\[ \begin{aligned} \sqrt{\frac{324}{361}} &= \frac{\sqrt{324}}{\sqrt{361}} \\ \\ &= \frac{\sqrt{18^2}}{\sqrt{19^2}} \\ \\ &= \frac{18}{19} \end{aligned} \]

Answer \( \displaystyle \color{red}\sqrt{\frac{324}{361}} = \frac{18}{19} \)

\[ \begin{aligned} \sqrt{\frac{441}{961}} &= \frac{\sqrt{441}}{\sqrt{961}} \\ \\ &= \frac{\sqrt{21^2}}{\sqrt{31^2}} \\ \\ &= \frac{21}{31} \end{aligned} \]

Answer \( \displaystyle \color{red} \sqrt{\frac{441}{961}} = \frac{21}{31} \)

\[ \begin{array}{c|c} \begin{aligned} & = \sqrt{5\frac{19}{25}} \\ \\ &= \sqrt{\frac{144}{25}} \\ \\ &= \frac{\sqrt{144}}{\sqrt{25}} \\ \\ &= \frac{\sqrt{12^2}}{\sqrt{5^2}} \\ \\ &= \frac{12}{5} \\ \\ &= 2\frac{2}{5} \end{aligned} & \begin{aligned} \begin{array}{r} 25 \\ \times 5 \\ \hline 125\\ + 19\\ \hline 144 \\ \hline \\ \\ \\ \begin{array}{l} \hspace{.5cm}2 \\ 5 \enclose{longdiv}{12} \\ \hspace{0.1cm} -10\\ \hline \hspace{0.5cm} 2 \\ \hline \\ \end{array} \end{array} \end{aligned} \end{array} \]

Answer \( \displaystyle \color{red} \sqrt{5\frac{19}{25}} = 2\frac{2}{5} \)

\[ \begin{array}{c|c} \begin{aligned} & = \sqrt{21\frac{51}{169}} \\ \\ &= \sqrt{\frac{3600}{169}} \\ \\ &= \frac{\sqrt{3600}}{\sqrt{169}} \\ \\ &= \frac{\sqrt{60^2}}{\sqrt{13^2}} \\ \\ &= \frac{60}{13} \\ \\ &= 4\frac{8}{13} \end{aligned} & \begin{aligned} \begin{array}{r} 169 \\ \times 21 \\ \hline 169 \\ +3380\\ \hline 3549 \\ \hline +51 \\ \hline \color{green}3600 \\ \hline \\ \\ \\ \\ \begin{array}{l} \hspace{.7cm}4 \\ 13 \enclose{longdiv}{60} \\ \hspace{0.3cm} -52\\ \hline \hspace{0.8cm} 8 \\ \hline \\ \end{array} \end{array} \end{aligned} \end{array} \]

Answer \( \displaystyle \color{red} \sqrt{21\frac{51}{169}} = 4\frac{8}{13} \)

\[ \begin{aligned} & =\sqrt{\frac{5625}{441}} \\ \\ & =\sqrt{\frac{\cancel{5625}^{625}}{\cancel{441}_{49}}} \\ \\ & =\sqrt{\frac{625}{49}} \\ \\ & =\sqrt{\frac{25^2}{7^2}} \\ \\ & =\frac{25}{7} \\ \\ & = 3\frac{4}{7} \\ \\ \end{aligned} \]

Answer \( \displaystyle \color{red} \sqrt{\frac{5625}{441}} = 3\frac{4}{7} \)

\[ \begin{array}{c|c} \begin{aligned} & = \sqrt{7 \frac{18}{49}} \\ \\ &= \sqrt{\frac{361}{49}} \\ \\ &= \frac{\sqrt{361}}{\sqrt{49}} \\ \\ &= \frac{\sqrt{19^2}}{\sqrt{7^2}} \\ \\ & =\frac{19}{7} \\ \\ & = 2\frac{5}{7} \\ \\ \end{aligned} & \begin{aligned} \begin{array}{r} 49 \\ \times 7 \\ \hline 343 \\ \hline + 18 \\ \hline \color{green}361 \\ \hline \\ \\ \\ \\ \begin{array}{l} \hspace{.5cm}2 \\ 7 \enclose{longdiv}{19} \\ \hspace{0.1cm} -14\\ \hline \hspace{0.6cm} 5 \\ \hline \\ \end{array} \end{array} \end{aligned} \end{array} \]

Answer \( \displaystyle \color{red} \sqrt{7\frac{18}{49}} = 2\frac{5}{7} \)

\[ \begin{array}{c|c} \begin{aligned} & = \sqrt{23 \frac{394}{729}} \\ \\ &= \sqrt{\frac{17161}{729}} \\ \\ &= \frac{\sqrt{17161}}{\sqrt{729}} \\ \\ \end{aligned} & \begin{aligned} \begin{array}{r} 729 \\ \times 23 \\ \hline 2187 \\ +14580 \\ \hline 16767 \\ \hline + 394 \\ \hline \color{green}17161 \\ \hline \\ \end{array} \end{aligned} \end{array} \] \[ \begin{array}{r|l} & \phantom{0}131 \\ \hline 1 & \phantom{-}\overline{1} \,\, \overline{71} \,\, \overline{61} \\ & -1 \\ \hline 23 & \phantom{-}071 \\ & -\phantom{0}69 \\ \hline 261 & \phantom{-00}261 \\ & -\phantom{00}261 \\ \hline & \phantom{0000}0 \\ \\ \end{array} \] \[ \begin{array}{c|c} \begin{aligned} & \color{green} \sqrt{17161} = 131 \\ \\ &= \sqrt{\frac{17161}{729}} \\ \\ &= \frac{\sqrt{131^2}}{\sqrt{27^2}} \\ \\ &= \frac{131}{27} \\ \\ &= 4\frac{23}{27} \end{aligned} & \begin{array}{l} \hspace{.9cm}4 \\ 27 \enclose{longdiv}{131} \\ \hspace{0.3cm} -108\\ \hline \hspace{0.8cm} 23 \\ \hline \\ \end{array} \end{array} \]

Answer \( \displaystyle \color{red} \sqrt{23 \frac{394}{729}} = 4\frac{23}{27} \)

\[ \begin{array}{c|c} \begin{aligned} & = \sqrt{35 \frac{85}{1444}} \\ \\ &= \sqrt{\frac{50625}{1444}} \\ \\ &= \frac{\sqrt{50625}}{\sqrt{1444}} \\ \\ \end{aligned} & \begin{aligned} \begin{array}{r} 1444 \\ \times 35 \\ \hline 7220 \\ +43320 \\ \hline 50540 \\ \hline + 85 \\ \hline \color{green}50625 \\ \hline \\ \end{array} \end{aligned} \end{array} \] \[ \begin{array}{c c} \begin{array}{c|c} 3 & 50625 \\ \hline 3 & 16875 \\ \hline 3 & 5625 \\ \hline 3 & 1875 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array} & \quad \begin{array}{c|c} 2 & 1444 \\ \hline 2 & 722 \\ \hline 19 & 361 \\ \hline 19 & 19 \\ \hline & 1 \\ \end{array} \end{array} \] \begin{aligned} &= \frac{\sqrt{3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 \times 5}}{\sqrt{2 \times 2 \times 19 \times 19}} \\ \\ &= \frac{{3 \times 3 \times 5 \times 5}}{{2 \times 19}} \\ \\ &= \frac{225}{38} \\ \\ &= 5\frac{35}{38} \\ \\ \end{aligned}

Answer \( \displaystyle \color{red} \sqrt{35 \frac{85}{1444}} = 5\frac{35}{38} \)

\[ \color{green}\text{Method - 1} \] \[ \begin{array}{r|l} & \phantom{00}0.03 \\ \hline 0 & \phantom{-} {0} \, . \overline{00} \, \overline{09} \\ & -0 . 00\\ \hline 03 & \phantom{-0000}09 \\ & \phantom{0000}-9 \\ \hline & \phantom{000000.}0 \end{array} \] \[ \begin{aligned} \\ &\color{green}\text{Method - 2} \\ \\ \sqrt{0.0009} & = \sqrt{\frac{9}{10000}} \\ \\ & = \frac{\sqrt9}{\sqrt{10000}} \\ \\ & = \frac{\sqrt{3^2}}{\sqrt{100^2}} \\ \\ & = \frac{3}{100} \\ \\ \sqrt{0.0009} & = 0.03 \end{aligned} \]

Answer \( \color{red}\sqrt{0.0009} = 0.03 \)

\[ \color{green}\text{Method - 1} \] \[ \begin{array}{r|l} & \phantom{00}0.09 \\ \hline 0 & \phantom{-} {0} \, . \overline{00} \, \overline{81} \\ & -0 . 00\\ \hline 09 & \phantom{-0000}81 \\ & \phantom{000}-81 \\ \hline & \phantom{000000}0 \end{array} \] \[ \begin{aligned} \\ &\color{green}\text{Method - 2} \\ \\ \sqrt{0.0081} & = \sqrt{\frac{81}{10000}} \\ \\ & = \frac{\sqrt{81}}{\sqrt{10000}} \\ \\ & = \frac{\sqrt{9^2}}{\sqrt{100^2}} \\ \\ & = \frac{9}{100} \\ \\ \sqrt{0.0081} & = 0.09 \end{aligned} \]

Answer \( \color{red}\sqrt{0.0081} = 0.09 \)

\[ \color{green}\text{Method - 1} \] \[ \begin{array}{r|l} & \phantom{00}0.111 \\ \hline 0 & \phantom{-} {0} \, . \overline{01} \, \overline{23} \, \overline{21} \\ & -0 \\ \hline 01 & \phantom{-00}01 \\ & \phantom{0}-01 \\ \hline 21 & \phantom{-000}023 \\ & \phantom{000}-21 \\ \hline 221 & \phantom{000-0}221 \\ & \phantom{0111}-221 \\ \hline & \phantom{0110-1}0 \\ \end{array} \] \[ \begin{aligned} \\ &\color{green}\text{Method - 2} \\ \\ \sqrt{0.012321} & = \sqrt{\frac{12321}{1000000}} \\ \\ & = \frac{\sqrt{12321}}{\sqrt{1000000}} \\ \\ & = \frac{\sqrt{111^2}}{\sqrt{1000^2}} \\ \\ & = \frac{111}{1000} \\ \\ \sqrt{0.012321} & = 0.111 \end{aligned} \]

Answer \( \color{red}\sqrt{0.012321} = 0.111 \)

\[ \color{green}\text{Method - 1} \] \[ \begin{array}{r|l} & \phantom{00}2.7 \\ \hline 2 & \phantom{-} {7} . \overline{29} \\ & -4 \\ \hline 47 & \phantom{-}329 \\ & -\phantom{}329 \\ \hline & \phantom{0-}0 \end{array} \] \[ \begin{aligned} \\ &\color{green}\text{Method - 2} \\ \\ \sqrt{7.29} & = \sqrt{\frac{729}{100}} \\ \\ & = \frac{\sqrt{729}}{\sqrt{100}} \\ \\ & = \frac{\sqrt{27^2}}{\sqrt{10^2}} \\ \\ & = \frac{27}{10} \\ \\ \sqrt{7.29} & = 2.7 \end{aligned} \]

Answer \( \color{red}\sqrt{7.29} = 2.7 \)

\[ \begin{array}{r|l} & \phantom{00}0.231 \\ \hline 0 & \phantom{-} {0} . \overline{05} \, \overline{33} \, \overline{61} \\ & -0 \\ \hline 2 & \phantom{-00}05 \\ & \phantom{0}-04 \\ \hline 43 & \phantom{0-0}133 \\ & \phantom{00}-129 \\ \hline 461 & \phantom{00000-}461 \\ & \phantom{0000}-461 \\ \hline & \phantom{000000-}0 \\ \end{array} \]

Answer \( \color{red}\sqrt{0.053361} = 0.231 \)

\[ \begin{array}{r|l} & \phantom{00}0.0231 \\ \hline 0 & \phantom{-} {0} . \overline{00} \, \overline{05} \, \overline{33} \, \overline{61} \\ & -0 \\ \hline 2 & \phantom{-0000}05 \\ & \phantom{000}-04 \\ \hline 43 & \phantom{000-0}133 \\ & \phantom{0000}-129 \\ \hline 461 & \phantom{00000-0}461 \\ & \phantom{000000}-461 \\ \hline & \phantom{00000000-}0 \\ \end{array} \]

Answer \( \color{red}\sqrt{0.00053361} = 0.0231 \)

\[ \begin{array}{r|l} & \phantom{0}12.25 \\ \hline 1 & \phantom{-}\overline1 \, \overline{50} . \overline{06} \, \overline{25} \\ & -1 \\ \hline 22 & \phantom{00}050 \\ & \phantom{000}44 \\ \hline 242 & \phantom{0000}606 \\ & \phantom{-}-484 \\ \hline 2445& \phantom{0000}12225 \\ & \phantom{-}-12225 \\ \hline & \phantom{---}0 \\ \end{array} \]

Answer \( \color{red}\sqrt{150.0625} = 12.25 \)

\[ \begin{array}{r|l} & \phantom{00}0.612 \\ \hline 0 & \phantom{-} {0} . \overline{37} \, \overline{45} \, \overline{44} \\ & -0 \\ \hline 6 & \phantom{-00}37 \\ & \phantom{0}-36 \\ \hline 121 & \phantom{0-0}145 \\ & \phantom{00}-121 \\ \hline 1222 & \phantom{0000-}2444 \\ & \phantom{000}-2444 \\ \hline & \phantom{000000-}0 \\ \end{array} \]

Answer \( \color{red}\sqrt{0.374544} = 0.612 \)

\[ \begin{array}{r|l} & \phantom{0}24.7 \\ \hline 2 & \phantom{-} \overline{6} \, \overline{10} . \overline{09} \\ & -4 \\ \hline 44 & \phantom{00}210 \\ & -\phantom{.}176 \\ \hline 487 & \phantom{--}3409 \\ & \phantom{0}-3409 \\ \hline & \phantom{--0}0 \\ \end{array} \]

Answer \( \color{red}\sqrt{610.09} = 24.7 \)

\[ \begin{array}{r|l} & \phantom{0}2.6457 \\ \hline 2 & \phantom{-}7 . \overline{00} \, \overline{00} \, \overline{00} \, \overline{00} \\ & -4 \\ \hline 46 & \phantom{-}300 \\ & -\phantom{}276 \\ \hline 524 & \phantom{0-}2400 \\ & -\phantom{0}2096 \\ \hline 5285 & \phantom{00-}30400 \\ & - \phantom{00}26425 \\ \hline 52907 & \phantom{000-}397500 \\ & -\phantom{000}370349 \\ \hline & \phantom{000000}27151 \\ \end{array} \] \[ \begin{aligned} \sqrt{7} & = 2.6457 \\ \sqrt{7} & = 2.646 \\ \end{aligned} \]

Answer \( \color{red}\sqrt{7} = 2.646 \)

\[ \begin{array}{r|l} & \phantom{0}1.5811 \\ \hline 1 & \phantom{-}2 . \overline{50} \, \overline{00} \, \overline{00} \, \overline{00} \\ & -1 \\ \hline 25 & \phantom{-}150 \\ & -\phantom{}125 \\ \hline 308 & \phantom{0-}2500 \\ & -\phantom{0}2464 \\ \hline 3161 & \phantom{00-}3600 \\ & -\phantom{00}3161 \\ \hline 31621 & \phantom{000-}43900 \\ & -\phantom{000}31621 \\ \hline & \phantom{0-0}12279 \\ \end{array} \] \[ \begin{aligned} \sqrt{2.5} & = 1.5811 \\ \sqrt{2.5} & = 1.581 \\ \end{aligned} \]

Answer \( \color{red}\sqrt{2.5} = 1.581 \)

\[ \begin{array}{c|c} \begin{aligned} & =\sqrt{2\frac{1}{12}} \\ \\ & =\sqrt{\frac{25}{12}} \\ \\ & =\sqrt{2.08333} \\ \\ \end{aligned} & \begin{aligned} \begin{array}{l} \hspace{.5cm}2.0833 \\ 12 \enclose{longdiv}{25.0000} \\ \hspace{0.3cm} -24\\ \hline \hspace{0.9cm} 100 \\ \hspace{0.8cm} -96 \\ \hline \hspace{1.3cm} 40 \\ \hspace{1cm} -36 \\ \hline \hspace{1.5cm} 40 \\ \hspace{1.2cm} -36 \\ \hline \hspace{1.8cm} 4 \\ \hline \end{array} \end{aligned} \\\\ \end{array} \] \[ \begin{array}{r|l} & \phantom{0}1.4433 \\ \hline 1 & \phantom{-}2 . \overline{08} \, \overline{33} \, \overline{33} \, \overline{33} \\ & -1 \\ \hline 24 & \phantom{-}108 \\ & -\phantom{0}96 \\ \hline 284 & \phantom{-0}1233 \\ & -\phantom{0}1136 \\ \hline 2883 & \phantom{-000}9733 \\ & -\phantom{000}8649 \\ \hline 28863 & \phantom{-000}108433 \\ & -\phantom{0000}86589 \\ \hline & \phantom{000000}21844 \\ \end{array} \] \[ \begin{aligned} \sqrt{2 \frac{1}{12}} & = 1.4433 \\ \\ \sqrt{2 \frac{1}{12}} & = 1.443 \\ \end{aligned} \]

Answer \( \displaystyle \color{red}\sqrt{2 \frac{1}{12}} = 1.443 \)

\[ \begin{aligned} & = \sqrt{367\frac{2}{7}} \\ \\ & = \sqrt{\frac{2571}{7}} \\ \\ & = \sqrt{367.28571428} \\ \\ \end{aligned} \] \[ \begin{aligned} \begin{array}{l} \hspace{.5cm}367.28571428 \\ 7 \enclose{longdiv}{2571.0000000} \\ \hspace{0.1cm} -21\\ \hline \hspace{0.7cm} 47 \\ \hspace{0.4cm} -42 \\ \hline \hspace{1.cm} 51 \\ \hspace{.7cm} -49 \\ \hline \hspace{1.3cm} 20 \\ \hspace{1.cm} -14 \\ \hline \hspace{1.6cm} 60 \\ \hspace{1.3cm} -56 \\ \hline \hspace{1.8cm} 40 \\ \hspace{1.5cm} -35 \\ \hline \hspace{2.0cm} 50 \\ \hspace{1.7cm} -49 \\ \hline \hspace{2.2cm} 10 \\ \hspace{2.1cm} -7 \\ \hline \hspace{2.5cm} 30 \\ \hspace{2.2cm} -28 \\ \hline \hspace{2.8cm} 20 \\ \hspace{2.5cm} -14 \\ \hline \hspace{3.cm} 60 \\ \hspace{2.7cm} -56 \\ \hline \hspace{3.1cm} 4 \\ \hline \end{array} \end{aligned}\\\\ \] \[ \begin{aligned} \\ \color{green}\sqrt{367\frac{2}{7}} = \sqrt{367.28571428} \\ \\ \end{aligned} \] \[ \begin{array}{r|l} & \phantom{0}19.1646 \\ \hline 1 & \phantom{-}\overline{3} \, \overline{67} . \overline{28} \, \overline{57} \, \overline{14} \, \overline{28} \\ & -1 \\ \hline 29 & \phantom{-}267 \\ & -261 \\ \hline 381 & \phantom{-0}628 \\ & -\phantom{0}381 \\ \hline 3826 & \phantom{-0}24757 \\ & -\phantom{0}22956 \\ \hline 38324 & \phantom{000}180114 \\ & -\phantom{-}153296 \\ \hline 383286 & \phantom{0000}2681828 \\ & - \phantom{-0}2299716 \\ \hline & - \phantom{-00}382112 \\ \end{array} \] \[ \begin{aligned} \\ \sqrt{367 \frac{2}{7}} & = 19.1646 \\ \\ \sqrt{367 \frac{2}{7}} & = 19.165 \\ \\ \end{aligned} \]

Answer \( \displaystyle \color{red}\sqrt{367 \frac{2}{7}} = 19.165 \)

\[ \begin{aligned} &\text{The perfect square near 90 are 81 and 100.}\\ & 81 < 90 < 100 \\ & 9^2 < 90 < 10^2 \\ &\text{The answer lies between 9 and 10.}\\ & (9.4)^2 = 88.36 < 90 \\ & (9.5)^2 = 90.25 > 90 \\ \\ & 88.36 < 90 < 90.25 \\ & (9.4)^2 < 90 < (9.5)^2 \\ \\ & \implies 90.25 \text{ is much closer to 90} \\ &\therefore \color{green}\sqrt{90} \text{ is approximately } 9.5\\ \end{aligned} \]

Answer \( \color{red}\sqrt{90} = 9.5 \, approximately \)

\[ \begin{aligned} &\text{The perfect square near 150 are 144 and 169.}\\ & 144 < 150 < 169 \\ & 12^2 < 150 < 13^2 \\ &\text{The answer lies between 12 and 13.}\\ & (12.2)^2 = 148.84 < 150 \\ & (12.3)^2 = 151.29 > 150 \\ \\ & 148.84 < 150 < 151.29 \\ & (12.2)^2 < 150 < (12.3)^2 \\ \\ & \implies 148.84 \text{ is much closer to 150} \\ &\therefore \color{green}\sqrt{150} \text{ is approximately } 12.2\\ \end{aligned} \]

Answer \( \color{red}\sqrt{150} = 12.2 \, approximately \)

\[ \begin{aligned} &\text{The perfect square near 600 are 576 and 625.}\\ & 576 < 600 < 625 \\ & 24^2 < 600 < 25^2 \\ &\text{The answer lies between 24 and 25.}\\ & (24.4)^2 = 595.36 < 600 \\ & (24.5)^2 = 600.25 > 600 \\ \\ & 595.36 < 600 < 600.25 \\ & (24.4)^2 < 600 < (24.5)^2 \\ \\ & \implies 600.25 \text{ is much closer to 600} \\ &\therefore \color{green}\sqrt{600} \text{ is approximately } 24.5\\ \end{aligned} \]

Answer \( \color{red}\sqrt{600} = 24.5 \, approximately \)

\[ \begin{aligned} &\text{The perfect square near 1000 are 961 and 1024.}\\ & 961 < 1000 < 1024 \\ & 31^2 < 1000 < 32^2 \\ &\text{The answer lies between 31 and 32.}\\ & (31.6)^2 = 998.56 < 1000 \\ & (31.7)^2 = 1004.89 > 1000 \\ \\ & 998.56 < 1000 < 1004.89 \\ & (31.6)^2 < 1000 < (31.7)^2 \\ \\ & \implies 998.56 \text{ is much closer to 1000} \\ &\therefore \color{green}\sqrt{1000} \text{ is approximately } 31.6\\ \end{aligned} \]

Answer \( \color{red}\sqrt{1000} = 31.6 \, approximately \)

Solution

\[ \begin{aligned} \text{Total area of the cloth} &= 9 \, m^2 \\ \\ \text{Number of scarves} &= 16 \\ \\ \text{Area of 1 small scarf} &= \frac{9}{16} \, m^2 \\ \\ Side \times Side &= \frac{9}{16} \, m^2 \\ \\ (Side)^2 &= \frac{9}{16} \\ \\ Side &= \sqrt{\frac{9}{16}} \\ \\ Side &= \sqrt{\frac{3^2}{4^2}} \\ \\ Side &= {\frac{3}{4}} \, m \\ \\ \text{Length of the scarf} &= {\frac{3}{4}} \, m \\ \\ \end{aligned} \]

Answer Length of the side of the scarf \( = \displaystyle \color{red} {\frac{3}{4}} \, m \)

Solution

\[ \begin{aligned} & \text{Total area of the plot} = 800 \, m^2 \\ & \text{Side} \times \text{Side} = 800 \, m^2 \\ & (\text{Side})^2 = 800 \\ & \text{Side} = \sqrt{800} \\ \\ & \text{The perfect squares near 800 are 784 and 841} \\ & 784 < 800 < 841 \\ & 28^2 < 800 < 29^2 \\ \\ & \text{The answer lies between 28 and 29.} \\ \\ & (28.2)^2 = 795.24 < 800 \\ & (28.3)^2 = 800.89 > 800 \\ \\ & 795.24 < 800 < 800.89 \\ \\ & (28.2)^2 < 800 < (28.3)^2 \\ \\ & \implies 800.89 \text{ is much closer to 800} \\ &\therefore \color{green}\sqrt{800} \text{ is approximately } 28.3\\ \\ \end{aligned} \]

Answer Estimated length of the side of the plot \( = \color{red} 28.3 \, m \, approximately \)