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1) Write the pythagorean triplet whose one member is 16.

2) Find the cube root of 8000

3) Find the smallest square number that is divisible by each of the numbers 8,15,20

4) Find the smallest whole number of 252 by which it should be multiplied so as get a perfect square number.Also find the square root of the new number.

5) Find the square root of

1 . 51.84

2 5776

For any natural number n>1; 2n, n2-1 and n2+1 forms a pythagorean triplet.
Taking n2+1 = 16, then n2 = 15
Here the value of n will not be an integer.
Taking n2-1 = 16, then n2 = 17
Here also the value of n will not be an integer.
Taking 2n = 16 ⇒ n = 8
Thus, n2+1 = 82+1 = 65 and n2-1 = 82-1 = 63
Therefore the pythagorean triplet is 16, 63 and 65.

 

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 2) 20

othr answ. u'll find in sum nice mathematician's bk like rd sharma, dn kundra etc.

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thank u shalina(chechi)

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 1) the three members are 9,16 and 25. (it is the squares of the famous first pythagorean triple 3,4,5).

 

2) 20

 

3) useful clue -  for the number to be divisible by 15 and 20, it SHOULD be divisible by 10. so, it shud end in 0 and shud be a multiple of 10.

 

4) the smallest number thats is divisible by 252 n is a square number is 1764. its sq. root is 42.

 

5)  1) 7.2

    2) 76

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 1) 9,16 and 25.

2) 20

4) smallest number divisible by 252  =1764 

square root = 42.

5) 1)7,2

  2) 76

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dont mind 

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tumne toh homework kar dala! 

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