Tag: squares and square roots

Questions Related to squares and square roots

Find the square root correct upto $2$ decimals $60.92$.

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

  1. $7.92$

  2. $7.81$

  3. $7.27$

  4. $7.56$

Show answer
Correct Option: B
Explanation:
$\sqrt{6092}$

$=\sqrt{\dfrac{6092}{100}}$

$=\dfrac{78.05}{10}$

$=7.81$

The positive square root of $( \sqrt { 48 } - \sqrt { 45 } )$ is _________.

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

  1. $\frac { \sqrt [ 4 ] { 3 } } { \sqrt { 2 } } ( \sqrt { 5 } - \sqrt { 3 } )$

  2. $\frac { \sqrt [ 4 ] { 3 } } { 2 } ( \sqrt { 5 } - \sqrt { 3 } )$

  3. $\frac { \sqrt { 2 } } { \sqrt [ 4 ] { 3 } } ( \sqrt { 5 } - \sqrt { 3 } )$

  4. $\frac { \sqrt [ 4 ] { 3 } } { \sqrt { 2 } } ( \sqrt { 5 } + \sqrt { 3 } )$

Show answer
Correct Option: A
Explanation:

Solving $(\sqrt{48}-\sqrt{45})$

$4\sqrt{3}-3\sqrt{5}$

$\dfrac{\sqrt{3}}{2}(8-2\sqrt{15})$

$\dfrac{\sqrt{3}}{2}(3+5-2\sqrt{15})$

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

$\dfrac{\sqrt3}{2}(\sqrt5-\sqrt3)^2$

$Now \ Finding\ Square \ Root$

$\pm{ \dfrac{3^{\frac{1}{4}}}{\sqrt2}(\sqrt5-\sqrt3)}$

$So \ it's \ positive \ root \ is \ $$ \dfrac{3^{\frac{1}{4}}}{\sqrt2}(\sqrt5-\sqrt3)$
Correct Answer is $A$

If $\sqrt{2}=1.414$ then the value of $\sqrt{8}$ is

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

  1. $2.828$

  2. $1.828$

  3. $2.282$

  4. $2.288$

Show answer
Correct Option: A
Explanation:
$\sqrt{2}=1.414$(given)
$\sqrt{8}=\sqrt{2\times 2\times 2}=2\sqrt{2}=2\times 1.414=2.828$
$\therefore \sqrt{8}=2.828$

Find the square root of 
$5-2\sqrt{6}$

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

  1. $\sqrt{13}-\sqrt{2}$

  2. $\sqrt{3}-\sqrt{2}$

  3. $\sqrt{5}-\sqrt{3}$

  4. $\sqrt{5}-\sqrt{2}$

Show answer
Correct Option: B
Explanation:

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


$=({\sqrt{3}})^2+({\sqrt{2}})^2-2\sqrt{3}\times \sqrt {2}$

Using $(a-b)^2=a^2+b^2-2ab$


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

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

$So, \  the \  square \  root \  of \  (5-2\sqrt{6})=\sqrt{3}-\sqrt{2}$

$\sqrt{(a - b)^2} + \sqrt{(b - a)^2}$ is

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

  1. Always zero

  2. Never zero

  3. Positive if and only if a > b

  4. Positive only if a $\ne$ b

Show answer
Correct Option: D
Explanation:

$\sqrt{(a - b)^2} + \sqrt{(b - a)^2}$
$= |a - b| + |b - a|$

Now, If $a > b$
$= a - b + a - b$
$= 2a - 2b$...+ ve

If $b > a$
$= b - a + b - a$
$= 2b - 2a$...+ ve
Therefore, if $a \ne b$ then the given equation is always positive.
Hence, option 'D' is correct.

The value of $\displaystyle \frac{1\, +\, \sqrt{0.01}}{1\, -\, \sqrt{0.1}}$ is close to .......... .

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

  1. 0.6

  2. 1.1

  3. 1.6

  4. 1.7

Show answer
Correct Option: C
Explanation:

$\displaystyle \frac{1\, +\, \sqrt{0.01}}{1\, -\, \sqrt{0.1}}\, =\, \displaystyle \frac{1\, +\, 0.1}{1\, -\, 0.32}\, =\, \displaystyle \frac{1.1}{0.68}\, =\, 1.6$

If $\sqrt{6}\, =\, 2.55,$ then the value of $\displaystyle {\sqrt{\frac{2}{3}\, +\, 3\frac{3}{2}}}$ is

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

  1. 4.48

  2. 4.49

  3. 4.50

  4. None of these

Show answer
Correct Option: D
Explanation:
$ {\sqrt{\cfrac{2}{3} + 3\cfrac{3}{2}}}$
$=  {\cfrac{\sqrt{2}}{\sqrt{3}} \times \cfrac{\sqrt{3}}{\sqrt{3}} + 3 \times \cfrac{\sqrt{3}}{\sqrt{2}} \times \cfrac{\sqrt{2}}{\sqrt{2}}}$
$=  {\cfrac{\sqrt{6}}{3} + \cfrac{3\sqrt{6}}{2} = \cfrac{2.55}{3} + \cfrac{3 \times 2.55}{2}}$
$=  {\cfrac{2.55}{3} + \cfrac{7.65}{2} = \cfrac{5.10 + 22.95}{6}}$
$=  \cfrac{28.05}{6} = 4.675$

$\displaystyle {\sqrt{\frac{4}{3}}\, -\, \sqrt{\frac{3}{4}}\, =\, ?}$

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

  1. $\displaystyle \frac{1}{2\sqrt{3}}$

  2. $\displaystyle - \frac{1}{2\sqrt{3}}$

  3. 1

  4. $\displaystyle \frac{5\sqrt{3}}{6}$

Show answer
Correct Option: A
Explanation:

$\displaystyle {\frac{\sqrt{4}}{\sqrt{3}} - \frac{\sqrt{3}}{\sqrt{4}} = \frac{2}{\sqrt{3}} - \frac{\sqrt{3}}{2} = \frac{4 - 3}{2\sqrt{3}} = \frac{1}{2\sqrt{3}}}$

If $\sqrt{75.24\, +\, x}\, =\, 8.71,$ then the value of x is

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

  1. 0.6241

  2. 6.241

  3. 62.41

  4. None of these

Show answer
Correct Option: A
Explanation:
$\sqrt{75.24\, +\, x}\, =\, 8.71$        ...Given
Squaring on both sides, we get
$75.24 + x = 8.71 \times 8.71$
$\Rightarrow x=8.71^2-75.24$
$ \Rightarrow x = 0.6241$

If $\sqrt{3}\, =\, 1.732,$ then the approximate value of $\displaystyle \frac{1}{\sqrt{3}}$ is

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

  1. 0.617

  2. 0.313

  3. 0.577

  4. 0.173

Show answer
Correct Option: C
Explanation:

$ {\cfrac{1}{\sqrt{3}} = \cfrac{1}{\sqrt{3}} \times \cfrac{\sqrt{1}}{\sqrt{3}} = \cfrac{\sqrt{1}}{3} = \cfrac{1.732}{3} = 0.577}$