Exercise 3.1
Question 1
(i) 729
Sol :
Prime fractorisation
$\begin{array}{r|l}3&729\\ \hline 3& 243 \\ \hline 3&81\\ \hline 3& 27\\ \hline 3& 9 \\ \hline 3&3\\ \hline&1\end{array}$
⇒729=3×3×3×3×3×3
∴$729=27^{2}$
because 729 can be expressed as product of pairs of equal prime factors.
(ii) 5488
Sol:
$\begin{array}{r|l}2& 5488\\ \hline 2& 2744 \\ \hline 2&1372\\ \hline 2& 686\\ \hline 7& 343 \\ \hline 7&49\\ \hline 7&7\\ \hline&1\end{array}$
⇒5488= 2×2×2×2×7×7×7
Since 7 left unpaired 5488 in not a perfect square
(iii)1024
Sol:
$\begin{array}{r|l}2&1024\\ \hline 2& 512 \\ \hline 2&256\\ \hline 2& 128\\ \hline 2& 64 \\ \hline 2&32\\ \hline2&16 \\ \hline 2&8\\ \hline 2&4\\ \hline 2&2\\ \hline&1\end{array}$
⇒1024= 2x2x2x2x2x2x2x2x2x2x2
since 1024 is expressed as the product of pairs of equal prime number so it is a perfect square
(iv) 243
Sol:
$\begin{array}{r|l}3&243\\ \hline 3& 81 \\ \hline 3&27\\ \hline 3& 9\\ \hline 3& 3 \\ \hline&1\end{array}$
$243=3 \times 3 \times 3 \times 3 \times 3$
As ' 3 ' let unpaired, So 243 is not a square (perfect)
Question 2
(i) 1296
$\begin{array}{r|l}2&1246\\ \hline 2& 648 \\ \hline 2&324\\ \hline 2& 162\\ \hline 3& 81 \\ \hline 3&27\\ \hline3&9 \\ \hline 3&3\\ \hline&1\end{array}$
⇒1296=2x2x2x2x3x3x3x3
since 1296 is expressed as the product of pairs of equal prime numbers so it is a perfect square
1296= $2^{2} \times 2^{2} \times 3^{2} \times 3^{2}$
1296= $(2 \times 2 \times 3 \times 3)^{2}=36^{2}$
∴ 1296 is square of 36
(ii) 1784
Sol:
$\begin{array}{r|l}2&1784\\ \hline 2& 892 \\ \hline 2&446\\ \hline 223& 223\\ \hline&1\end{array}$
1784= $=2 \times 2 \times 2 \times 223$
As 1784 can not be expressed as product of pairs of
equal prime factors, so is not a perfect square
(iii) 3025
Sol:
$\begin{array}{r|l}5&3025\\ \hline 5& 605 \\ \hline 11&121\\ \hline&1\end{array}$
$3025=5 \times 5 \times 11 \times 11$
Since 3025 can be expressed as the product of pairs of
equal prime factors.
$3025=(5 \times 11)^{2}=55^{2}$
Hence, 55 is a number whose square is 3025
(iv) 3969
Sol:
$\begin{array}{r|l}3&3969\\ \hline 3& 1323 \\ \hline 3&441\\ \hline 3& 147\\ \hline 3& 49 \\ \hline 7&7\\ \hline&1\end{array}$
$3969=3 \times 3 \times 3 \times 3 \times 7 \times 7$
3969 Can be expressed as product of pairs of
equal prime numbers.
$3969=3^{4} \times 3^{2} \times 7^{2}$
$3969=(3 \times 3 \times 7)^{2}$
$3969=63^{2}$
Hence, 63 is the number whose square is 3969
Question 3
1008
Sol: $\begin{array}{r|l}2& 1008\\ \hline 2& 504 \\ \hline 2&252\\ \hline 2& 126\\ \hline 7& 63 \\ \hline 3&9\\ \hline 3&3\\ \hline&1\end{array}$
$1008=2 \times 2 \times 2 \times 2 \times 7 \times 3 \times 3$
Since' 7 ' is left unpaired, so to make 1008 a
Perfect square it should be muttiplied by 7
Question 4
5808
Sol:
⇒ $\begin{array}{r|l}2& 5808\\ \hline 2& 2904 \\ \hline 2&1452\\ \hline 2& 726\\ \hline 3& 363 \\ \hline 11&121\\ \hline 11&11\\ \hline&1\end{array}$
$5808=2 \times 2 \times 2 \times 2 \times 3 \times 11 \times 11$
Since ' 3' let unpaired. To make 5808 a perfect square
it should be divided by '3'.
so divide 5808 by '3'
$\frac{5808}{3}=\frac{2 \times 2 \times 2 \times 2 \times 3 \times 11 \times 11}{3}$
$1936=(2 \times 2 \times 11)^{2}=(44)^{2}$
so 44 is a number whose square is 1936
Exercise 3.2
Question 1
(i) 2
(ii) 13
(iii) 27
(iv) 88
(v) 243
Question 2
(i) 1
(ii) 4
(iii) 1
(iv) 9
(v) 6
(vi) 5
(vii) 9
(viii) 4
(ix) 0
(x) 6
Question 3
(i) 567
567 has '7' in its unit's place. a perfect square
Should have 1,4,5,6,9,0 in it's unit's place.
so 567 is not a persect square.
(ii) 2453
2453 has '3' in it's unit's place. But a pertect square
should have 0,1,4,5,6,9 in it's unit's place.
So 2453 is not a persect square.
(iii) 5298
5298 has 8 in it's unit's place. But a perfect square
should have 0,1,4,5,6,9 in it's unit's place.
so 5298 is not a persect square.
(iv)4692
46292 has 2 in it's unit's place. But a perfect square
Should have 0,1,4,5,6,9 in it's unit's place
so 46292 is not a pertect square.
(v) 74000
74000 has 0 in it's unit's place but it has
odd no.of zero's and 740 is not a perfect square
so 74000 is not a perfeet square.
Question 4
(i) 573
square of 573 is a odd number because ,If a number
has 3 in the units place , then its square and in '9'
(ii) 4096
Square of 4096 is a even number because, If a rumber
has ' 6 ' in the units place, Then its square ends in ' 6 '
iii) 8267
Squar of 8267 is a odd number becouse, If a number
has 7 in the Units place, Then its square ends in ' 9
iv) 37916
square of 37916 is a even number becaise if a number
has ' 6 ' in the Units place, then it square ends in ' 6 '
Question 5
i. 12 and 13
There are 2n non-square numbers betweew the squares of
two conseclutive numbers n and n+1
∴ natural numbers between 12 and $(12+1)=2 \times 12=24$
hence , there are 24 natural number between $12^{2}$ and $13^{2}$
(ii) 90 and 91
There are 2n non-square rumbers betweow the Squares
of two consective numbers n and n+1
∴ Natural numbers between 90 and 91=2 \times 90= 180
Hence, There are 180 natural numbers betwecn $90^{2}$ and $91^{2}$
Question 6
(i) $1+3+5+7+9+11+13=7^{2}=49$
(ii) $1+3+5+7+9+11+13+15+17+\cdots+29=15^{2}=225$
sum of first 'n' odd numbers = $n^{2}$
Question 7
(i) 64
$64-1=63 ; 63-3=60 ; 60-5=55$
$55-7=48: \quad 48-9=39 ; \quad 39-11=28$
$28-13=15 ; \quad 15-15=0$
∴ $64=1+3+5+7+9+11+13+15=8^{2}$
(ii) 121
$121-1=120 ; 120-3=117 ; \quad 117-5=112 ; 112-7=105 ;$
$105-9=96 ; 96-11=85 ; 85-13=72 ; 72-15=57 ;$
$57-17=40 ; 40-19=21 ; \quad 21-21=0$
∴ $121=1+3+5+7+9+11+13+15+17+19+21=11^{2}$
Question 8
(i) $19^{2}=361$
"we can exrpress the square of any odd number greater
than 1 as the sum of two consective natural numbers."
First number $=\frac{19^{2}-1}{2}=180$
Second number $=\frac{19^{2}+1}{2}=181$
$19^{2}=361=180+181$
(ii) $33^{2}=1089$
First number $=\frac{33^{2}-1}{2}=544$
Second number $=\frac{33^{2}+1}{2}=545$
$33^{2}=1089=544+545$
(iii) $47^{2}=2209$
First number $=\frac{47^{2}-1}{2}=1104$
Second number $=\frac{47^{2}+1}{2}=1105$
$47^{2}=2209=1104+1105$
Question 9
(i) $31^{2}=(30+1)^{2}=(30+1)(30+1)$
$=30(30+1)+1(30+1)$
$=900+30+30+1$
$31^{2}=961$
(ii) $42^{2}=(40+2)^{2}=(40+2)(40+2)$
$=40(40+2)+2(40+2)$
$=1600+80+80+4$
$42^{2}=1764$
(iii) $86^{2}=(80+6)^{2}=(80+6)(80+6)$
$=80(80+6)+6(80+6)$
$=6400+480+480+36$
$86^{2}=7396$
(iv) $94^{2}=(90+4)^{2}=(90+4)(90+4)$
$=90(90+4)+4(90+4)$
$=8100+360+360+16$
$94^{2}=8836$
Question 10
(i) 45
Comparing with a5 where a = 4
$4^{-1}(a 5)^{2}=a(a+1)$ hundreds +25
$45^{r}=4(4+1)$ hundreds +25
$=20$ hundreds +25
$45^{2}=2025$
(ii) 305
Comparing with a5 where a =30
$\left(a_{5}\right)^{2}=a(a+1)$ hundreds +25
$(305)^{2}=30(30+1)$ hundreds +25
⇒930 hundred +25
$(305)^{2}$= 93025
(iii) 525
Comparing with a5 where a = 52
$(a 5)^{2}=a(a+1)$ hundreds +25
$(525)^{2}=52(52+1)$ humdreds +25
⇒ 2756 hundreds +25
⇒$(525)^{2}=275625$
Question 11
(i) 8
Given number = 8
⇒let us assume $m^{2}-1=8$
⇒$m^{2}=9$
⇒m = 3
Remaining two numbers of pythagorean triplet are
$m^{2}+1,2 m$
$3^{2}+1, 2 \times 3$
10 , 6
The required triplet (6,8,10) with one number
(ii) 15
Givew number =15
Let us assume $m^{2}-1=15$
$m^{2}=16$
m = 4
Remaining two numbers of Pythagorean triplet are
$m^{2}+1,2 m$
$16+1 \quad ,2 \times 4$
$17 \quad , 8$
∴ The required triplet (8,15,17) with one number as 15
(iii) 63
Givew number 63
Let us assume $m^{2}-1=63$
$m^{2}=64$
$m=8$
Remaining two numbers of Pythagorean triplet are
$m^{2}+1,2 m$
$8^{2}+1 \quad 2 \times 8$
$65-16$
∴ We required triplet (16,63,65) with one rumber '63'
(iv) 80
given number 80
let us aasume $m^{2}-1=80 \Rightarrow m^{2}=81$
$m=9$
Remainig two numbers of Pythagorean triplet are
$m^{2}+1,2 m$
$q^{2}+1,2 \times 9$
82,18
∴ The required triplet (18,80,82) wits one number '80'
Question 12
$21^{2}=$ 441
$201^{2}=$ 40401
$2001^{2}=$ 4004001
$20001^{2}=$ 40004001
$200001^{2}=$ 4000400001
Question 14
$7^{2}=$ 49
$67^{2}=$ 4489
$667^{2}=$ 444889
$6667^{2}=$ 44448889
$66667^{2}=$ 4444488889
$666667^{2}=$ 444444888889
Exercise 3.3
Question 1
(i) 121
given number = 121
$121-1=120 ; 120-3=117 ; 117-5=112 ; 112-7=105$
$105-9=96 ; 96-11=85 ; \quad 85-13=72 ; 72-15=57$
$57-17=40 ; \quad 40-19=21 ; \quad 21-21=0$
∴ $121$ is a perfect squars
we have done '11 Substractions
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
$36-1=35 ; 35-3=32 ; 32-5=27 ; 27-7=20$
$20-9=11 ; \quad 11-11=0$
$\ \quad 36$ is a perbect square
we have done 6 subtractions
Hence, square root of 36 i'e $\sqrt{36}$= 6
iv) 90
Givew 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
(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
Givew 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
Givew 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 x19
$\sqrt{9025}=\sqrt{5^{2} \times 19^{2}}$
$\sqrt{9025}=95$
Question 3
(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
(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) Givew 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 '
$3042 \times 2=2 \times 2 \times 3 \times 3 \times 13 \times 13$
$$6084=2^{2} \times 3^{2} \times 13^{2}$
$\sqrt{6084}=2 \times 3 \times 13=78$
(v) 6300
Given number =6300
Expressiny in prime factors
$\begin{array}{r|l}2 & 6300 \\ \hline 2 & 3150 \\ \hline 5 & 1575 \\ \hline 5 & 315 \\ \hline 7 & 63 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline&1\end{array}$
$6300=2 \times 2 \times 5 \times 5 \times 7 \times 3 \times 3$
since '7' left unparied , so to make 6300 a perfect square , it should be multiplied by '7'
$6300 \times 7=2 \times 2 \times 5 \times 5 \times 7 \times 7 \times 3 \times 3$
$\sqrt{44100}=\sqrt{2^{2} \times 5^{2} \times 7^{2} \times 3^{2}}$
$\sqrt{44100}=2 \times 5 \times 7 \times 3$
$\sqrt{44100}=210$
Question 5
(i) Given number 1872 expressing in prime factors
$\begin{array}{r|l}2 &1872 \\ \hline 2 & 936 \\ \hline 2 & 468 \\ \hline 2&234 \\ \hline 3 & 117 \\\hline 3 & 39 \\ \hline 13 & 13 \\ \hline & 1\end{array}$
Since 13 left unpaired, so to make 1872 a perfect square it should be divided by 13
$\frac{1872}{13}=\frac{2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 13}{13}$
$144=2 \times 2 \times 2 \times 2 \times 3 \times 3$
$\sqrt{144}=\sqrt{2^{2} \times 2^{2} \times 3^{2}}$
$\sqrt{144}=2 \times 2 \times 3=12$
(ii) Given number 2592 expressing in prime number
$\begin{array}{r|l}2 &2592 \\ \hline 2 & 1296 \\ \hline 2 & 648 \\ \hline 2&324 \\ \hline 2 & 162 \\ \hline 2 & 81 \\ \hline 3 & 27 \\ \hline3 & 9 \\ \hline3 & 3 \\ \hline & 1\end{array}$
2592 = $2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$
Since ' 2 ' lets unpaired, so to
make 2592 a prefect square, it
Should be divided ' 2 '
$\frac{2592}{2}=\frac{2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3}{2}$
sqrt{2^{2}
$1296=2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$$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
givew no.of rows equal to no of column
ஃ no of rows = x
Total students equal to 2000 but 64 studers
could not be accomndated 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 interms 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
$61347 \div 3=\frac{3 \times 11 \times 11 \times 13 \times 13}{3}$
$20449=11 \times 11 \times 13 \times 13$
$\sqrt{20449}=\sqrt{11^{2} \times 13^{2}}$
$\sqrt{20449}=11 \times 13={143}$
(6) 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.ot 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.ot plants in a row $=65$
Question 10
Let no of students = x
given contribution of such student = no of students
∴ contribution of each students = $\bar{₹} x$
Total collected for Pcnic =22304
∴ $x \times x=2304$
$x^{2}=2304$
$x=\sqrt{2304}$
$x=48$
Question 11
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
EXERCISE 3.4
Question 1
(i) Given number 2401
(diagram should be added )
$\therefore \sqrt{2401}=49$
(ii) 4489
(diagram should be added )
$\therefore \sqrt{4489}=67$
(iii) 106929
(diagram should be added )
$\therefore \sqrt{106929}=327$
(iv) given number 167281
(diagram should be added)
$\therefore \sqrt{167281}=409$
v) givew number 53824
$\therefore \sqrt{213444}=462$
$\therefore \sqrt{53824}=232$
vi) given number 213444
$\therefore \sqrt{213444}=462$
$\therefore \sqrt{213444}=462$
Question 2
i) Given number 81=2 (even)
⇒ The number of digits in its square root = $\frac{2}{2}=1$
ii) Given number $169=3($ odd $)$
⇒The number of digits in its square root $=\frac{3+1}{2}=2$
iii) Given number $4761=4$ (even)
⇒the number of digits in its square root $=\frac{4}{2}=2$
iv) Given number $27889=5$ (odd)
⇒ The number of digits in its square root $=\frac{5+1}{2}=3$
V) Givew number $525625=6($ even $)$
∴ The number of digits in its square root $\frac{6}{2}=3$
Question 3
Sol :
(i) Given number 51.84
(diagram should be added )
$\therefore \sqrt{51.84}=7.2$
(ii) 42.25
(diagram should be added )
$\sqrt{42.25}=6.5
(iii) Given number 18.4041
(diagram should be added )
$\sqrt{18.4041}=4.29$
(iv) Given number 5.774409
(diagram should be added )
$\therefore \sqrt{5.774409}=2.403$
Question 4
(i) 645.8
(diagram should be added )
$\therefore \sqrt{645.8}=25.412 \approx 25.41$ (correct to 2 decimals )
(ii) 107.45
(diagram should be added )
$\sqrt{107 \cdot 45}=10.365 \approx 10.36$
(iii) Given number 5.462
(diagram should be added )
$\therefore \sqrt{5.462}=2.337 \approx 2.34$ (corrected to '2 'decimals)
(iv) Given number 2
(diagram should be added )
$\sqrt{2}=1.414 \approx 1.41$
(v) Given number 3
(diagram should be added )
$\sqrt{3}=1.732 \approx 1.73$ (Corrected to 2 decimals $)$
Question 5
(i) $\frac{841}{1521}$
(diagram should be added )
$=\frac{\sqrt{841}}{\sqrt{1521}}=\frac{29}{39}$
(ii) $8 \frac{257}{529}=\frac{4489}{529}$
(diagram should be added )
$\sqrt{8 \frac{257}{529}}=\sqrt{\frac{4489}{529}}$
$\sqrt{8 \frac{257}{529}}=\frac{67}{23}$
(iii) $16 \frac{169}{441}=\frac{7225}{441}$
(diagram should be added )
$\sqrt{16 \frac{169}{441}}=\frac{\sqrt{7225}}{\sqrt{441}}=\frac{85}{21}$
Question 6
(i) Given nunmber 2000
⇒(diagram should be added )
⇒ Hence , the least number that must be subtracted from
2000 so as to make it a perfect square is 64
∴ Requred perfect square numbers =2000 - 64
= $1936=44^{2}$
(ii) Given number 984
⇒(diagram should be added )
⇒ Hence , the least number that must be subtracted
from 984 so as to make it a perfect square is 23
∴ Requred perfect square numbers = 984 - 23 = 961 = $=31^{2}$
iii) Givew number 8934
⇒(diagram should be added )
⇒ Hence, the least number that must be subtracted
from 8934 so as to make it a perfect square in 98
∴ The required Square number 8934-98 = 8836=$94^{2}$
iv) Givew number 11021
⇒(diagram should be added )
⇒ Hence , the least number that must be subtracted
from 11021 so as to make it a perfect square is 205
The required square number 11021 - 205 = 10816 = $104^{2}$
Question 7
(i) Given number 1750
⇒(diagram should be added )
⇒ 1750\rangle$(41)^{2} \Rightarrow$ Remainder $=69$
⇒$(42)^{2}=1764$
⇒ $\therefore$ Required number =1764-1750=14
⇒ Hence, the least number that must be added to 1750
So as to make it a perfect square is 14
(ii) Givew number 6412
⇒(diagram should be added )
⇒$6412>(80)^{2}$
=$81^{2}=6561$
⇒ $\therefore$ Required number $=6561-6412=149$
⇒ Hence, the least number That must be added to 6412
So as to make it a perfect square is 149
(iii) givew number 6598
⇒(diagram should be added )
⇒ $6598>(81)^{2}$
=$(82)^{2}=6724$
$\therefore$ Required number $=6(82)^{2}-6598=126$
⇒ hence , the minimum number that must be added to 6598 so as to make it a perfect square is 126
(iv) Givew number 8000
⇒(diagram should be added )
⇒ $8000>89^{2}$
⇒$90^{2}=8100$
⇒ $\therefore$ Required number $=90^{2}-8000=100$
⇒ hence , the minimum number that must be added to
8000 so as to make it a perfect square is 100
Question 8
Smallest four digit number = 1000
⇒(diagram should be added )
⇒ $1000>31^{2}$
⇒ $32^{2}$ will be next perfect square
⇒ $32^{2}=1024$
⇒ Hence , 1024 is smallest four digit number which perfect square
Question 9
Greatest six digit number = 999999
⇒ (diagram should be added )
⇒ To make 999999 a perfect square , we have to subtract 1998 from 999999
⇒ The required number = 998001
⇒ hence , 998001 is greatest six digit number which is a perfect square
Question 10
(i) AB = 14 cm
Bc = 48 cm
according to pythagorus theorem
⇒ $A C^{2}=A B^{2}+B C^{2}$
⇒$14^{2}+48^{2}$
⇒ $A C^{2}=2500$
⇒ $A C=\sqrt{2500}$
⇒ $A C=50 \mathrm{~cm}$
(ii) $A C=37 \mathrm{~cm}, B C=35 \mathrm{Cm}, A B=?$
⇒ According to pythagorus theorem
⇒ $A C^{2}=A B^{2}+B C^{2}$
⇒ $37^{2}=A B^{2}+35^{2}$
⇒ $1369=A B^{2}+1225$
⇒ $A B^{2}=144$
⇒ $A B=12 \mathrm{~cm}$
Question 11
Total plants = 1400
let no . of rows = x
no. of columns = x
⇒ (diagram should be added )
$x^{2}=1400$
$1400>(37)^{2}$
$38^{2}=1444$
So To make 1400 a perfect square, we have add
minimum of 44
$\therefore 44$ plants needed more.
Question 12
⇒ Total no of students = 1000
⇒ let no of row = no of columns = x
⇒ Total students rows x columns = 1000
⇒ (diagram should be added )
⇒ $x \times x=1000$
$x^{2}=1000$
$x=\sqrt{1000}$
So Remainder $=39$
⇒ hence 39 children will be left out
Question 13
⇒ (diagram should be added )
⇒ Distance that amit walk while retuning
⇒ AC
In $\triangle A B C$
⇒ According to pythagorus theorem
⇒ $A C=A B^{2}+B C^{2}$
⇒$A C^{2}=16^{2}+63^{2}$
⇒$A C^{2}=4225$
⇒$A C=65 m$
ஃ Hence amit walks 65c while returing to his house
Question 14
Sol: (diagram should be added )
⇒ Length of ladder = 6m
height of wall = 4.8m
In $\triangle A B C$
According Pythagorus tbeorem
⇒ $A C^{2}=A B^{n}+B C^{2}$
$B^{2}=4 \cdot 8^{2}+B C^{2}$
$B C^{N}=12.96$
$B C=\sqrt{12.96}$
$B C=3.6 \mathrm{~m}$
⇒ Hence, Distance between wall ond foot of ladder
is 3.6 m
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