Question

(a) Find one Pythagorean triplet with 12 as one of the members. (b) Find the square root of 7056 using prime factorisation method.​

Answer 1

Step-by-step explanation:

Given :-

a) 12

b) 7056

To find:-

(a) Find one Pythagorean triplet with 12 as one of the members. (b) Find the square root of 7056 using prime factorisation method.

Solution :-

a) Given number = 12

Let n = 12

12 is an even number

If the number "n" is even then the Pythagorean Triplet is n, (n/2)²-1, (n/2)²+1.

=>(n/2)²-1=( 12/2)²- 1 = 6²-1 = 36-1 = 35

=> (n/2)²+1 = (12/2)²+1 = 6²+1 = 36+1 = 37

The Pythagorean Triplets (12,35,37)

b) Given number = 7056

7056 = 2×3528

7056 = 2×2×1764

7056 = 2×2×2×882

7056 = 2×2×2×2×441

7056 = 2×2×2×2×3×147

7056 = 2×2×2×2×3×3×49

7056 = 2×2×2×2×3×3×7×7

therefore,

√7056 = √[2×2×2×2×3×3×7×7]

√7056 = √[(2×2)×(2×2)×(3×3)×(7×7)]

√7056 = 2×2×3×7

√7056 = 84

Answer:-

a) The Pythagorean Triplet is (12,35,37)

b) √7056 = 84

Used formulae:-

• If the number "n" is even then the Pythagorean Triplet is

[ n, (n/2)²-1, (n/2)²+1]

• Prime Factorization method

Answer 2

answer is

(a) 12,35,37

(b) 84

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