Thursday, May 28, 2020

NCERT Solutions for class 8 Maths chapter 6 Square and square roots exercise 6.4

Exercise 6.4



Introduction Ex 6.4, Squares & Square Root, NCERT Class 8th Maths Solutions



Q 1 - Ex 6.4 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6


Question 1
Find the square root of each of the following numbers by division method.
(i) 2304
Sol :
(i) The square root of 2304 can be calculated as follows.
$\begin{array}{r|l}
& 48 \\
\hline 4 &\phantom{-} \overline{23}~\overline{04} \\
& -16 \\
\hline 88 & \phantom{-}704 \\
& \phantom{-}704 \\
\hline & \phantom{-}0
\end{array}$
∴ $\sqrt{2304}=48$

(ii) 4489
Sol :
(ii) The square root of 4489 can be calculated as follows.
$\begin{array}{r|l}

& 67 \\

\hline{6} & \phantom{-}\overline{44} ~\overline{89} \\ & -36 \\

\hline {127} & \phantom{-}889 \\

& \phantom{-}889 \\

\hline & \phantom{-}0


\end{array}$

∴ $\sqrt{4489}=67$

(iii) 3481
Sol :
(iii) The square root of 3481 can be calculated as follows.
$\begin{array}{r|l}

& 59 \\

\hline 5& \phantom{-} \overline{81} \\

& -25 \\

\hline 109& \phantom{-}981 \\

& \phantom{-}981 \\

\hline & \phantom{-}0


\end{array}$

Therefore, $\sqrt{3481}=59$

(iv) 529
Sol :
(iv) The square root of 529 can be calculated as follows.
$\begin{array}{r|l}& 23 \\\hline 2& \phantom{-} \overline{5}~\overline{29} \\& -4 \\

\hline 43& \phantom{-}129 \\& \phantom{-}129 \\\hline & \phantom{-}0


\end{array}$

∴ $\sqrt{529}=23$


(v) 3249
Sol :
(v) The square root of 3249 can be calculated as follows.
$\begin{array}{r|l}& 57 \\\hline 5& \phantom{-} \overline{32}~\overline{49} \\& -25 \\

\hline 107& \phantom{-}749 \\& \phantom{-}749 \\\hline & \phantom{-}0


\end{array}$

∴ $\sqrt{3249}=57$

(vi) 1369
Sol :
(vi) The square root of 1369 can be calculated as follows.
$\begin{array}{r|l}& 37 \\\hline 3& \phantom{-} \overline{13}~\overline{69} \\& -9 \\

\hline 67& \phantom{-}469 \\& \phantom{-}469 \\\hline & \phantom{-}0


\end{array}$

∴ $\sqrt{1369}=37$


(vii) 5776
Sol :
(vii) The square root of 5776 can be calculated as follows.
$\begin{array}{r|l}& 76 \\\hline 7& \phantom{-} \overline{57}~\overline{76} \\& -49 \\

\hline 146& \phantom{-}876 \\& \phantom{-}876 \\\hline & \phantom{-}0


\end{array}$

∴ $\sqrt{5776}=76$

(viii) 7921
Sol :
(viii) The square root of 7921 can be calculated as follows.
$\begin{array}{r|l}& 89 \\\hline 8& \phantom{-} \overline{79}~\overline{21} \\& -64 \\

\hline 169& \phantom{-}1521 \\& \phantom{-}1521 \\\hline & \phantom{-}0


\end{array}$

∴ $\sqrt{7921}=89$


(ix) 576
Sol :
(ix) The square root of 576 can be calculated as follows.
$\begin{array}{r|l}& 24 \\\hline 2& \phantom{-} \overline{5}~\overline{76} \\& -4 \\

\hline 44& \phantom{-}176 \\& \phantom{-}176 \\\hline & \phantom{-}0


\end{array}$

∴ $\sqrt{576}=24$

(x) 1024
Sol :
(x) The square root of 1024 can be calculated as follows.
$\begin{array}{r|l}& 32 \\\hline 3& \phantom{-} \overline{10}~\overline{24} \\& -9 \\

\hline 62& \phantom{-}124 \\& \phantom{-}124 \\\hline & \phantom{-}0


\end{array}$

∴ $\sqrt{1024}=32$


(xi) 3136
Sol :
(xi) The square root of 3136 can be calculated as follows.
$\begin{array}{r|l}& 56 \\\hline 5& \phantom{-} \overline{31}~\overline{36} \\& -25 \\

\hline 106& \phantom{-}636 \\& \phantom{-}636 \\\hline & \phantom{-}0


\end{array}$

∴ $\sqrt{3136}=56$

(xii) 900
Sol :
(xii) The square root of 900 can be calculated as follows.
$\begin{array}{r|l}& 30 \\\hline 3& \phantom{-} \overline{9}~\overline{00} \\& -9 \\

\hline 60& \phantom{-}00 \\& \phantom{-}00 \\\hline & \phantom{-}0


\end{array}$

∴ $\sqrt{900}=30$


Q 2 - Ex 6.4 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6

Question 2
Find the number of digits in the square root of each of the following numbers (without any calculation).
(i) 64
Sol :
(i) By placing bars, we obtain
$64=\overline{64}$
Since there is only one bar, the square root of 64 will have only one digit in it.

(ii) 144
Sol :
(ii) By placing bars, we obtain
$144=\overline{1}~ \overline{44}$
Since there are two bars, the square root of 144 will have 2 digits in it.


(iii) 4489
Sol :
(iii) By placing bars, we obtain
$4489=\overline{44}~ \overline{89}$
Since there are two bars, the square root of 4489 will have 2 digits in it.

(iv) 27225
Sol :
(iv) By placing bars, we obtain
$27225=\overline{2}~ \overline{72}~ \overline{25}$
Since there are three bars, the square root of 27225 will have three digits in it.

(v) 390625
Sol :
(v) By placing the bars, we obtain
$390625=\overline{39}~\overline{06}~\overline{25}$
Since there are three bars, the square root of 390625 will have 3 digits in it.


Q 3 - Ex 6.4 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6



Question 3
Find the square root of the following decimal numbers.
(i) 2.56
Sol :
(i) The square root of 2.56 can be calculated as follows.
$\begin{array}{r|l}& 1.6 \\\hline 1& \phantom{-} \overline{2.}~\overline{56} \\& -1 \\

\hline 26& \phantom{-}156 \\& \phantom{-}156 \\\hline & \phantom{-}0


\end{array}$

∴ $\sqrt{2.56}=1.6$

(ii) 7.29
Sol :
(ii) The square root of 7.29 can be calculated as follows.
$\begin{array}{r|l}& 2.7 \\\hline 2& \phantom{-} \overline{7}.~\overline{29} \\& -4\\

\hline 47& \phantom{-}329 \\& \phantom{-}329 \\\hline & \phantom{-}0


\end{array}$

∴ $\sqrt{7.29}=2.7$


(iii) 51.84
Sol :
(iii) The square root of 51.84 can be calculated as follows.
$\begin{array}{r|l}& 7.2 \\\hline 7& \phantom{-} \overline{51}.~\overline{84} \\& -49\\

\hline 142& \phantom{-}284 \\& \phantom{-}284 \\\hline & \phantom{-}0


\end{array}$

∴ $\sqrt{51.84}=7.2$

(iv) 42.25
Sol :
(iv) The square root of 42.25 can be calculated as follows.
$\begin{array}{r|l}& 6.5 \\\hline 6& \phantom{-} \overline{42}.~\overline{25} \\& -36\\

\hline 125& \phantom{-}625 \\& \phantom{-}625 \\\hline & \phantom{-}0


\end{array}$

∴ $\sqrt{42.25}=6.5$


(v) 31.36
Sol :
(v) The square root of 31.36 can be calculated as follows.
$\begin{array}{r|l}& 5.6 \\\hline 5& \phantom{-} \overline{31}.~\overline{36} \\& -25\\

\hline 106& \phantom{-}636 \\& \phantom{-}636 \\\hline & \phantom{-}0


\end{array}$

∴ $\sqrt{31.36}=5.6$


Q 4 - Ex 6.4 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6




Question 4
Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 402
Sol :
(i) The square root of 402 can be calculated by long division method as follows.
$\begin{array}{r|l}& 20 \\\hline 2& \phantom{-} \overline{4}~\overline{02} \\& -4\\

\hline 40& \phantom{-}02 \\& \phantom{-}00 \\\hline & \phantom{-}2


\end{array}$The remainder is 2. It represents that the square of 20 is less than 402 by 2. Therefore, a perfect square will be obtained by subtracting 2 from the given number 402.
Therefore, required perfect square = 402 − 2 = 400
And, $\sqrt{400}=20$

(ii) 1989
Sol :
(ii) The square root of 1989 can be calculated by long division method as follows.
$\begin{array}{r|l}& 44 \\\hline 4& \phantom{-} \overline{19}~\overline{89} \\& -16\\

\hline 84& \phantom{-}389 \\& \phantom{-}389 \\\hline & \phantom{-}53


\end{array}$

The remainder is 53. It represents that the square of 44 is less than 1989 by 53. Therefore, a perfect square will be obtained by subtracting 53 from the given number 1989.
Therefore, required perfect square = 1989 − 53 = 1936
And,$\sqrt{1936}=44$

(iii) 3250
Sol :
(iii) The square root of 3250 can be calculated by long division method as follows.
$\begin{array}{r|l}& 57 \\\hline 5& \phantom{-} \overline{32}~\overline{50} \\& -25\\

\hline 107& \phantom{-}750 \\& \phantom{-}750 \\\hline & \phantom{-}1


\end{array}$

The remainder is 1. It represents that the square of 57 is less than 3250 by 1. Therefore, a perfect square can be obtained by subtracting 1 from the given number 3250.
Therefore, required perfect square = 3250 − 1 = 3249
And, $\sqrt{3249}=57$

(iv) 825
Sol :
(iv) The square root of 825 can be calculated by long division method as follows.
$\begin{array}{r|l}& 28 \\\hline 2& \phantom{-} \overline{8}~\overline{25} \\& -4\\

\hline 48& \phantom{-}425 \\& \phantom{-}384 \\\hline & \phantom{-}41


\end{array}$

The remainder is 41. It represents that the square of 28 is less than 825 by 41. Therefore, a perfect square can be calculated by subtracting 41 from the given number 825.
Therefore, required perfect square = 825 − 41 = 784
And, $\sqrt{784}=28$

(v) 4000
Sol :
(v) The square root of 4000 can be calculated by long division method as follows.
$\begin{array}{r|l}& 63 \\\hline 6& \phantom{-} \overline{40}~\overline{00} \\& -36\\

\hline 123& \phantom{-}400 \\& \phantom{-}369 \\\hline & \phantom{-}31


\end{array}$

The remainder is 31. It represents that the square of 63 is less than 4000 by 31. Therefore, a perfect square can be obtained by subtracting 31 from the given number 4000.
Therefore, required perfect square = 4000 − 31 = 3969
And,$\sqrt{3969}=63$


Q 5 - Ex 6.4 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6



Question 5
Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 525
Sol :
(i) The square root of 525 can be calculated by long division method as follows.
$\begin{array}{r|l}& 22 \\\hline 2& \phantom{-} \overline{5}~\overline{25} \\& -4\\

\hline 142& \phantom{-}125 \\& \phantom{-}84 \\\hline & \phantom{-}41


\end{array}$

The remainder is 41.
It represents that the square of 22 is less than 525.
Next number is 23 and 232 = 529
Hence, number to be added to 525 = 232 − 525 = 529 − 525 = 4
The required perfect square is 529 and $\sqrt{529}=23$

(ii) 1750
Sol :
(ii) The square root of 1750 can be calculated by long division method as follows.
$\begin{array}{r|l}& 41 \\\hline 4& \phantom{-} \overline{17}~\overline{50} \\& -16\\

\hline 81& \phantom{-}150 \\& \phantom{-}81 \\\hline & \phantom{-}69


\end{array}$The remainder is 69.
It represents that the square of 41 is less than 1750.
The next number is 42 and 422 = 1764
Hence, number to be added to 1750 = 422 − 1750 = 1764 − 1750 = 14
The required perfect square is 1764 and $\sqrt{1764}=42$


(iii) 252
Sol :
(iii) The square root of 252 can be calculated by long division method as follows.
$\begin{array}{r|l}& 15 \\\hline 1& \phantom{-} \overline{2}~\overline{52} \\& -1\\

\hline 25& \phantom{-}152 \\& \phantom{-}125 \\\hline & \phantom{-}27


\end{array}$

The remainder is 27. It represents that the square of 15 is less than 252.
The next number is 16 and 162 = 256
Hence, number to be added to 252 = 162 − 252 = 256 − 252 = 4
The required perfect square is 256 and $\sqrt{256}=16$

(iv) 1825
Sol :
(iv) The square root of 1825 can be calculated by long division method as follows.
$\begin{array}{r|l}& 42 \\\hline 4& \phantom{-} \overline{18}~\overline{25} \\& -1\\

\hline 82& \phantom{-}225 \\& \phantom{-}164 \\\hline & \phantom{-}61


\end{array}$

The remainder is 61. It represents that the square of 42 is less than 1825.
The next number is 43 and 432 = 1849
Hence, number to be added to 1825 = 432 − 1825 = 1849 − 1825 = 24
The required perfect square is 1849 and $\sqrt{1849}=43$


(v) 6412
Sol :
(v) The square root of 6412 can be calculated by long division method as follows.
$\begin{array}{r|l}

& 80 \\

\hline 8&\phantom{-} 6412 \\

& -64 \\

\hline {160} &\phantom{-} 012 \\

&\phantom{-} 0 \\

\hline & \phantom{-}12

\end{array}$
The remainder is 12.
It represents that the square of 80 is less than 6412.
The next number is 81 and 812 = 6561
Hence, number to be added to 6412 = 812 − 6412 = 6561 − 6412 = 149
The required perfect square is 6561 and $\sqrt{6561}=81$


Q 6 - Ex 6.4 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6




Question 6
Find the length of the side of a square whose area is 441 m2.
Sol :
Let the length of the side of the square be x m.
Area of square = (x)2 = 441 m2
$x=\sqrt{441}$
The square root of 441 can be calculated as follows.
$\begin{array}{r|l}

& 21 \\

\hline 2& \phantom{-}\overline{4}~\overline{41} \\

& -4 \\

\hline 41 & \phantom{-}41 \\

&\phantom{-} 41 \\

\hline & \phantom{-}0 \\





\end{array}$

∴ x=21m
Hence, the length of the side of the square is 21 m.


Q 7 - Ex 6.4 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6



Question 7
In a right triangle ABC, ∠B = 90°.
(a) If AB = 6 cm, BC = 8 cm, find AC
(b) If AC = 13 cm, BC = 5 cm, find AB
Sol :
(a) ΔABC is right-angled at B.
Therefore, by applying Pythagoras theorem, we obtain
AC2 = AB2 + BC2
AC2 = (6 cm)2 + (8 cm)2
AC2 = (36 + 64) cm2 =100 cm2
AC $=(\sqrt{100}) \mathrm{cm}=(\sqrt{10 \times 10}) \mathrm{cm}$
AC = 10 cm
(b) ΔABC is right-angled at B.
Therefore, by applying Pythagoras theorem, we obtain
AC2 = AB2 + BC2
(13 cm)2 = (AB)2 + (5 cm)2
AB2 = (13 cm)2 − (5 cm)2 = (169 − 25) cm2 = 144 cm2
AB $=(\sqrt{144}) \mathrm{cm}=(\sqrt{12 \times 12}) \mathrm{cm}$
AB = 12 cm


Q 8 - Ex 6.4 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6

Question 8
A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.
Sol :
It is given that the gardener has 1000 plants. The number of rows and the number of columns is the same.
We have to find the number of more plants that should be there, so that when the gardener plants them, the number of rows and columns are same.
That is, the number which should be added to 1000 to make it a perfect square has to be calculated.
The square root of 1000 can be calculated by long division method as follows.
$\begin{array}{r|l}

& 31 \\

\hline 3& \phantom{-}\overline{10}~\overline{ 00} \\& -9 \\

\hline {61} & \phantom{-}100 \\

& \phantom{-}61 \\

\hline & \phantom{-}39


\end{array}$

The remainder is 39. It represents that the square of 31 is less than 1000.
The next number is 32 and 322 = 1024
Hence, number to be added to 1000 to make it a perfect square
= 322 − 1000 = 1024 − 1000 = 24
Thus, the required number of plants is 24.


Q 9 - Ex 6.4 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6

Question 9
These are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement?
Sol :
It is given that there are 500 children in the school. They have to stand for a P.T. drill such that the number of rows is equal to the number of columns.
The number of children who will be left out in this arrangement has to be calculated. That is, the number which should be subtracted from 500 to make it a perfect square has to be calculated.
The square root of 500 can be calculated by long division method as follows.
$\begin{array}{c|c}

& 22 \\

\hline {2} & \begin{aligned}\overline{5}& 00 \\ -4 \end{aligned}\\

\hline {42} & 100 \\& 84 \\

\hline & 16

\end{array}$

The remainder is 16.
It shows that the square of 22 is less than 500 by 16. Therefore, if we subtract 16 from 500, we will obtain a perfect square.
Required perfect square = 500 − 16 = 484
Thus, the number of children who will be left out is 16.

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