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Squares and Square Roots Test - 27

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Squares and Square Roots Test - 27
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  • Question 1
    1 / -0
    Write a Pythagorean triplet whose one number is 8
    Solution
    $$8^2=64$$
    $$\Rightarrow$$  $$100-64=36$$
    $$\Rightarrow$$  $$\sqrt{36}=6$$
    $$\Rightarrow$$  $$\sqrt{100}=10$$
    Let
    Hypotenuse $$=10$$            [ Since its greater than $$6$$ and $$8$$  ]
    Base $$=8$$ and 
    Perpendicular $$=6$$

    $$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$$                [ Pythagoras theorem ]

    $$\Rightarrow$$  $$(10)^2=(6)^2+(8)^2$$
    $$\Rightarrow$$  $$100=36+64$$
    $$\Rightarrow$$  $$100=100$$
    $$\therefore$$   Pythagorean triplet are $$6,8,10$$

  • Question 2
    1 / -0
    What will be the unit digits of the square of the following numbers?
    $$39$$, $$297$$, $$5125$$, $$7286$$
    Solution
    $$(a)$$The numbers with unit's digits $$1$$ or $$9$$ has its square as $$1$$
    $$\therefore\,$$ the unit's digit of square of $$39$$ is $$1$$
    $$(b)$$The numbers with unit's digits $$3$$ or $$7$$ has its square as $$9$$
    $$\therefore\,$$ the unit's digit of square of $$297$$ is $$9$$
    $$(c)$$The numbers with unit's digits $$5$$  has its square as $$5$$
    $$\therefore\,$$ the unit's digit of square of $$5125$$ is $$5$$
    $$(d)$$The numbers with unit's digits $$4$$ or $$6$$ has its square as $$6$$
    $$\therefore\,$$ the unit's digit of square of $$7286$$ is $$6$$
  • Question 3
    1 / -0
    Find the square root of $$126736$$ by long division method.
  • Question 4
    1 / -0
    Find the square root of which of the following numbers will be the least :
    Solution
    $$A.$$

    $$7\dfrac{58}{81}=\dfrac{625}{81}$$

    $$\Rightarrow$$  $$\sqrt{\dfrac{625}{81}}=\dfrac{25}{9}=2.77$$

    $$B.$$

    $$11\dfrac{14}{25}=\dfrac{289}{25}$$

    $$\Rightarrow$$  $$\sqrt{\dfrac{289}{25}}=\dfrac{17}{5}=3.4$$

    $$C.$$

    $$10\dfrac{1}{36}=\dfrac{361}{36}$$

    $$\Rightarrow$$  $$\sqrt{\dfrac{361}{36}}=\dfrac{19}{6}=3.16$$

    $$D.$$

    $$0.3481=\dfrac{3481}{10000}$$

    $$\Rightarrow$$  $$\sqrt{\dfrac{3481}{10000}}=\dfrac{59}{100}=0.59$$

    $$\therefore$$  We can see, $$0.3481$$  has least square root.

  • Question 5
    1 / -0
    Find value of $$\sqrt{5184}$$.
    Solution
    $$\sqrt{5184}\\=\sqrt{2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3\times 3}\\=\sqrt{2^6\times 3^4}\\=2^3\times3^2\\=8\times9=72$$
  • Question 6
    1 / -0
    Find the square root by prime factorisation: 
    i) $$196$$
    ii) $$225$$
    Solution
    (1)

    $$196=2\times 2\times 7\times 7$$

    $$\sqrt{196}=\sqrt{2\times 2\times 7\times 7}$$

    $$=\sqrt{2^{2}\times 7^{2}}$$

    $$=2\times 7$$
    $$=14$$

    (2)

    $$225=3\times 3\times 5\times 5$$

    $$\sqrt{225}=\sqrt{3\times 3\times 5\times 5}$$

    $$=\sqrt{3^{2}\times 5^{2}}$$

    $$=3\times 5$$
    $$=15$$
  • Question 7
    1 / -0
    Find the value of $$\sqrt{66049}$$.
    Solution
    The number $$66049$$ can only be divided by $$257$$
    $$66049=257\times 257$$
    $$66049=(257)^2$$
    $$\sqrt{66049}=\sqrt{257^2}$$
    $$\therefore$$  $$\sqrt{66049}=257$$
  • Question 8
    1 / -0
    A pythagorean triplet whose two numbers are 8 and 10, then third number is
    Solution
    Require pythagorean triplet is (6,8,10) as $$6^2+8^2 = 36+64= 100 = 10^2 $$
  • Question 9
    1 / -0
    Which of the following is a perfect square number?
    Solution

  • Question 10
    1 / -0
    Which of the following is pythagorean triplet?
    Solution
    $$\textbf{Step-Check option one by one using condition of pythagorian triplet}$$
               $$\text{$$a,b$$ and $$c$$ are called pythagorean triplet if}$$ $$a^2+b^2=c^2$$
              $$\text{Clearly,}$$ $$3^2+4^2=9+16=25=5^2$$
              $$5^2+12^2=25+144=169=13^2$$
              $$7^2+24^2=49+576=625=25^2$$
              $$\text{Here all three of the given options are satisfy the pythagorean triplet.}$$
    $$\textbf{Hence option D is correct}$$
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