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

Case I. In ∆ABC

Consider AB = 2.5 cm, BC = 4.8 cm and AC = 5.2 cm
AB + BC = 2.5 cm + 4.8 cm = 7.3 cm
Since, 7.3 > 5.2 , AB + BC > AC
Therefore, the sum of any two sides of a triangle is greater than the third side.

Case II. In ∆PQR,

Consider PQ = 2 cm, QR = 2.5 cm and PR = 3.5 cm
PQ + QR = 2 cm + 2.5 cm = 4.5 cm
Since, 4.5 > 3.5, PQ + QR > PR
Therefore, the sum of any two sides of a triangle is greater than the third side.

Case III. In ∆XYZ,

Consider XY = 5 cm, YZ = 3 cm and ZX = 6.8 cm
XY + YZ = 5 cm + 3 cm = 8 cm
Since, 8 > 6.8, XY + YZ > ZX
Therefore, the sum of any two sides of a triangle is greater than the third side.

Case IV. In ∆MNS,

Consider MN = 2.7 cm, NS = 4 cm and MS = 4.7 cm
MN + NS = 2.7 cm + 4 cm = 6.7 cm
Since, 6.7 >4.7, MN + NS > MS
Hence, the sum of any two sides of a triangle is greater than the third side.

Case V. In ∆KLM,

Consider KL = 3.5 cm, LM = 3.5 cm and KM = 3.5 cm
KL + LM = 3.5 cm + 3.5 cm = 7 cm
7 cm > 3.5 cm, KL + LM > KM
Therefore, the sum of any two sides of a triangle is always greater than the third side.