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Squares and Square Roots Test - 20

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Squares and Square Roots Test - 20
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  • Question 1
    1 / -0
    Find the square root of $$3136$$ by division method.
    Solution
    $$\therefore \sqrt{3136} = 56$$

  • Question 2
    1 / -0
    Find the square root of $$576$$ by division method.
    Solution
    We will use long division method
    In this, first group the numbers into two starting from unit place, these groups are called periods.
    Use the largest number whose square is equal or less than the first period number
    Subtract the product of divisor and quotient from the first period and bring down the second period to the right of remainder. This will become new divident.
    In this way we can find out the square root .
    Hence, Option B is correct.

  • Question 3
    1 / -0
    Find the square root of the following number by division method.
    3481
    Solution
    We will use long division method
    In this, first group the numbers into two starting from unit place, these groups are called periods.
    Use the largest number whose square is equal or less than the first period number
    Subtract the product of divisor and quotient from the first period and bring down the second period to the right of remainder. This will become new divident.
    In this way we can find out the square root .
    Hence, Option D is correct.

  • Question 4
    1 / -0
    Find the square root of the following number by division method:
    4489
    Solution
    We will use long division method
    In this, first group the numbers into two starting from unit place, these groups are called periods.
    Use the largest number whose square is equal or less than the first period number
    Subtract the product of divisor and quotient from the first period and bring down the second period to the right of remainder. This will become new divident.
    In this way we can find out the square root .
    Hence, Option B is correct.

  • Question 5
    1 / -0
    Find the least number which must be subtracted from each of the following numbers so as to get a perfect square Also find the square root of the perfect square so obtained 
    $$(i) 402, (ii) 1989, (iii) 3250, (iv) 825, (v) 4000$$
    Solution
    For $$402$$
    Nearest perfect square is $$400$$
    $$\sqrt{400}$$ $$=$$ $$20$$
    $$402$$ $$-$$ $$400$$ $$=$$ $$2$$
    For $$1989$$
    Nearest perfect square is $$1936$$
    $$\sqrt{1936}$$ $$=$$ $$44$$
    $$1936$$ $$-$$ $$1989$$ $$=$$ $$53$$
    For $$3250$$
    Nearest perfect square is $$3249$$
    $$\sqrt{3249}$$ $$=$$ $$57$$
    $$3250$$ $$-$$ $$3249$$ $$=$$ $$1$$
    For $$825$$
    Nearest perfect square is $$784$$
    $$\sqrt{784}$$ $$=$$ $$28$$
    $$825$$ $$-$$ $$784$$ $$=$$ $$41$$
    For $$4000$$
    Nearest perfect square is $$3969$$
    $$\sqrt{3969}$$ $$=$$ $$63$$
    $$4000$$ $$-$$ $$3969$$ $$=$$ $$31$$
    From this, Option B is correct answer.
  • Question 6
    1 / -0
    Find square root of 8281.
    Solution
    $$\therefore \sqrt{8281} = 91$$

  • Question 7
    1 / -0
    Find the smallest number by which  $$9408$$  must be divided so that the quotient is a perfect square. Find the square root of the quotient.
    Solution
    We have 9408=  $$ \displaystyle \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times 3\times \underline{7\times 7}  $$
    If we divide 9408 by the factor 3 then 
     $$ \displaystyle  9408\div 3=3136=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{7\times 7} $$ which is a perfect square (because $$ 3136 = (56)^2$$)
    And  $$ \displaystyle  \sqrt{3136}=2\times 2\times 2\times 7=56 $$
  • Question 8
    1 / -0
    Find a Pythagorean triplet in which one member is 12 

    Solution
    If we take m$$\displaystyle ^{2}$$-1=12                  
    Then m$$\displaystyle ^{2}$$=12+1=13                        
    Then the value of m will not be an integer 
    So we try to take $$m^{2}+1=12\ Again\ m \displaystyle ^{2}=11$$ will not give an integer value for m 
    So let us take 2m = 12
    then m=6
    Thus $$\displaystyle m^{2}-1=36-1=35$$  and   $$\displaystyle m^{2}+1=36+1=37$$
    Therefore the required triplet is 12, 35, 37 
    Note : All Phythagorean triplets may not be obtained using this form For example another triplet 5, 12, 13 also has 12 as a member 
  • Question 9
    1 / -0
    Find the square root of : (i) $$729$$ (ii) $$1296$$
    Solution
    Prime factorization of $$729$$ is
    $$729=3\times 3\times 3\times 3\times 3\times 3=3^6$$
    Prime factorization of $$1296$$ is
    $$1296=2\times 2\times 2 \times 2\times 3\times 3\times 3\times 3=2^4\times 3^4$$

    $$\sqrt{729}=\sqrt{3^6}=3^3=27$$
    $$\sqrt{1296}=\sqrt{2\times2\times 2\times 2\times 3\times 3\times 3\times 3}=2\times 2\times 3\times 3=36$$
  • Question 10
    1 / -0
    Find the square root of 6400.
    Solution
    Write 6400=  $$ \displaystyle \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{5\times 5}  $$
    Therefore  $$ \displaystyle \sqrt{6400}=2\times 2\times 2\times 2\times 5=80 $$
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