ML AGGARWAL CLASS 8 CHAPTER 3 SQUARES AND ROOTS Exercise 3.3

 Exercise 3.3

Question 1

Sol :

(i) 121 

given number = 121

121-1=120 ; 120-3=117 ; 117-5=112 ; 112-7=105

105-9=96 ; 96-11=85 ; 85-13=72 ; 72-15=57

57-17=40 ; 40-19=21 ; 21-21=0

∴ 121 is a perfect square 

we have done 11 Subtractions

Hence, square root of 121 is $11 \Rightarrow \sqrt{121}$ = 11

(ii) 55 

given number = 55

55-1=54 ; 54-3=51 ; 51-5=46 ; 46-7=39

39-9=30 ; 30-11=29 ; 29-13=16 ; 16-15=1

1-17=-16

∴ 55 is not a perfect square   

(iii) 36 

Given number  =36

36-1=35 ; 35-3=32 ; 32-5=27 ; 27-7=20

20-9=11 ; 11-11=0

36 is a perfect square

we have done 6 subtractions

Hence, square root of 36 i.e $\sqrt{36}$= 6

(iv) 90

Given number = 90

90-1=89 ; 89-3=86 ; 86-5=81 ; 81-7=74: 74-9=65

65-11=54 ; 54-13=41 ; 41-15=25 ; 25-15=10 ; 10-17=-7

∴ 90 is not a pertect square

 Question 2

Sol :

(i) 784

Given number = 784 

⇒ $\begin{array}{r|l}2&784\\ \hline 2& 392 \\ \hline 2&196\\ \hline 2& 98\\ \hline 7& 49 \\ \hline 7&7\\ \hline&1\end{array}$

784 = 2 x 2 x 2 x 2 x 7 x7

$\sqrt{784}=\sqrt{2^{2} \times 2 \times 7^{2}}$

$\sqrt{784}=2 \times 2 \times 7=28$

(ii) 441

Given number = 441 

 $\begin{array}{r|l}3&441\\ \hline 3& 147\\ \hline 7&49\\ \hline 7& 7\\  \hline&1\end{array}$

441= 3 x 3x 7 x 7 

$\sqrt{441}=\sqrt{3^{2} \times 7^{2}}$

$\sqrt{441}=3 \times 7=21$

(iii) 1849

Given number = 1849

 $\begin{array}{r|l}43&1849\\ \hline 43& 43\\ \hline&1\end{array}$

1849 = 43 x 43

$\sqrt{1849}=\sqrt{43 \times 43}$

$\sqrt{1849}=43$

(iv) 4356

Given number = 4356 

 $\begin{array}{r|l}2&4356\\ \hline 2&2178 \\ \hline 3&1089\\ \hline 3& 368\\ \hline 11& 121 \\ \hline 11&11\\ \hline&1\end{array}$

4356 = 2 x 2 x 3 x 3 x 11 x 11

$\sqrt{4356}=\sqrt{2^{2} \times 3^{2} \times 11^{2}}$

$\sqrt{4356}=2 \times 3 \times 11=66$

(v) 6241 

Given number = 6241

$\begin{array}{r|l}79&6241\\ \hline 79& 79\\ \hline&1\end{array}$

 6241 = 79 x 79

$\sqrt{6241}=\sqrt{79^{2}}=79$

(vi) 8836 

Given number = 8836

 $\begin{array}{r|l}2&8836\\ \hline 2& 4418\\ \hline 47&2209\\ \hline 47&47\\  \hline&1\end{array}$

8836 = 2 x 2 x 47 x 47

$\sqrt{8836}=\sqrt{2^{2} \times 47^{2}}$

$\sqrt{8836}=2 \times 47=94$

(vi)  8281 

Given number = 8281 

 $\begin{array}{r|l}7&8281\\ \hline 7& 1183\\ \hline 13&169\\ \hline 13&13\\  \hline&1\end{array}$

8281 =7 x 7 x 13 x 13 

$\sqrt{8281}=\sqrt{7^{2} \times 13^{2}}$

$\sqrt{8281}=7 \times 13=91$ 

(viii) 9025

 $\begin{array}{r|l}5&9025\\ \hline 5& 1805\\ \hline 19&361\\ \hline 19&19\\  \hline&1\end{array}$

9025 = 5 x 5 x 19 x 19

$\sqrt{9025}=\sqrt{5^{2} \times 19^{2}}$

$\sqrt{9025}=95$

 Question 3

Sol :

(i)  $9 \frac{67}{121}=\frac{1156}{121}$

Sol: 

 ⇒ $\begin{array}{r|l}2&1156\\ \hline 2& 578\\ \hline 17&289\\ \hline 17&17\\  \hline&1\end{array}$

⇒$1156=2 \times 2 \times 17 \times 17$

⇒$9 \frac{67}{121}=\frac{2 \times 2 \times 17 \times 17}{11 \times 11}$

⇒$\sqrt{9 \frac{67}{121}}=$ $\sqrt{\frac{2 \times 2 \times 17 \times 17}{11 \times 11}}$

$=\frac{2 \times 17}{11}$

⇒ $\sqrt{9 \frac{67}{121}}=\frac{34}{11}$

(ii) $17 \frac{13}{36}=\frac{625}{36}$

Sol:

$\begin{array}{r|l}5&625\\ \hline 5& 125\\ \hline 5&25\\ \hline 5&5\\  \hline&1\end{array}$

$625=5 \times 5 \times 5 \times 15$

$17 \frac{13}{36}=\frac{5 \times 5 \times 5 \times 5}{6 \times 6}$

$\sqrt{17 \frac{13}{36}}=\sqrt{\frac{5 \times 5 \times 5 \times 5}{6 \times 6}}$

$=\frac{5 \times 5}{6}$

$\sqrt{17 \frac{13}{36}}$ = $\frac{25}{6}$

(iii) $1.96=\frac{196}{100}$

Sol:

$1.96=\frac{2 \times 2 \times 7 \times 7}{10 \times 10}$

$\begin{array}{r|l}2&196\\ \hline 2& 98\\ \hline 7&49\\ \hline 7&7\\  \hline&1\end{array}$

$196=2 \times 2 \times 7 \times 7$

$1.96=\frac{2 \times 2 \times 7 \times 7}{10 \times 10}$

$\sqrt{1.96}=\frac{2 \times 7}{10}=1.4$

(iv)  0.0064

Sol: 

$0.0064=\frac{64}{10000}$

$\begin{array}{r|l}2&64\\ \hline 2&32 \\ \hline 2&16\\ \hline 2& 8\\ \hline 2& 4 \\ \hline 2&1\\ \hline&1\end{array}$

$0.0064=\frac{2 \times 2 \times 2 \times 2 \times 2 \times 2}{10 \times 10 \times 10 \times 10}$

$\sqrt{0.0064}=\sqrt{\frac{2 \times 2 \times 2 \times 2 \times 2 \times 2}{10 \times 10 \times 10 \times 10}}$

$=\frac{2 \times 2 \times 2}{10 \times 10}=0.08$

$\sqrt{0.0064}=0.08$

Question 4

Sol :

(i) Given number =588

Expressing in prime factors

$\begin{array}{r|l}2&588\\ \hline 2&294 \\ \hline 7&147\\ \hline 3& 21\\ \hline 7& 7 \\ \hline&1\end{array}$

$588=2 \times 2 \times 7 \times 7 \times 3$

Since ' 3 ' left unpaired, so to

make   588 it should multiplied 

by ' 3 '

 

$588 \times 3=2 \times 2 \times 7 \times 7 \times 3 \times 3$

$1764=2^{2} \times 7 \times 3^{2}$

$\sqrt{1764}=\sqrt{2^{2} \times 7^{2}+3^{2}}$

$\sqrt{1764}=2 \times 7 \times 3=42$

(ii) Given number =720

Expressing in prime factors

 $\begin{array}{r|l}2& 720\\ \hline 2& 360 \\ \hline 2&180\\ \hline 2& 90\\ \hline 3& 45 \\ \hline 3&15\\  \hline 5&5\\ \hline&1\end{array}$

$720=2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5$

Since 5 left unpaired, to make 

720 perfect square it should be

multiplied by 5

$3600=2^{2} \times 3^{2} \times 5^{2} \times 2^{2}$

$\sqrt{3600}= \sqrt{2^{2} \times 2^{2} \times 3^{2} \times 5^{2}}$

$\sqrt{3600}=2 \times 2 \times 3 \times 5=60$

(iii)  Sol:

Given number 2178

Expressing in prime factors

$\begin{array}{r|l}2&2178\\ \hline 3&1089 \\ \hline 3&363\\ \hline 11& 121\\ \hline 11& 11 \\ \hline&1\end{array}$

\

$2178=2 \times 3 \times 3 \times 11 \times 11$

since 2' left unpaired to make 2178  a product square it should be multiplied by 2

$\begin{aligned} 2178 \times 2 &=2 \times 2 \times 3 \times 3 \times 11 \times 11 \\ 4356 &=2^{2} \times 3^{2} \times 11^{2} \end{aligned}$

$\sqrt{4356}=\sqrt{2^{7} \times 3^{2} \times 11^{2}}$

= 2 x 3 x 11

$\sqrt{4356}=66$

(iv) Given number =3042

Expressing in prime factors 

$\begin{array}{r|l}2&2178\\ \hline 3&1089 \\ \hline 3&363\\ \hline 11& 121\\ \hline 11& 11 \\ \hline&1\end{array}$

$3042=2 \times 3 \times 3 \times 13 \times 13$

 

since '2' left unpaired so to make 3042 a perfect square it should be multiplied by ' 2 '

$3042 \times 2=2 \times 2 \times 3 \times 3 \times 13 \times 13$

$6084=2^{2} \times 3^{2} \times 13^{2}$

$\sqrt{6084}=2 \times 3 \times 13=78$

(v) 6300

Given number =6300

Express in y in prime factors

$\begin{array}{r|l}2 & 6300 \\ \hline 2 & 3150 \\ \hline 5 & 1575 \\ \hline 5 & 315 \\ \hline 7 & 63 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline&1\end{array}$

$6300=2 \times 2 \times 5 \times 5 \times 7 \times 3 \times 3$ 

 since '7' left unpaired , so to make 6300 a perfect square , it should be multiplied by '7' 

$6300 \times 7=2 \times 2 \times 5 \times 5 \times 7 \times 7 \times 3 \times 3$

$\sqrt{44100}=\sqrt{2^{2} \times 5^{2} \times 7^{2} \times 3^{2}}$

$\sqrt{44100}=2 \times 5 \times 7 \times 3$

$\sqrt{44100}=210$

Question 5

Sol :

(i)   Given number 1872 expressing in prime factors 

$\begin{array}{r|l}2 &1872 \\ \hline 2 & 936 \\ \hline 2 & 468 \\ \hline 2&234 \\ \hline 3 & 117 \\\hline 3 & 39 \\ \hline 13 & 13 \\ \hline & 1\end{array}$ 

Since 13 left unpaired, so to make 1872 a perfect square it should be divided by 13

$\frac{1872}{13}=\frac{2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 13}{13}$

$144=2 \times 2 \times 2 \times 2 \times 3 \times 3$

$\sqrt{144}=\sqrt{2^{2} \times 2^{2} \times 3^{2}}$

$\sqrt{144}=2 \times 2 \times 3=12$

(ii)  Given number 2592  expressing in prime number

$\begin{array}{r|l}2 &2592 \\ \hline 2 & 1296 \\ \hline 2 & 648 \\ \hline 2&324 \\ \hline 2 & 162 \\ \hline 2 & 81 \\ \hline 3 & 27 \\ \hline3 & 9 \\ \hline3 & 3 \\ \hline & 1\end{array}$ 

2592 =  $2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$

Since ' 2 ' lets unpaired, so to

make 2592 a prefect square, it

Should be divided ' 2 '

$\frac{2592}{2}=\frac{2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3}{2}$

sqrt{2^{2}

$1296=2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$$1

$\sqrt{1296}=\sqrt{2^{2} \times 2^{2} \times 3^{2} \times 3^{2}}$

$\sqrt{1296}=36$

(iii)  Given number 3380 expressing interms of prime factors 

$\begin{array}{r|l}2&3380\\ \hline 2&1690 \\ \hline 5&845\\ \hline 13& 169\\ \hline 13& 13 \\ \hline&1\end{array}$

since '5' is left unpaired so to make a perfect square it should be divided by 5 

$3380 \div 5=\frac{2 \times 2 \times 5 \times 13 \times 13}{5}$

$676=2 \times 2 \times 13 \times 13$

$\sqrt{676}=\sqrt{2 \times 2 \times 13 \times 13}$

$\sqrt{676}=2 \times 13 \in 26$

(iv) 16244

Expressing in terms of prime factors

16244

$\begin{array}{r|l}2 &16244 \\ \hline 2 & 8122 \\ \hline11 & 4061 \\ \hline 3&371 \\ \hline 3 & 117 \\ \hline 3 & 39 \\ \hline 13 & 13 \\ \hline & 1\end{array}$ 

Question 8 

Sol: let breadth of rectangle = x .m 

given 

Length of rectangle is equal to 4 times the breadth 

∴ L = 4 . x 

area of rectangle = 1936 

l x b = 1936 

$\begin{aligned} 4 x \times x &=1936 \\ x^{2} &=\frac{1936}{4} \\ x^{r} &=484 \\ x &=22 \mathrm{~m} \end{aligned}$

hence breadth of rectangle = x = 22m 

length of rectangle $=4 x=4 \times 22=88 \mathrm{~m}$

Question 9

Sol :

let the no.of columns  = x

given no.of rows equal to no of column 

ஃ no of rows = x

Total students equal to 2000 but 64 students

could not be accommodated in These rows columns

$\begin{aligned} \therefore \quad x \times x &=2000-64 \\ x^{2} &=1936 \\ x &=\sqrt{1936} \\ x &=44 \end{aligned}$

hence , no of rows = 44

$\begin{array}{r|l}2 &1936 \\ \hline 2 & 868 \\ \hline2 &434 \\ \hline 2&242 \\ \hline 11 & 121 \\ \hline 11 & 11\\ \hline & 1\end{array}$

$1936=2 \times 2 \times 2 \times 2 \times 11 \times 11$

$\sqrt{1936}=2 \times 2 \times 11=44$

(v) Given number 61347 

expressing in terms of prime factors 

$\begin{array}{r|l}3&61347\\ \hline 11&20449 \\ \hline 11&1859\\ \hline 13& 169\\ \hline 13& 13 \\ \hline&1\end{array}$

61347 = $=3 \times 11 \times 11 \times 13 \times 13$

since 3 left unpaired so to make 61347 a perfect square it should be divided by 3

$61347 \div 3=\frac{3 \times 11 \times 11 \times 13 \times 13}{3}$

$20449=11 \times 11 \times 13 \times 13$

$\sqrt{20449}=\sqrt{11^{2} \times 13^{2}}$

$\sqrt{20449}=11 \times 13={143}$

(vi) let no of rows of plants in garden = x 

given each row contains as many plants as the no of rows 

no of plants in each row = x 

total no of plants = $x \times x=x^{2}$

given Total no.of plants in garden =4225

$\begin{aligned} \therefore \quad x^{2} &=4225 \\ x &=\sqrt{4225} \\ & x=65 \end{aligned}$

hence, no.of rows in garden = 65

no.of plants in a row $=65$

Question 10 

Sol :

Let no of students = x 

given contribution of such student = no of students 

∴ contribution of each students = $\bar{₹} x$

Total collected for Picnic =22304

∴ $x \times x=2304$

$x^{2}=2304$

$x=\sqrt{2304}$

x=48

Question 11

Sol :

Let number is 15 times the other 

second number = 15.x

product of two number = 7260 

$\begin{aligned} 15 x \cdot x &=7260 \\ x^{2} &=\frac{7260}{15} \\ x^{2} &=484 \\ x &=\sqrt{484} \\ x &=22 \end{aligned}$

$\begin{array}{r|l}2&484\\ \hline 2&242 \\ \hline 11&121\\ \hline 11& 11\\ \hline&1\end{array}$

$484=2 \times 2 \times 11 \times 11$

$\sqrt{484}=2 \times 11=22$

Question 12 

Sol :

Given number are in ratio of 2 : 3 : 5 

let number be 2x , 3x , 5x 

given sum of squares of numbers $=950$

$(2 x)^{2}+(3 x)^{2}+(5 x)^{2}=950$

$4 x^{2}+9 x^{2}+25 x^{2}=950$

$x^{2}=\frac{950}{38}$

$x^{2}=25$

x=5

$\begin{aligned} \text { Hence, } & \text { Numbers are } 2 x, 3 x, 5 x \\ & 10,15,25 \end{aligned}$

 

Question 13

Sol:

Perimeters of two Squares $=60 \mathrm{~m}, 144 \mathrm{~m}$

$P_{1}=60 \mathrm{~m} ; P_{2}=144 \mathrm{~m}$

Perimeter let length of sides of square are = $x_{1}, x_{2}$

$\begin{aligned} \therefore P_{1} &=4 x_{1} \\ 60 &=4 x_{1} \\ x_{1} &=15 \mathrm{~m} \end{aligned}$

$P_{2}=4 x_{2}$

$144=4 x_{2}$

$x_{2}=36 m$

$\begin{aligned} \text { Areas } A_{1} &=x_{1}^{2} \\ A_{1} &=15^{2} \\ A_{1} &=225 \mathrm{~m}^{2} \end{aligned}$

Area $\begin{aligned} A_{2} &=x_{2}^{2} \\ A_{2} &=36^{2} \\ A_{2} &=1296 \mathrm{~m}^{2} \end{aligned}$

let a square of side x with area ' $A_{1}+A_{2}$'

$A=A_{1}+A_{2}$

$x^{2}=225+1296$

$x^{2}=1521$

$\begin{array}{r|l}3&1521\\ \hline 3&507 \\ \hline 13&169\\ \hline 13& 13\\ \hline&1\end{array}$

$\begin{aligned} 1521 &=3 \times 3 \times 13 \times 13 \\ \sqrt{1521} &=3 \times 13=39 \end{aligned}$

$x^{2}=1521$

$x_{2} \sqrt{1521}$

$x=39 \mathrm{~m}$

Perimeter of square 

$\begin{aligned} P &=4 x \\ &=4 \times 39 \\ P &=156 \mathrm{~m} \end{aligned}$

∴ Hence, perimeter =156 m