Tag: square root of perfect square

Questions Related to square root of perfect square

The value of $\sqrt{\dfrac{64x^{2}}{49y^{2}}}$ is

  1. $\dfrac {8x}{7y}$

  2. $\dfrac {8}{7y}$

  3. $\dfrac {8x^{4}}{7y^{2}}$

  4. $none$


Correct Option: A
Explanation:

$\sqrt{\dfrac{64x^{2}}{49y^{2}}}$


$\Rightarrow$  $\sqrt{\dfrac{(8x)^2}{(7y)^2}}$
Taking square root on both sides,

$\Rightarrow$  $\dfrac{8x}{7y}$

$\therefore$  $\sqrt{\dfrac{64x^{2}}{49y^{2}}}=\dfrac{8x}{7y}$

Find the square root of $7-4 \sqrt{3}$

  1. $2+ \sqrt{3}$

  2. $5- \sqrt{3}$

  3. $2- \sqrt{3}$

  4. $5+\sqrt{3}$


Correct Option: C
Explanation:
$7-4\sqrt 3$
$=7-2\sqrt {2\times 3}$
$=7-2\sqrt {12}$
$=(\sqrt 3)^2-2\sqrt {3\times 4}+(\sqrt 4)$
$=(-\sqrt {3}+\sqrt {4})^2$
$\therefore \ \sqrt {7-}4\sqrt {3}=2-\sqrt {3}$
$\boxed {2-\sqrt 3}$

$(1\frac{7}{9})^{-1/2} =$

  1. $\frac{4}{3}$

  2. $\frac{3}{4}$

  3. $-\frac{4}{3}$

  4. $-\frac{3}{4}$


Correct Option: B
Explanation:

$(1\frac{7}{9})^{-1/2} =(\frac{16}{9})^{-1/2}=(\frac{4}{3})^{-1}=\frac{3}{4}$

If $\displaystyle \sqrt{1+\frac{25}{144}}=\frac{x}{12}$ then x equals

  1. 1

  2. 11

  3. 13

  4. 7


Correct Option: C
Explanation:
$ \sqrt{1+\cfrac{25}{144}}=\cfrac{x}{12}$
$\Rightarrow \sqrt{\cfrac{169}{144}}=\cfrac{x}{12}$
$ \Rightarrow \cfrac{x}{12}=\cfrac{13}{12}$
$\Rightarrow x=\cfrac{13}{12}\times 12$
$\Rightarrow x=13 $

Find the square root of $\displaystyle 9\frac { 49 }{ 64 } $.


  1. $\displaystyle 3\frac { 1 }{ 8 } $.

  2. $\displaystyle 4\frac { 3 }{ 8 } $.

  3. $\displaystyle 3\frac { 7 }{ 8 } $.

  4. $\displaystyle 4\frac { 1}{ 8 } $.


Correct Option: A
Explanation:

$\sqrt { 9\displaystyle\frac { 49 }{ 64 }  } =\sqrt {\displaystyle \frac { 625 }{ 64 }  } =\displaystyle\frac { \sqrt { 625 }  }{ \sqrt { 64 }  } =\displaystyle\frac { 25 }{ 8 } =3\displaystyle\frac { 1 }{ 8 } $.

Evaluate: $\displaystyle\sqrt{5 \left(2\frac{3}{4}\, -\, \frac{3}{10}\right)}$ is $\displaystyle 3\frac{1}{m}$

$m$ is 

  1. $2$

  2. $5$

  3. $4$

  4. $6$


Correct Option: A
Explanation:

  $\sqrt{5 \left(2\dfrac{3}{4}\, -\, \dfrac{3}{10}\right)}$
$=\sqrt { 5\left( \dfrac { 11 }{ 4 } \, -\, \dfrac { 3 }{ 10 }  \right)  } $
$=\sqrt { 5\left( \dfrac { 55-6 }{ 20 } \,  \right)  } $
$ =\sqrt { \left( \dfrac { 49 }{ 4 } \,  \right)  } $
$=\dfrac{7}{2}$
$=3\dfrac{1}{2}$

Evaluate:$\displaystyle\sqrt{(0.5)^3\, \times\, 6\, \times\, 3^5}$

  1. $13.5$

  2. $13.6$

  3. $13.7$

  4. $13.8$


Correct Option: A
Explanation:

$\sqrt { { 0.5 }^{ 3 }\times6\times{ 3 }^{ 5 } } \ =\sqrt { { 0.5 }^{ 2 }\times2\times0.5\times{ 3 }^{ 6 } } \ =0.5*3^{ 3 }\quad =13.5$

Find the square root of: $\displaystyle27\frac{9}{16}$

  1. $\displaystyle2\frac{1}{4}$

  2. $\displaystyle7\frac{1}{4}$

  3. $\displaystyle3\frac{1}{4}$

  4. $\displaystyle5\frac{1}{4}$


Correct Option: D
Explanation:

Given number is $ 27\dfrac {9}{16} = \dfrac {441}{16} $

Square root of $ \dfrac {441}{16} =  \dfrac { \sqrt {441}}{\sqrt {16}} = \dfrac {21}{4} = 5 \dfrac {1}{4} $

Evaluate: $\sqrt{\displaystyle \frac{1}{16}\, +\, \displaystyle \frac{1}{9}}$

  1. $\displaystyle \frac{7}{12}$

  2. $\displaystyle \frac{25}{144}$

  3. $\displaystyle \frac{5}{12}$

  4. None of these


Correct Option: C
Explanation:

$\sqrt{\displaystyle \frac{1}{16}\, +\, \displaystyle \frac{1}{9}}\,=\sqrt{\displaystyle \frac{(1\times9)+(1\times16)}{9\times16}} =\, \sqrt{\displaystyle \frac{25}{144}}\, =\, \displaystyle \frac{5}{12}$

The square root of $71\, \times\, 72\, \times\, 73\, \times\, 74\, +\, 1$ is :

  1. 9,375

  2. 9,625

  3. 5,625

  4. 5,255


Correct Option: D
Explanation:

$\text{Here. consider the fact that the product of 4 consecutive numbers + 1 is perfect square.}$

$So, let\, x = 71$

⇒ $(71)(72)(73)*(74) + 1 = x(x + 1) (x + 2) (x + 3) +1$ 

⇒$ (x² + 3x)(x² + 3x + 2) + 1$

⇒$ (x² + 3x)² + 2(x² + 3x) + 1$

⇒ $(x² + 3x + 1)^2$

⇒ $Square \,root \,of \,(x² + 3x + 1)² = x² + 3x + 1$

$Here, x = 71$

$Therefore, square root is = (71)² + (3*71) + 1$

⇒ $5041 + 213 + 1$ 

= $5255$