Tag: square root of non perfect squares

Quizzes (0) Questions (61)

Questions Related to square root of non perfect squares

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

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

  1. 4.48

  2. 4.49

  3. 4.50

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$ {\sqrt{\cfrac{2}{3} + 3\cfrac{3}{2}}}$
$=  {\cfrac{\sqrt{2}}{\sqrt{3}} \times \cfrac{\sqrt{3}}{\sqrt{3}} + 3 \times \cfrac{\sqrt{3}}{\sqrt{2}} \times \cfrac{\sqrt{2}}{\sqrt{2}}}$
$=  {\cfrac{\sqrt{6}}{3} + \cfrac{3\sqrt{6}}{2} = \cfrac{2.55}{3} + \cfrac{3 \times 2.55}{2}}$
$=  {\cfrac{2.55}{3} + \cfrac{7.65}{2} = \cfrac{5.10 + 22.95}{6}}$
$=  \cfrac{28.05}{6} = 4.675$
Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

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

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

  3. 1
  4. $\displaystyle \frac{5\sqrt{3}}{6}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

  1. 0

  2. 1

  3. 2
  4. $2^{31/32}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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

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

  1. 3.63, 4.60

  2. 3.63, 4.69

  3. 3.53, 4.69

  4. 3.63, 4.19
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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 $10$, correct to four places of decimal.

  1. 3.4623

  2. 3.1023

  3. 3.1693

  4. 3.1623
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $3.1623$

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

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

  1. $\displaystyle 7\frac{5}{12}$
  2. $\displaystyle 7\frac{7}{12}$
  3. $\displaystyle 5\frac{5}{12}$
  4. $\displaystyle 5\frac{7}{12}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 =$\frac{13\times 5}{12}=\frac{65}{12}$=$5\frac{5}{12}$