Tag: estimating square roots

Questions Related to estimating square roots

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}$

$\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}\, =\, ?$

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

  1. 0

  2. 1

  3. 2

  4. $2^{31/32}$

Show answer
Correct Option: D
Explanation:

$\sqrt{2\, \times\, \sqrt{2\, \times\, \sqrt{2\, \times\, \sqrt{2\, \times\, 2^{1/2}}}}}$

$=\, \sqrt{2\, \times\, \sqrt{2\, \times\, \sqrt{(2\, \times\, 2^{3/4})}}}$
$=\, \sqrt{2\, \times\, \sqrt{2\, \times\, 2^{7/8}}}\, =\, \sqrt{2\, \times\, 2^{15/16}}\, =\, 2^{31/32}$

By using the table for square root find the value of $\sqrt{7}$.

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

  1. $2.652$

  2. $2.746$

  3. $2.646$

  4. $2.616$

Show answer
Correct Option: C
Explanation:

The sqaure root of $\sqrt 7$ is $2.646$

By using the table for square root find the value of
$13.21$
$21.97$

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

  1. 3.63, 4.60

  2. 3.63, 4.69

  3. 3.53, 4.69

  4. 3.63, 4.19

Show answer
Correct Option: B
Explanation:

(i)

From square root table, Square root of 13.21 is:

 √13.21 = 3.6345

Therefore,

The square root of 13.21 is 3.63

(ii) From square root table, Square root of 21.97 is:

 √21.97 = 4.687

Therefore,

The square root of 21.97 is 4.69

Find the square root of $10$, correct to four places of decimal.

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

  1. 3.4623

  2. 3.1023

  3. 3.1693

  4. 3.1623

Show answer
Correct Option: D
Explanation:
$3.16227$
$3$$+3$ $10$$9$
$61$$+1$ $100$$61$
$626$$+6$ $3900$$3756$
$6322$$+2$ $14400$$12644$
$63242$$+2$-------------$632447$ $175600$$126484$---------------$4911600$$4427129$

$\sqrt{10}=3.16227\simeq 3.1623$
$\therefore$ The square root of $10$ correct to four places of decimal is $3.1623$

The square root of $\displaystyle \frac{\left ( 3\frac{1}{4} \right )^{4}-\left ( 4\frac{1}{3} \right )^{4}}{\left ( 3\frac{1}{4} \right )^{2}-\left ( 4\frac{1}{3} \right )^{2}}$ is

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

  1. $\displaystyle 7\frac{5}{12}$

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

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

  4. $\displaystyle 5\frac{7}{12}$

Show answer
Correct Option: C
Explanation:

$\frac{\left ( 3\tfrac{1}{4} \right )^{4}-\left ( 4\tfrac{1}{3} \right )^{4}}{\left ( 3\tfrac{1}{4} \right )^{2}-\left ( 4\tfrac{1}{3} \right )^{2}}$

=$\frac{\left [ \left ( 3\tfrac{1}{4} \right )^{2}+\left ( 4\tfrac{1}{3} \right )^{2} \right ]\left [ \left ( 3\tfrac{1}{4} \right )^{2}-\left ( 4\tfrac{1}{3} \right )^{2} \right ]}{\left ( 3\tfrac{1}{4} \right )^{2}-\left ( 4\tfrac{1}{3} \right )^{2}}$
=$\left ( 3\tfrac{1}{4} \right )^{2}+\left ( 4\tfrac{1}{3} \right )^{2}$
=$\left ( \frac{13}{16} \right )^{2}+\left ( \frac{13}{9} \right )^{2}=169\times \left ( \frac{9+16}{144} \right )=169\times\frac{25}{144}$ 
Then squire root =$\frac{13\times 5}{12}=\frac{65}{12}$=$5\frac{5}{12}$