Draw five triangles and measure their sides. Check in each case, in the sum of length of their any two sides is always less than the third side.

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Solution

i) In above $△ l e A B = 3 c m$
$B C = 3 c m$
$C A = 3 c m$
So $A B + B C > C A$
$A B + C A > B C$
$B C + C A > A B$
ii) In above $△ l e A B = 6 c m$
$B C = 5 c m$
$B C = 5.3 c m$
So $A B + B C > C A$
$A B + C A > B C$
$A B + C A > A B$
iii) In above $△ A B C A B = 4 c m$
$A C = 3 c m$
$B C = 5 c m$
So $A B + A C > B C$
$A B + B C > A C$
$B C + C A > A B$
iv) In above $△ A B C A B = 7 c m$
$B C = 7 c m$
$A C = 4 c m$
So $A B + B C > B A$
$A B + B C > A C$
$B C + C A > A B$
v) In above $△ A B C A B = 10 c m$
$A C = 4 c m$
$B C = 5 c m$
So $A B + B C > A C$
$A B + A C > B C$
$B C + C A > A B$
$A C + C B > A B$
In the above following any two sides always less than the third side.

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