Question

If a and b are natural numbers and a > b If (a² + b²), (a² – b²) and 2ab are the sides of the triangle, then prove that the triangle is right-angled. Find out two Pythagorean triplets by taking suitable values of a and b.

Answer 1

a² + b², a² – b², 2ab are sides of triangle.

By Pythagoras' theorem,

(a² + b²), (a² – b²) + (2ab)²

a⁴ + b⁴ + 2a²b² = a⁴ + b⁴ – 2a²b² + 4a²b²

a⁴ + b⁴ + 2a²b² = a⁴ + b⁴ + 2a²b²

As L.H.S. = R.H.S.

∴ Triangle is a right-angle triangle as it follows Pythagorean triplets

As a > b   .....[Given]

Let a = 4, b = 3

a² + b² = 4² + 3² = 16 + 9 = 25

a² – b² = 16 – 9 = 7

2ab = 2 × 4 × 3 = 24

∴ (25, 7, 24) is Pythagorean triplet.

Let a = 2, b = 1

a² + b² = 2² + 1² = 4 + 1 = 5

a² – b² = 2² – 1² = 4 – 1 = 3

2ab = 2 × 2 × 1 = 4

∴ (5, 3, 4) is a Pythagorean triplet.