Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem,
(Hypotenuse)² = (Base)² + (Perpendicular)²
(i) 2.5 cm, 6.5 cm, 6 cm
In ∆ABC, (AC)² = (AB)² + (BC)²
L.H.S. = (6.5)² = 42.25 cm
R.H.S. = (6)² + (2.5)² = 36 + 6.25 = 42.25 cm
Since, L.H.S. = R.H.S.
Therefore, the given sides are of the right angled triangle.
Right angle lies on the opposite to the greater side 6.5 cm, i.e., at B.

(ii) 2 cm, 2 cm, 5 cm
In the given triangle, (5)² = (2)² + (2)²
L.H.S. = (5)² = 25
R.H.S. = (2)² + (2)² = 4+4=8
Since, L.H.S. ≠ R.H.S.
Therefore, the given sides are not of the right angled triangle.

(iii) 1.5 cm, 2 cm, 2.5 cm
In ∆ PQR, (PR)² = (PQ)² + (RQ)²
L.H.S. = (2.5)² = 6.25
R.H.S. = (1.5)² + (2)² = 2.25 + 4 = 6.25 cm
Since, L.H.S. = R.H.S.
Therefore, the given sides are of the right angled triangle.
Right angle lies on the opposite to the greater side 2.5 cm, i.e., at Q.

Class 7 Maths Chapter 6 Exercise 6.5

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https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/