Exercise 3.3
Question 1
By repeated subtraction of odd numbers starting from 1, find whether the following numbers are perfect squares or not? If the number is a perfect square then find its square root:
(i) 121
(ii) 55
(iii) 36
(iv) 90
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 perfect square
Question 2
Find the square roots of the following numbers by prime factorisation method:
(i) 784
(ii) 441
(iii) 1849
(iv) 4356
(v) 6241
(vi) 8836
(vii) 8281
(viii) 9025
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
Find the square roots of the following numbers by prime factorisation method:
(i) $9\dfrac{67}{121}$
(ii) $17\dfrac{13}{36}$
(iii) 1.96
(iv) 0.0064
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
For each of the following numbers, find the smallest natural number by which it should be multiplied so as to get a perfect square. Also, find the square root of the square number so obtained:
(i) 588
(ii) 720
(iii) 2178
(iv) 3042
(v) 6300
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
For each of the following numbers, find the smallest natural number by which it should be divided so that this quotient is a perfect square. Also, find the square root of the square number so obtained:
(i) 1872
(ii) 2592
(iii) 3380
(iv) 16224
(v) 61347
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 6
Find the smallest square number that is divisible by each of the following numbers:
(i) 3, 6, 10, 15
(ii) 6, 9, 27, 36
(iii) 4, 7, 8, 16
Sol :
(i) 3, 6 , 10, 15
The number which is divisible by
3, 6, 10, 15=The LCM of3,6,10,15
$\begin{array}{r|l}2 & 3, 6,10,15 \\ \hline 3 & 3,3,5,15 \\ \hline 5 & 1,1,5,5 \\ \hline & 1,1,1,1\end{array}$
=2×3×5=30
and smallest square number which is divisible by 30 = 30×30 = 900
(ii) 6, 9, 27, 36
The number which is divisible by 6, 9, 27, 36 is their LCM
$\begin{array}{r|l}2 & 6, 9, 27, 36 \\ \hline 2 & 3,9,27,18 \\ \hline 3& 3,9,27,9 \\ \hline 3& 1,3,9,3 \\ \hline 3 & 1,1,3,1 \\ \hline & 1,1,1,\end{array}$
=3×3×2×2×3=108
and the smallest square
=108×3=324
(iii) 4, 7, 8, 16
The number which is divisible by 4, 7, 8, 16 = LCM of their numbers
$\begin{array}{r|l} 2 & 4,7,8,16 \\ \hline 2 & 2,7,4,8 \\ \hline 2 & 1,7,2,4 \\ \hline 2 & 1,7,1,2 \\ \hline 7 & 1,7,1,1 \\ \hline & 1,1,1,1 \end{array}$
=2×2×2×2×7=112
The smallest square =112×7 =784
Question 7
4225 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Sol :
Total number of plants = 4225
∵The number of rows = Number of the plant in each row.
Number of rows = Square root of 4225
$=\sqrt{5 \times 5 \times 13 \times 13}$
=5×13 =65
Number of rows=65
and number of plants in each row=65
Question 8
The area of a rectangle is 1936 sq. m. If the length of the rectangle is 4 times its breadth, find the dimensions of the rectangle.
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
In a school a P.T. teacher wants to arrange 2000 students in the form of rows and columns for P.T. display. If the number of rows is equal to number of columns and 64 students could not be accommodated in this arrangement. Find the number of rows.
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
In a school, the students of class VIII collected ₹2304 for a picnic. Each student contributed as mdny rupees as the number of students in the class. Find the number of students in the class.
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
The product of two numbers is 7260. If one number is 15 times the other number, find the numbers.
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
Find three positive numbers in the ratio 2 : 3 : 5, the sum of whose squares is 950.
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
The perimeter of two squares is 60 metres and 144 metres respectively. Find the perimeter of another square equal in area to the sum of the first two squares.
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
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