In the previous section we saw how to use 'scale' to represent a long iron rod on a small sheet of paper. In this section, we will see more applications of 'scale'.
We have seen how to draw an iron rod. It has only one dimension, which is the length. But most objects have two dimensions: Length and Width. Examples are: Floor of a room, Area of a play ground, Surface of a Lake etc., We must be able to make drawings of such ‘areas’ also.
Consider a frame ABCD shown in fig below. It is made by welding 3 different segments of iron rods. Segments AB, BC, and CD.
The longest segment AB has a length of 250 cm. The segment BC has a length of 150 cm, and is at a right angle to AB. And finally, CD has a length of 100 cm, and is at an angle of 70o to BC. Obviously, it is a large object, and we can not draw it to same size. We must use a scale and make the drawing smaller. Let us use 1:50. We must use this 1:50 for all the segments of the frame. So:
Line ab on paper which represents AB will have a length equal to L/S = 250/50 = 5 cm
Line bc on paper which represents BC will have a length equal to L/S = 150/50 = 3 cm
Line cd on paper which represents CD will have a length equal to L/S = 100/50 = 2 cm
So the final drawing on paper will be as shown below:
Note that the scale is not applied to angles. So 90o and 70o remains the same. Let us make a table of two quantities: (1) The actual lengths L and (2) The corresponding lengths on paper l. The table will be as shown below:
We find that l/L is a constant. Note that the constant 0.02 is the reciprocal of S (1/S = 1/50 =0.02). It is a case of direct proportion. So, if more segments are added to the actual frame, their actual lengths L, and their corresponding lengths l on paper will have appropriate places in the table. From that table, we can find the length l that has to be used to represent any new segment which is added.
For example, suppose a new segment of 400 cm is added to the frame. This new length of 400 cm, and it's corresponding length l1 will have a place in the table. So the table can be modified as:
This l1 and 400 will give the same constant 0.02. So we can write:
l1/400 = 0.02 ⇒ l1 = 0.02 × 400 = 8 cm. Thus we represent the new segment with a line 8 cm long.
From the above discussion, we can make some important conclusions:
We have seen that for a particular map or drawing, any segment on it, and it's corresponding actual length form a pair. This pair can be placed in a column in the table. So there are numerous columns possible, each column containing a pair.
• Each column will give the same constant.
• And this constant is equal to the reciprocal of 'S'.
• So one table having a particular constant belongs to a particular map.
• That table cannot be used for another map having a different scale. Because it will be having a different constant
Another method:AB, BC, and CD give us 3 columns to fill in the table. But there is one more. We are making the drawing at 1:50 scale. So if there is any segment with a length of 50 cm in the actual frame, then that segment will be represented by a 1 cm line in the drawing. With just this one information we can fill up the first column of the table, and then any other length can be calculated. Such a table to calculate the length for the 400 cm segment is given below:
The column (iii) is filled up using the scale 1:S. For our present problem S is 50. It gives a constant of 0.02. So l1/400 should also give the same constant. Thus we have:
l1/400 = 0.02 ⇒ l1 = 0.02 × 400 = 8 cm
Another important conclusion:
We have seen how to draw an iron rod. It has only one dimension, which is the length. But most objects have two dimensions: Length and Width. Examples are: Floor of a room, Area of a play ground, Surface of a Lake etc., We must be able to make drawings of such ‘areas’ also.
Consider a frame ABCD shown in fig below. It is made by welding 3 different segments of iron rods. Segments AB, BC, and CD.
 |
| Fig.3.9 |
The longest segment AB has a length of 250 cm. The segment BC has a length of 150 cm, and is at a right angle to AB. And finally, CD has a length of 100 cm, and is at an angle of 70o to BC. Obviously, it is a large object, and we can not draw it to same size. We must use a scale and make the drawing smaller. Let us use 1:50. We must use this 1:50 for all the segments of the frame. So:
Line ab on paper which represents AB will have a length equal to L/S = 250/50 = 5 cm
Line bc on paper which represents BC will have a length equal to L/S = 150/50 = 3 cm
Line cd on paper which represents CD will have a length equal to L/S = 100/50 = 2 cm
So the final drawing on paper will be as shown below:
 |
| Fig.3.10 |
Note that the scale is not applied to angles. So 90o and 70o remains the same. Let us make a table of two quantities: (1) The actual lengths L and (2) The corresponding lengths on paper l. The table will be as shown below:
We find that l/L is a constant. Note that the constant 0.02 is the reciprocal of S (1/S = 1/50 =0.02). It is a case of direct proportion. So, if more segments are added to the actual frame, their actual lengths L, and their corresponding lengths l on paper will have appropriate places in the table. From that table, we can find the length l that has to be used to represent any new segment which is added.
For example, suppose a new segment of 400 cm is added to the frame. This new length of 400 cm, and it's corresponding length l1 will have a place in the table. So the table can be modified as:
This l1 and 400 will give the same constant 0.02. So we can write:
l1/400 = 0.02 ⇒ l1 = 0.02 × 400 = 8 cm. Thus we represent the new segment with a line 8 cm long.
From the above discussion, we can make some important conclusions:
We have seen that for a particular map or drawing, any segment on it, and it's corresponding actual length form a pair. This pair can be placed in a column in the table. So there are numerous columns possible, each column containing a pair.
• Each column will give the same constant.
• And this constant is equal to the reciprocal of 'S'.
• So one table having a particular constant belongs to a particular map.
• That table cannot be used for another map having a different scale. Because it will be having a different constant
Another method:AB, BC, and CD give us 3 columns to fill in the table. But there is one more. We are making the drawing at 1:50 scale. So if there is any segment with a length of 50 cm in the actual frame, then that segment will be represented by a 1 cm line in the drawing. With just this one information we can fill up the first column of the table, and then any other length can be calculated. Such a table to calculate the length for the 400 cm segment is given below:
The column (iii) is filled up using the scale 1:S. For our present problem S is 50. It gives a constant of 0.02. So l1/400 should also give the same constant. Thus we have:
l1/400 = 0.02 ⇒ l1 = 0.02 × 400 = 8 cm
Another important conclusion:
We find from table that l/L is a constant. So we must use a constant scale for the whole portion of a drawing. We must not use one scale 1:S1 for some segments and another scale 1:S2 for other segments. In that case, the drawing will not be proportional to the actual object.
Now we will see some solved examples based on the above discussion.
Now we will see some solved examples based on the above discussion.
Solved example 3.9The scale indicated on a map is 1 cm = 12 km. On that map, the distance between two cities is found to be 3.6 cm. What is the actual distance between the two cities.
Solution:
The scale of the map is 1 cm = 12 km. If the actual distance between any two places is 12 km, then the distance between them on the map will be 1 cm. We can use this information to fill one column of the table.
The whole map would be drawn with a constant scale. So 3.6 and L1 should also give the same ratio of 0.08. Thus we can write:
3.6/L1 = 0.08 ⇒ L1 = 3.6/0.08 = 45.0 km
Solved example 3.10
The actual distance between two cities A and B is 480 km. On a map, the distance between them is 6 cm. On the same map, the distance between another two cities C and D is 4 cm. (i) What is the actual distance between C and D? (ii) What is the scale of the map?
Solution:
For A and B, L = 480 km and l = 6cm. So l/L = 6/480 = 0.0125
(i) For C and D, l1 is given as 4. We have to find L1. This L1 and 4 have appropriate places in the table. They must give the same ratio 0.0125
(ii) Let the scale be 1 cm = 'S' km. So, if the actual distance between any two places is 'S' km, then the distance between them on the map will be 1 cm. So this 'S' km and 1 cm should have appropriate places in the table. And they should give the same ratio 0.0125
With this two information, we can form the table:
(i) we can write:
4.0/L1 = 0.0125 ⇒ L1 = 4.0/0.0125 = 320.0 km
(ii) we can write:
1.0/S = 0.0125 ⇒ S = 1.0/0.0125 = 80.0 km
So the scale of the map is 1 cm = 80 km.
We can also use the symbol ':' to write the scale. But in that method, we do not indicate the unit. That means, the unit should be same on both the sides. We will convert 80 km to cm. The steps are as follows:
• 1 km = 1000 m • 1 m = 100 cm • So 1 km = 1000 × 100 = 100000 cm • Thus 80 km = 8000000 cm.
So the scale is 1:8000000
In the next section we will see a few more solved examples.
Solution:
The scale of the map is 1 cm = 12 km. If the actual distance between any two places is 12 km, then the distance between them on the map will be 1 cm. We can use this information to fill one column of the table.
The whole map would be drawn with a constant scale. So 3.6 and L1 should also give the same ratio of 0.08. Thus we can write:
3.6/L1 = 0.08 ⇒ L1 = 3.6/0.08 = 45.0 km
Solved example 3.10
The actual distance between two cities A and B is 480 km. On a map, the distance between them is 6 cm. On the same map, the distance between another two cities C and D is 4 cm. (i) What is the actual distance between C and D? (ii) What is the scale of the map?
Solution:
For A and B, L = 480 km and l = 6cm. So l/L = 6/480 = 0.0125
(i) For C and D, l1 is given as 4. We have to find L1. This L1 and 4 have appropriate places in the table. They must give the same ratio 0.0125
(ii) Let the scale be 1 cm = 'S' km. So, if the actual distance between any two places is 'S' km, then the distance between them on the map will be 1 cm. So this 'S' km and 1 cm should have appropriate places in the table. And they should give the same ratio 0.0125
With this two information, we can form the table:
(i) we can write:
4.0/L1 = 0.0125 ⇒ L1 = 4.0/0.0125 = 320.0 km
(ii) we can write:
1.0/S = 0.0125 ⇒ S = 1.0/0.0125 = 80.0 km
So the scale of the map is 1 cm = 80 km.
We can also use the symbol ':' to write the scale. But in that method, we do not indicate the unit. That means, the unit should be same on both the sides. We will convert 80 km to cm. The steps are as follows:
• 1 km = 1000 m • 1 m = 100 cm • So 1 km = 1000 × 100 = 100000 cm • Thus 80 km = 8000000 cm.
So the scale is 1:8000000
In the next section we will see a few more solved examples.
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