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Question -

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



Answer -

Case 1. In triangle ABC

AB= 2.5 cm
BC = 4.8 cm and
AC = 5.2 cm
AB + BC = 2.5 cm + 4.8 cm
= 7.3 cm
As 7.3 > 5.2
∴ AB + BC > AC
Hence, the sum of any two sides of a triangle is greater than the third side.
Case 2. In triangle PQR
PQ = 2 cm
QR = 2.5 cm
PR = 3.5 cm
PQ + QR = 2 cm + 2.5 cm
= 4.5 cm
As 4.5 > 3.5
∴ PQ + QR > PR
Hence, the sum of any two sides of a triangle is greater than the third side.
Case 3. In triangle XYZ
XY = 5 cm
YZ = 3 cm
ZX = 6.8 cm
XY + YZ = 5 cm + 3 cm
= 8 cm
As 8 > 6.8
∴ XY + YZ > ZX
Hence, the sum of any two sides of a triangle is greater than the third side.
Case 4. In triangle MNS
MN = 2.7 cm
NS = 4 cm
MS = 4.7 cm
MN + NS = 2.7 cm + 4 cm
6.7 cm
As 6.7 > 4.7
∴ MN + NS > MS
Hence, the sum of any two sides of a triangle is greater than the third side.
Case 5. In triangle KLM
KL = 3.5 cm
LM = 3.5 cm
KM = 3.5 cm
KL + LM = 3.5 cm + 3.5 cm
= 7 cm
As 7 cm > 3.5 cm
∴ KL + LM > KM
Hence, the sum of any two sides of a triangle is greater than the third side.
Therefore we conclude that the sum of any two sides of a triangle is always greater than the third side.


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