In the previous section we saw Equilateral and Isosceles triangles. In this section, we will discuss about the 'Sum of two sides of a triangle'.
Let us do an experiment: We are going to try to make a triangle with three segments 'a', 'b' and 'c'. Segment 'a' has a length of 8 cm, 'b' has a length of 6 cm and 'c' 2 cm. Segment 'a' is fixed as the base. This is shown in the fig.9.15 below.
But look at the animation below. It shows the various positions of ‘b’ and ‘c’. The base ‘a’ is fixed.
No matter what direction we draw ‘b’ and ‘c’, we are not able to complete the triangle. This is because ‘b’ and ‘c’ never meet. Why is this so? Let us analyse:
Fig.9.17(ii) below shows ‘b’ and ‘c’ aligned with ‘a’. Not just aligned, ‘b’ and ‘c’ are directly above ‘a’, and also, the vertices are the same. This is the shortest possible route.
Even in this position, ‘b’ and ‘c’ does not meet. This is because the sum of ‘b’ and ‘c’ which is equal to (4 + 2) = 6, is less than ‘a’ which is equal to 8 cm. That means, lengths of ‘b’ and ‘c’ are shorter than required.
Let us increase the lengths. This is shown in fig.(iii). Now the lengths are such that they just meet. When they just meet, the ‘point of meeting’ lies on ‘a’ itself. To form a triangle, the meeting point must lie outside ‘a’. That means, the sum of ‘b’ and ‘c’ being equal to ‘a’ is not good enough. The sum must be greater than ‘a’. If the sum is a little greater than ‘a’, then we get a triangle with small height as in fig(iv). If the sum is far greater than ‘a’, then we get a triangle with a larger height as in fig.(v)
So we get a condition for 'a given set of 3 sides to form a triangle'. It is: (b + c) must be greater than 'a'. This is same as saying: ‘a’ must be less than (b + c).
In the above experiment, we took ‘a’ as the base. We could do the same experiment with ‘b’ as the base. In that case, ‘b’ must be less than (a + c). Similarly, if ‘c’ is taken as the base, then c must be less than (a + b). Thus three combinations are possible, and the condition should be satisfied for all the 3 combinations:
• 'a' must be less than ('b' + 'c')
• 'b' must be less than ('a' + 'c')
• 'c' must be less than ('a' + 'b')
In general, we can write:
Condition 1:
Any one side of a triangle should be less than the sum of the other two sides.
If we are given three sides, we can check the 'condition 1' for each of the three combinations. If all the 3 combinations satisfy the condition, then a triangle will be formed.
Now let us see another condition:
The fig.9.18(i) given below shows a triangle ABC. Here, 'a' will surely be less than (b + c). Other wise this triangle would not be formed. Now we are going to align the side AB with AC, keeping the vertex A the same. This is shown in fig(ii).
So in fig(ii), the yellow side is directly above the green side. The exposed green portion is the difference between ‘b’ and ‘c’. We will reduce ‘a’. This is shown in fig(iii). We can see that after reduction, ‘a’, which is shown by the red line will never meet the yellow line. And so, a triangle will not be formed. So let us write down the steps to arrive at a condition:
• In fig(iii), we find that the red side does not meet the yellow side
• This is because, the red side is shorter than the exposed green portion
• But this exposed green portion is the difference between ‘b’ and ‘c’
• So it follows that the red side ‘a’ should be greater than the difference between ‘b’ and ‘c’
We have arrived at the above condition by fixing ‘b’ as the base. We could take any other side as the base. We will get similar results. We will write all those possible results:
• ‘a’ should be greater than the difference between ‘b’ and ‘c’
• ‘b’ should be greater than the difference between ‘a’ and ‘c’
• ‘c’ should be greater than the difference between ‘a’ and ‘b’
So we find that, here also there are 3 combinations possible. In general, we can write:
Condition 2:
Any one side of a triangle should be greater than the difference between the other two sides.
A real life situation where we will want to apply the above ‘findings’ is described below:
We are given three lengths. And we are asked to determine whether it is possible to form a triangle with those three lengths.
Solution: We will demonstrate the method of solution with the same lengths 8, 4 and 2 that we saw above. It is always convenient to do the calculations in the form of a table as shown below:
The formation of the above table is as follows:
• When we get the three sides, we split them into two groups: Group 1 and Group 2.
• Group 1 contains one side and Group 2 contains the other two sides.
• Three such combinations are possible.
• In each combination, we add the lengths in Group 2. This is entered in the column no.5.
• We also find the difference. This is entered in the column no.6.
• The sum is compared with the value in Group 1. The result of the comparison is entered in column 7. We must get a ‘yes’ in all the three combinations.
• The difference is compared with the value in Group 1. The result of this comparison is entered in column 8.
• We must get a ‘yes’ in all the three combinations. If we get 3 'yes' in the column no.7, we will surely get 3 'yes' in the column no.8 also.
• We will be able to form a triangle with the given three lengths, only if we get a total of 6 'yes'.
We will now see some solved examples:
Solved example 9.7
Check whether it is possible to form a triangle with sides 3, 4 and 5 cm.
Solution:
We have a total of 6 'yes'. So it is possible to form a triangle with sides 3, 4, and 5 cm
In the next section we will see more solved examples.
Let us do an experiment: We are going to try to make a triangle with three segments 'a', 'b' and 'c'. Segment 'a' has a length of 8 cm, 'b' has a length of 6 cm and 'c' 2 cm. Segment 'a' is fixed as the base. This is shown in the fig.9.15 below.
 |
| Fig.9.15 |
 |
| Fig.9.16 'b' and 'c' never meets |
Fig.9.17(ii) below shows ‘b’ and ‘c’ aligned with ‘a’. Not just aligned, ‘b’ and ‘c’ are directly above ‘a’, and also, the vertices are the same. This is the shortest possible route.
 |
| Fig.9.17 |
Let us increase the lengths. This is shown in fig.(iii). Now the lengths are such that they just meet. When they just meet, the ‘point of meeting’ lies on ‘a’ itself. To form a triangle, the meeting point must lie outside ‘a’. That means, the sum of ‘b’ and ‘c’ being equal to ‘a’ is not good enough. The sum must be greater than ‘a’. If the sum is a little greater than ‘a’, then we get a triangle with small height as in fig(iv). If the sum is far greater than ‘a’, then we get a triangle with a larger height as in fig.(v)
So we get a condition for 'a given set of 3 sides to form a triangle'. It is: (b + c) must be greater than 'a'. This is same as saying: ‘a’ must be less than (b + c).
In the above experiment, we took ‘a’ as the base. We could do the same experiment with ‘b’ as the base. In that case, ‘b’ must be less than (a + c). Similarly, if ‘c’ is taken as the base, then c must be less than (a + b). Thus three combinations are possible, and the condition should be satisfied for all the 3 combinations:
• 'a' must be less than ('b' + 'c')
• 'b' must be less than ('a' + 'c')
• 'c' must be less than ('a' + 'b')
In general, we can write:
Condition 1:
Any one side of a triangle should be less than the sum of the other two sides.
If we are given three sides, we can check the 'condition 1' for each of the three combinations. If all the 3 combinations satisfy the condition, then a triangle will be formed.
Now let us see another condition:
The fig.9.18(i) given below shows a triangle ABC. Here, 'a' will surely be less than (b + c). Other wise this triangle would not be formed. Now we are going to align the side AB with AC, keeping the vertex A the same. This is shown in fig(ii).
 |
| Fig.9.18 |
• In fig(iii), we find that the red side does not meet the yellow side
• This is because, the red side is shorter than the exposed green portion
• But this exposed green portion is the difference between ‘b’ and ‘c’
• So it follows that the red side ‘a’ should be greater than the difference between ‘b’ and ‘c’
We have arrived at the above condition by fixing ‘b’ as the base. We could take any other side as the base. We will get similar results. We will write all those possible results:
• ‘a’ should be greater than the difference between ‘b’ and ‘c’
• ‘b’ should be greater than the difference between ‘a’ and ‘c’
• ‘c’ should be greater than the difference between ‘a’ and ‘b’
So we find that, here also there are 3 combinations possible. In general, we can write:
Condition 2:
Any one side of a triangle should be greater than the difference between the other two sides.
A real life situation where we will want to apply the above ‘findings’ is described below:
We are given three lengths. And we are asked to determine whether it is possible to form a triangle with those three lengths.
Solution: We will demonstrate the method of solution with the same lengths 8, 4 and 2 that we saw above. It is always convenient to do the calculations in the form of a table as shown below:
The formation of the above table is as follows:
• When we get the three sides, we split them into two groups: Group 1 and Group 2.
• Group 1 contains one side and Group 2 contains the other two sides.
• Three such combinations are possible.
• In each combination, we add the lengths in Group 2. This is entered in the column no.5.
• We also find the difference. This is entered in the column no.6.
• The sum is compared with the value in Group 1. The result of the comparison is entered in column 7. We must get a ‘yes’ in all the three combinations.
• The difference is compared with the value in Group 1. The result of this comparison is entered in column 8.
• We must get a ‘yes’ in all the three combinations. If we get 3 'yes' in the column no.7, we will surely get 3 'yes' in the column no.8 also.
• We will be able to form a triangle with the given three lengths, only if we get a total of 6 'yes'.
We will now see some solved examples:
Solved example 9.7
Check whether it is possible to form a triangle with sides 3, 4 and 5 cm.
Solution:
In the next section we will see more solved examples.
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