Draw five triangles and measure their sides. Check in each case, if the sum of the length of any two sides is always less than the third side
Solution:
We will be using the concept of triangles to solve this.
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 greater than the third side. Hence, we conclude that the sum of any two sides of a triangle is never less than the third side.
NCERT Solutions for Class 6 Maths Chapter 5 Exercise 5.1 Question 7
Draw five triangles and measure their sides. Check in each case, if the sum of the length of any two sides is always less than the third side.
Summary:
Five triangles are drawn and their sides are measured and in each case, we see that the sum of any two sides of a triangle is greater than the third side.
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