3 and
3.64 ÷ 1.82 = 2]
• That is., A2B2
: B2C2
: C2D2
= 1:3:2
■ So we get the same ratio 1:3:2 in all the four cases. In fact, we will get the same ratio for any number of transverse cyan lines that we draw, in any direction that we like.
■ The distance between lines B
and C will always be 3 times the distance between Lines A and B
• Also, the distance between
lines C and D will always be 2 times the distance between Lines A and
B
■ For the four red parallel
lines in fig.18.4, the constant ratio is 1:3:2
■ We will write a general form applicable to any
ratio:
• Let the constant ratio be p:q:r
• p:q:r can be written as 1:(q⁄p):(r⁄p)
• The distance between lines B
and C will always be q⁄p
times the distance between Lines A and B
• Also, the distance between
lines C and D will always be r⁄p
times the distance between Lines A and B
This property can be used to
determine unknown quantities. Let us see a quick example:
In fig.18.6(a) below, AB, GH, EF
and CD are parallel. Determine the lengths BH and FC
Solution:
 |
| Fig.18.6 |
• Given that AB, GH, EF and CD are
parallel. So they can be considered as the red parallel lines that we
saw in fig.18.1. This is shown in fig.18.6(b).
• AD and BC are
transverse lines that cross the parallel lines. What ever be the
direction of the transverse lines, the ratio of the distances will be
a constant.
• So we can write: AG:GE:ED = BH:HF:FC = 2:3:1
• 2:3:1 can be written as 1: 3⁄2 : 1⁄2
• Thus we get: AG:GE:ED = BH:HF:FC = 2:3:1 = 1: 3⁄2 : 1⁄2
• That means: GE is 3⁄2 times AG.
Also, ED is 1⁄2 times AG.
• Thus we get: AG:GE:ED = BH:HF:FC = 2:3:1 = 1: 3⁄2 : 1⁄2
• That means: GE is 3⁄2 times AG.
Also, ED is 1⁄2 times AG.
• HF will be 3⁄2
times BH.
Also, FC will be 1⁄2 times BH.
Also, FC will be 1⁄2 times BH.
• So 3.95 = 3⁄2
× BH
♦ Thus BH = 3.95 × 2⁄3
= 2.63
■ We have AG:GE:ED = 2:3:1. This gives us the following information:
• If the total length from A to D is divided into 6 equal parts,
♦ AG will take up 2 parts
♦ GE will take up 3 parts
♦ DE will take up 1 part.
• BH:HF:FC is also the same 2:3:1
• So, if the total length from B to C is divided into 6 equal parts,
♦ BH will take up 2 parts
♦ HF will take up 3 parts
♦ FC will take up 1 part.
• Thus we get: 3 parts out of 6 equal parts of BC = 3.95
• That is: BC × 3⁄6 = 3.95 ⇒ BC = 3.95 × 6⁄3 = 7.9
• Now, 2 parts out of 6 equal parts of BC is BH
• That is: BC × 2⁄6 = BH ⇒ BH = 7.9 × 2⁄6 = 2.63
• Similarly, 1 part out of 6 equal parts of BC is FC
• That is: BC × 1⁄6 = FC ⇒ FC = 7.9 × 1⁄6 = 1.32.
• FC = 1⁄2 × BH
♦ Thus FC = 2.63 × 1⁄2 = 1.32.
Another method:■ We have AG:GE:ED = 2:3:1. This gives us the following information:
• If the total length from A to D is divided into 6 equal parts,
♦ AG will take up 2 parts
♦ GE will take up 3 parts
♦ DE will take up 1 part.
• BH:HF:FC is also the same 2:3:1
• So, if the total length from B to C is divided into 6 equal parts,
♦ BH will take up 2 parts
♦ HF will take up 3 parts
♦ FC will take up 1 part.
• Thus we get: 3 parts out of 6 equal parts of BC = 3.95
• That is: BC × 3⁄6 = 3.95 ⇒ BC = 3.95 × 6⁄3 = 7.9
• Now, 2 parts out of 6 equal parts of BC is BH
• That is: BC × 2⁄6 = BH ⇒ BH = 7.9 × 2⁄6 = 2.63
• Similarly, 1 part out of 6 equal parts of BC is FC
• That is: BC × 1⁄6 = FC ⇒ FC = 7.9 × 1⁄6 = 1.32.
So we find that the ratio is a constant for 4 parallel lines also. The proof can be written in the same way as we did for 3 parallel lines.
In the examples that we saw so far, the red parallel lines were all horizontal. But the property will work in any direction. That is., the parallel lines can be slanting as in fig.18.7(i) below, or even vertical as in fig.18.7(ii)
The discussion that we had so
far in this chapter can be written in the form of a theorem. We will
write it in a step by step manner:
In the examples that we saw so far, the red parallel lines were all horizontal. But the property will work in any direction. That is., the parallel lines can be slanting as in fig.18.7(i) below, or even vertical as in fig.18.7(ii)
 |
| Fig.18.7 |
Theorem 18.1
1. We have a set of cyan
coloured lines
2. Three or more red parallel
lines cut through the cyan lines
3. Take the ratio of the
distances cut in any one cyan line.
4. This ratio will be same for
all the cyan lines
Let us see some important
points that has to be noted in the above steps:
• In (1), the minimum number
of cyan lines must be 2. Then only we can apply the theorem.
♦ But there is no upper
limit. There can be any number of cyan lines greater than 2
• In (2), the minimum number
of red parallel lines is specified as 3. This is because, to take a
ratio, we need a minimum of two distances, and to cut two distances,
we must have a minimum of 3 lines
♦ But there is no upper
limit. There can be any number of red parallel lines greater than 3
Now we will discuss a special
case of the above theorem.
Consider any one cyan line.
The red parallel lines cut distances on this cyan line. Suppose that
all the distances cut on that single line are equal.
Now, when does such an equal
division on a single line occur?
When the distances between the
red parallel lines are equal, the distances cut on any one cyan line
will also be equal. Then we can write the ratio a:b:c as a:a:a.
That is., a:b:c = a:a:a
a:a:a can be simply written as
1:1:1
Now, we know that a:b:c =
p:q:r = l:m:n
So we can write: a:b:c = p:q:r
= l:m:n = 1:1:1
That means, if the red
parallel lines cut equal distances on any one cyan line, they will
cut equal distances on all other cyan lines also. We will write it as
a theorem.
Theorem 18.2
1. We have a set of cyan
coloured lines
2. Three or more red parallel
lines cut through the cyan lines
3. The distances cut in any
one cyan line are equal.
4. Then the distances cut in
each of the other cyan lines will also be equal
An example is shown in fig.18.8 below:
 |
| Fig.18.8 |
• In the above fig., the first cyan line is
perpendicular to the red parallel lines. So the distance measured
along this cyan line will give us the actual distances between the
red parallel lines.
• We find that all those distances are 3. That
means, the red parallel lines are equally spaced.
• These equally spaced red
parallel lines will cut equal distances on all the cyan lines
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