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So, we can conclude some inequalities as:

1. In any triangle, the sum of the lengths of any two sides is always greater than the length of
the third side.
In DABC, AB + BC > CA, BC + CA > AB, CA + AB > BC.
2. The difference between the lengths of any two sides of a triangle is always less than the
length of the third side.

AB – BC < CA or BC – AB < CA
AB – CA < BC or CA – AB < BC

BC – CA < AB or CA – BC < AB
Example 8: Is a triangle possible whose sides have lengths Remember
5 cm, 6 cm and 8 cm? • A triangle cannot be formed if any of

Solution: To form a triangle, the sum of the lengths of any the three sides is longer than the sum
two sides should be greater than the length of the third side. of another two sides.
• A triangle cannot be formed if any
So, 5 cm + 6 cm = 11 cm, which is > 8 cm of the three sides is shorter than the
difference of another two.
5 cm + 8 cm = 13 cm, which is > 6 cm
6 cm + 8 cm = 14 cm, which is > 5 cm

Thus, a triangle is possible with length of sides 5 cm, 6 cm, and 8 cm.

Example 9: Can a triangle have sides of lengths 7 cm, 8 cm, and 10 cm?

Solution: To form a triangle the sum of the lengths of any two sides should be greater than the
length of the third side.
So, 7 cm + 8 cm = 15 cm, which is > 10 cm

8 cm + 10 cm = 18 cm, which is > 7 cm

7 cm + 10 cm = 17 cm, which is > 8 cm
Thus, a triangle is possible with length of sides 7 cm, 8 cm and 10 cm.

Example 10: The lengths of two sides of a triangle are 9 cm and 12 cm. Between which two numbers
can the length of the third side fall?

Solution: We know that the sum of two sides of a triangle is always greater than the third side.

Therefore, the third side has to be less than the sum of the given two sides.
Thus, the third side is less than 9 + 12 = 21 cm.

And, the third side cannot be less than or equal to the difference of the given two sides.

Thus, the third side has to be more than 12 – 9 = 3 cm.
Hence, the length of the third side could be any length greater than 3 cm and less than 21 cm.
A
Example 11: In the given figure, AM is a median of a DABC.

Is AB + BC + CA > 2 AM? Verify it. B M C

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