Tag: squares and square roots

Quizzes (7) Questions (116)

Questions Related to squares and square roots

Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

If sin$\theta -cosec  \theta =\sqrt{5},$ then the value of sin  $\theta  + cosec  \theta$ is:

  1. $\sqrt{3}$

  2. 1

  3. 3

  4. 9
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\Rightarrow \sin\theta-cosec\theta=\sqrt{5}$


$\Rightarrow \sin\theta-\dfrac{1}{\sin\theta}=\sqrt{5}$      $(\because cosec\theta=\dfrac{1}{\sin\theta})$

$\Rightarrow \sin^2\theta-\sqrt{5}\sin\theta-1=0$

Solving equation to get roots.

$\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{\pm3+\sqrt{5}}{2}$ (substitute values to get roots)

To find:-

$\sin\theta+cosec\theta$

Using $\dfrac{3+\sqrt{5}}{2}$

$=\dfrac{3+\sqrt{5}}{2}+\dfrac{2}{3+\sqrt{5}}$

$=\dfrac{(3+\sqrt{5})^2+4}{2(3+\sqrt{5})}$

$=\dfrac{9+4+5+6\sqrt{5}}{2(3+\sqrt{5})}$

$=\dfrac{6(3+\sqrt{5})}{2(3+\sqrt{5})}$

$=3$


Using $\dfrac{-3+\sqrt{5}}{2}$

$=\dfrac{-3+\sqrt{5}}{2}+\dfrac{2}{-3+\sqrt{5}}$

$=\dfrac{(-3+\sqrt{5})^2+4}{2(-3+\sqrt{5})}$

$=\dfrac{9+4+5-6\sqrt{5}}{2(-3+\sqrt{5})}$

$=\dfrac{-6(-3+\sqrt{5})}{2(-3+\sqrt{5})}$

$=-3$


According to option answer is $3$

Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

Evaluating the following :
$(3+\sqrt{2})^{5}-(3-\sqrt{2})^{5}$

  1. $1718\sqrt 3$
  2. $1718\sqrt 2$
  3. $1178\sqrt 3$
  4. $1178\sqrt 2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given term is $(3+\sqrt{2})^{5}-(3-\sqrt{2})^{5}$


$\Rightarrow 2\left[\ ^{5}C _{1}\times 3^{4}\times (\sqrt{2})^{1}+\ ^{5}C _{3}\times 3^{2}\times (\sqrt{2})^{3}+\ ^{5}C _{5}\times 3^{0}\times (\sqrt{2})^{5}\right]$

$\Rightarrow 2\left[5\times 81\times\sqrt{2}+10\times 9\times 2\sqrt{2}+4\sqrt{2}\right]$

$\Rightarrow 2\sqrt{2}(405+180+4)$

$\Rightarrow 1178\sqrt{2}$

Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

Evaluating the following :
$(1+2\sqrt{x})^{5}+(1-2\sqrt{x})^{5}$

  1. $2(1+40x^2+80x)$
  2. $2(1-40x+81x^2)$
  3. $2(1+40x+80x^2)$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given to evaluate is $(1+2\sqrt x)^5 +(1-2\sqrt x)^5$

$\Rightarrow 2[^{5}C _{0} (2\sqrt x)^0 +^5C _2 (2\sqrt x)^2 +^5C _4 (2\sqrt x)^4]$

$\Rightarrow 2[1+10\times 4x+5\times 16x^2]$

$\Rightarrow 2[1+40x+80x^2]$

Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

The number which exceeds its positive square root by $12$ is

  1. $9$
  2. $16$
  3. $25$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let the positive number be x according to question,

$\sqrt{x}+12=x$

$\Rightarrow \sqrt{x}=x-12$

Squaring both sides,

$\Rightarrow x=x^{2}-24x+144$

$x^{2}-25x+144=0$

$x^{2}-16x-9x+144=0$

$x(x-16)-9(x-16)=0$

$(x-9)(x-16)=0$

 $  (x-9)=0 $ or $    (x-16)=0 $

 $ x=9 $  or $ x=16 $

So, the number is $16$.
Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

Find the square root of which of the following numbers will be the least :

  1. $7\dfrac{58}{81}$
  2. $11\dfrac{14}{25}$
  3. $10\dfrac{1}{36}$
  4. $0.3481$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$A.$


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


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

$B.$

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

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

$C.$

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

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

$D.$

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

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

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

Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

Simlify: $\sqrt{\dfrac{-17}{144}-i}$

  1. $ \pm \left( {\dfrac{3}{2} - \dfrac{i}{3}} \right)$
  2. $ \pm \left( {\dfrac{3}{4} - \dfrac{{2i}}{3}} \right)$
  3. $ \pm \left( {\dfrac{3}{5} - \dfrac{{5i}}{6}} \right)$
  4. $ \pm \left( {\dfrac{2}{3} - \dfrac{{3i}}{4}} \right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that,

${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$

 

Now, let,

$ -2ab=-i $

$ ab=i $

 

Now, consider $\dfrac{-17}{144}$. We can write it as,

$\dfrac{-17}{144}=\dfrac{64-81}{9\times 16}=\dfrac{4}{9}-\dfrac{9}{16}$

 

Thus,

$ \sqrt{\dfrac{-17}{144}-i}=\sqrt{\dfrac{4}{9}-\dfrac{9}{16}-i} $

$ =\sqrt{{{\left( \dfrac{2}{3} \right)}^{2}}+{{\left( i \right)}^{2}}{{\left( \dfrac{3}{4} \right)}^{2}}-2\times \left( \dfrac{2}{3} \right)\times \left( \dfrac{3i}{4} \right)} $

$ =\sqrt{{{\left( \dfrac{2}{3}-\dfrac{3i}{4} \right)}^{2}}} $

$ =\pm \left( \dfrac{2}{3}-\dfrac{3i}{4} \right) $

 

Hence, this is the required result.

Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

The approximate value of$\sqrt { { \left( 1.97 \right)  }^{ 2 }{ \left( 4.02 \right)  }^{ 2 }{ \left( 3.98 \right)  }^{ 2 } }$

  1. $31.59 $
  2. $5.099$
  3. $5.009$
  4. $5.734$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have,

$ \sqrt{{{\left( 1.97 \right)}^{2}}{{\left( 4.02 \right)}^{2}}{{\left( 3.98 \right)}^{2}}} $

$ =1.97\times 4.02\times 3.98 $

$ =31.59 $

Hence, this is the answer.