Rs Aggarwal 2020 2021 Solutions for Class 8 Maths Chapter 3 Squares And Square Roots are provided here with simple step-by-step explanations. These solutions for Squares And Square Roots are extremely popular among Class 8 students for Maths Squares And Square Roots Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 8 Maths Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

#### Page No 42:

A perfect square can always be expressed as a product of equal factors.

(i)
Resolving into prime factors:
$441=49×9=7×7×3×3=7×3×7×3=21×21=\left(21{\right)}^{2}$

Thus, 441 is a perfect square.

(ii)
Resolving into prime factors:
$576=64×9=8×8×3×3=2×2×2×2×2×2×3×3=24×24=\left(24{\right)}^{2}$

Thus, 576 is a perfect square.

(iii)
Resolving into prime factors:
$11025=441×25=49×9×5×5=7×7×3×3×5×5=7×5×3×7×5×3=105×105=\left(105{\right)}^{2}$

Thus, 11025 is a perfect square.

(iv)
Resolving into prime factors:
$1176=7×168=7×21×8=7×7×3×2×2×2$

1176 cannot be expressed as a product of two equal numbers. Thus, 1176 is not a perfect square.

(v)
Resolving into prime factors:
$5625=225×25=9×25×25=3×3×5×5×5×5=3×5×5×3×5×5=75×75=\left(75{\right)}^{2}$

Thus, 5625 is a perfect square.

(vi)
Resolving into prime factors:
$9075=25×363=5×5×3×11×11=55×55×3$

9075 is not a product of two equal numbers. Thus, 9075 is not a perfect square.

(vii)
Resolving into prime factors:
$4225=25×169=5×5×13×13=5×13×5×13=65×65=\left(65{\right)}^{2}$

Thus, 4225 is a perfect square.

(viii)
Resolving into prime factors:
$1089=9×121=3×3×11×11=3×11×3×11=33×33=\left(33{\right)}^{2}$

Thus, 1089 is a perfect square.

#### Page No 42:

A perfect square is a product of two perfectly equal numbers.

(i)
Resolving into prime factors:

$1225=25×49=5×5×7×7=5×7×5×7=35×35=\left(35{\right)}^{2}$

Thus, 1225 is the perfect square of 35.

(ii)
Resolving into prime factors:
$2601=9×289=3×3×17×17=3×17×3×17=51×51=\left(51{\right)}^{2}$

Thus, 2601 is the perfect square of 51.

(iii)
Resolving into prime factors:

$5929=11×539=11×7×77=11×7×11×7=77×77=\left(77{\right)}^{2}$

Thus, 5929 is the perfect square of 77.

(iv)
Resolving into prime factors:
$7056=12×588=12×7×84=12×7×12×7=\left(12×7{\right)}^{2}=\left(84{\right)}^{2}$

Thus, 7056 is the perfect square of 84.

(v)
Resolving into prime factors:

$8281=49×169=7×7×13×13=7×13×7×13=\left(7×13{\right)}^{2}=\left(91{\right)}^{2}$

Thus, 8281 is the perfect square of 91.

#### Page No 42:

1. Resolving 3675 into prime factors:
$3675=3×5×5×7×7$

Thus, to get a perfect square, the given number should be multiplied by 3.

New number

Hence, the new number is the square of 105.

2. Resolving 2156 into prime factors:
$2156=2×2×7×7×11=\left({2}^{2}×{7}^{2}×11\right)$

Thus to get a perfect square, the given number should be multiplied by 11.

New number $=\left({2}^{2}×{7}^{2}×{11}^{2}\right)=\left(2×7×11{\right)}^{2}=\left(154{\right)}^{2}$

Hence, the new number is the square of 154.

3. Resolving 3332 into prime factors:
$3332=2×2×7×7×17={2}^{2}×{7}^{2}×17$

Thus, to get a perfect square, the given number should be multiplied by 17.

New number $=\left({2}^{2}×{7}^{2}×{17}^{2}\right)=\left(2×7×17{\right)}^{2}=\left(238{\right)}^{2}$

Hence, the new number is the square of 238.

4. Resolving 2925 into prime factors:
$2925=3×3×5×5×13={3}^{2}×{5}^{2}×13$

Thus, to get a perfect square, the given number should be multiplied by 13.

New number $=\left({3}^{2}×{5}^{2}×{13}^{2}\right)=\left(3×5×13{\right)}^{2}=\left(195{\right)}^{2}$

Hence, the number whose square is the new number is 195.

5. Resolving 9075 into prime factors:

​$9075=3×5×5×11×11=3×{5}^{2}×{11}^{2}$

Thus, to get a perfect square, the given number should be multiplied by 3.

New number $=\left({3}^{2}×{5}^{2}×{11}^{2}\right)=\left(3×5×11{\right)}^{2}=\left(165{\right)}^{2}$

Hence, the new number is the square of 165.

6. Resolving 7623 into prime factors:
​$7623=3×3×7×11×11={3}^{2}×7×{11}^{2}$

Thus, to get a perfect square, the given number should be multiplied by 7.

New number $=\left({3}^{2}×{7}^{2}×{11}^{2}\right)=\left(3×7×11{\right)}^{2}=\left(231{\right)}^{2}$

Hence, the number whose square is the new number is 231.

7. Resolving 3380 into prime factors:
​$3380=2×2×5×13×13={2}^{2}×5×{13}^{2}$

Thus, to get a perfect square, the given number should be multiplied by 5.

New number $=\left({2}^{2}×{5}^{2}×{13}^{2}\right)=\left(2×5×13{\right)}^{2}=\left(130{\right)}^{2}$

Hence, the new number is the square of 130.

8. Resolving 2475 into prime factors:
​$2475=3×3×5×5×11={3}^{2}×{5}^{2}×11$

Thus, to get a perfect square, the given number should be multiplied by 11.

New number $=\left({3}^{2}×{5}^{2}×{11}^{2}\right)=\left(3×5×11{\right)}^{2}=\left(165{\right)}^{2}$

Hence, the new number is the square of 165.

#### Page No 42:

(i) Resolving 1575 into prime factors:
$1575=3×3×5×5×7={3}^{2}×{5}^{2}×7$

Thus, to get a perfect square, the given number should be divided by 7

New number obtained$=\left({3}^{2}×{5}^{2}\right)=\left(3×5{\right)}^{2}=\left(15{\right)}^{2}$

Hence, the new number is the square of 15

(ii) Resolving 9075 into prime factors:

$9075=3×5×5×11×11=3×{5}^{2}×{11}^{2}$

Thus, to get a perfect square, the given number should be divided by 3

New number obtained$=\left({5}^{2}×{11}^{2}\right)=\left(5×11{\right)}^{2}=\left(55{\right)}^{2}$

Hence, the new number is the square of  55

(iii) Resolving 4851 into prime factors:

$4851=3×3×7×7×11={3}^{2}×{7}^{2}×11$

Thus, to get a perfect square, the given number should be divided by 11

New number obtained$=\left({3}^{2}×{7}^{2}\right)=\left(3×7{\right)}^{2}=\left(21{\right)}^{2}$

Hence, the new number is the square of 21

(iv) Resolving 3380 into prime factors:

$3380=2×2×5×13×13={2}^{2}×5×{13}^{2}$

Thus, to get a perfect square, the given number should be divided by 5

New number obtained$=\left({2}^{2}×{13}^{2}\right)=\left(2×13{\right)}^{2}=\left(26{\right)}^{2}$

Hence, the new number is the square of 26

(v) Resolving 4500 into prime factors:

$4500=2×2×3×3×5×5×5={2}^{2}×{3}^{2}×{5}^{2}×5$

Thus, to get a perfect square, the given number should be divided by 5

New number obtained$=\left({2}^{2}×{3}^{2}×{5}^{2}\right)=\left(2×3×5{\right)}^{2}=\left(30{\right)}^{2}$

Hence, the new number is the square of 30

(vi) Resolving 7776 into prime factors:

$7776=2×2×2×2×2×3×3×3×3×3={2}^{2}×{2}^{2}×2×{3}^{2}×{3}^{2}×3$

Thus, to get a perfect square, the given number should be divided by 6 whish is a product of 2 and 3

New number obtained$=\left({2}^{2}×{2}^{2}×{3}^{2}×{3}^{2}\right)=\left(2×2×3×3{\right)}^{2}=\left(36{\right)}^{2}$

Hence, the new number is the square of 36

(vii) Resolving 8820 into prime factors:

$8820=2×2×3×3×5×7×7={2}^{2}×{3}^{2}×5×{7}^{2}$

Thus, to get a perfect square, the given number should be divided by 5

New number obtained$=\left({2}^{2}×{3}^{2}×{7}^{2}\right)=\left(2×3×7{\right)}^{2}=\left(42{\right)}^{2}$

Hence, the new number is the square of 42

(viii) Resolving 4056 into prime factors:

$4056=2×2×2×3×13×13={2}^{2}×2×3×{13}^{2}$

Thus, to get a perfect square, the given number should be divided by 6, which is a product of 2 and 3

New number obtained$=\left({2}^{2}×{13}^{2}\right)=\left(2×13{\right)}^{2}=\left(26{\right)}^{2}$

Hence, the new number is the square of 26

#### Page No 42:

The first three digit number (100) is a perfect square. Its square root is 10.
The number before 10 is 9.
Square of 9
Thus, the largest 2 digit number that is a perfect square is 81.

#### Page No 42:

The largest 3 digit number is 999.

The number whose square is 999 is 31.61.

Thus, the square of any number greater than 31.61 will be a 4 digit number.
Therefore, the square of 31 will be the greatest 3 digit perfect square.

${31}^{2}=31×31=961$

#### Page No 45:

By observing the properties of square numbers, we can determine whether a given number is a square or not.

(i) 5372
A number that ends with 2 is not a perfect square.
Thus, the given number is not a perfect square.

(ii) 5963
A number that ends with 3 is not a perfect square.
Thus, the given number is not a perfect square.

(iii) 8457
A number that ends with 7 is not a perfect square.
Thus, the given number is not a perfect square.

(iv) 9468
A number ending with 8 is not a perfect square.
Thus, the given number is not a perfect square.

(v) 360
Any number ending with an odd number of zeroes is not a perfect square.
Hence, the given number is not a perfect square.

(vi) 64000
Any number ending with an odd number of zeroes is not a perfect square.
Hence, the given number is not a perfect square.

(vii) 2500000
Any number ending with an odd number of zeroes is not a perfect square.
Hence, the given number is not a perfect square.

#### Page No 45:

The square of an even number is always even.
Thus, even numbers in the given list of squares will be squares of even numbers.

(i) 196
This is an even number. Thus, it must be a square of an even number.

(ii) 441
This is an odd number. Thus, it is not a square of an even number.

(iii) 900
This is an even number. Thus, it must be a square of an even number.

(iv) 625
This is an odd number. Thus, it is not a square of an even number.

(v) 324
This is an even number. Thus, it is a square of an even number.

#### Page No 46:

According to the property of squares, the square of an odd number is also an odd number.
Using this property, we will determine which of the numbers in the given list of squares is a square of an odd number.

(i) 484.
This is an even number. Thus, it is not a square of an odd number.

(ii) 961
This is an odd number. Thus, it is a square of an odd number.

(iii) 7396
This is an even number. Thus, it is not a square of an odd number.

(iv) 8649
This is an odd number. Thus, it is a square of an odd number.

(v) 4225
This is an odd number. Thus, it is a square of an odd number.

#### Page No 46:

Sum of first n odd numbers

(i)

(ii) $\left(1+3+5+7+9+11+13+15+17+19\right)={10}^{2}=100$

(iii)

#### Page No 46:

Sum of first n odd natural numbers

(i) Expressing 81 as a sum of 9 odd numbers:
$81={\left(9\right)}^{2}\phantom{\rule{0ex}{0ex}}n=9\phantom{\rule{0ex}{0ex}}81=1+3+5+7+9+11+13+15+17$

(ii) Expressing 100 as a sum of 10 odd numbers:
$100={\left(10\right)}^{2}\phantom{\rule{0ex}{0ex}}n=10\phantom{\rule{0ex}{0ex}}100=1+3+5+7+9+11+13+15+17+19$

#### Page No 46:

For every number m > 1, the Pythagorean triplet is .

Using the above result:

(i)

Thus, the Pythagorean triplet is $\left[6,8,10\right]$.

(ii)

Thus, the Pythagorean triplet is $\left[14,48,50\right]$.

(iii)

Thus, the Pythagorean triplet is: $\left[16,63,65\right]$

(iv)

Thus, the Pythagorean triplet is $\left[20,99,101\right]$.

#### Page No 46:

Given:

(i) ${\left(38\right)}^{2}-{\left(37\right)}^{2}=38+37=75$

(ii) ${\left(75\right)}^{2}-{\left(74\right)}^{2}=75+74=149$

(iii) ${\left(92\right)}^{2}-{\left(91\right)}^{2}=92+91=183$

(iv) ${\left(105\right)}^{2}-{\left(104\right)}^{2}=105+104=209$

(v) ${\left(141\right)}^{2}-{\left(140\right)}^{2}=141+140=281$

(vi) ${\left(218\right)}^{2}-{\left(217\right)}^{2}=218+217=435$

#### Page No 46:

(i) ${310}^{2}={\left(300+10\right)}^{2}=\left({300}^{2}+2\left(300×10\right)+{10}^{2}\right)=90000+6000+100=96100$

(ii) ${508}^{2}={\left(500+8\right)}^{2}=\left({500}^{2}+2\left(500×8\right)+{8}^{2}\right)=250000+8000+64=258064$

(iii) ${630}^{2}={\left(600+30\right)}^{2}=\left({600}^{2}+2\left(600×30\right)+{30}^{2}\right)=360000+36000+900=396900$

#### Page No 46:

(i) ${\left(196\right)}^{2}={\left(200-4\right)}^{2}={200}^{2}-2\left(200×4\right)+{4}^{2}=40000-1600+16=38416$

(ii) ${\left(689\right)}^{2}={\left(700-11\right)}^{2}={700}^{2}-2\left(700×11\right)+{11}^{2}=490000-15400+121=474721$

(iii) ${\left(891\right)}^{2}={\left(900-9\right)}^{2}={900}^{2}-2\left(900×9\right)+{9}^{2}=810000-16200+81=793881$

#### Page No 46:

(i) $69×71=\left(70-1\right)×\left(70+1\right)=\left({70}^{2}-{1}^{2}\right)=4900-1=4899$

(ii) $94×106=\left(100-6\right)×\left(100+6\right)=\left({100}^{2}-{6}^{2}\right)=10000-36=9964$

#### Page No 46:

(i) ​$88×92=\left(90-2\right)×\left(90+2\right)=\left({90}^{2}-{2}^{2}\right)=8100-4=8096$

(ii) $78×82=\left(80-2\right)×\left(80+2\right)=\left({80}^{2}-{2}^{2}\right)=6400-4=6396$

#### Page No 46:

(i) The square of an even number is even.

(ii) The square of an odd number is odd.

(iii) The square of a proper fraction is smaller  than the given fraction.

(iv) ${n}^{2}=$the sum of first n odd natural numbers.

#### Page No 46:

(i) F
The number of digits in a square can also be odd. For example: 121

(ii) F
A prime number is one that is not divisible by any other number, except by itself and 1. Thus, square of any number cannot be a prime number.

(iii) F
Example: $4+9=13$
4 and 9 are perfect squares of 2 and 3, respectively. Their sum (13) is not a perfect square.

(iv) F
Example: $36-25=11$
36 and 25 are perfect squares. Their difference is 11, which is not a perfect square.

(v) T

#### Page No 48:

Using the column method:
∴  a = 2
b = 3

 ${a}^{2}$ 2ab ${b}^{2}$ 04 + 1= 5 12+0 = 12 9

∴ ${23}^{2}=529$

#### Page No 48:

Using the column method:
Here, a = 3 and b = 5

 ${a}^{2}$ 2ab ${b}^{2}$ $09$ +3 = 12 $30$ +2 = 32 25

∴

#### Page No 48:

Using the column method:

Here, a = 5
b = 2

 ${a}^{2}$ 2ab ${b}^{2}$ $25\phantom{\rule{0ex}{0ex}}+2$ =27 20 4

∴ ${52}^{2}=2704$

#### Page No 48:

Using column method:

Here,

 ${a}^{2}$ 2ab ${b}^{2}$ $81\phantom{\rule{0ex}{0ex}}+11$ =92 =111 36

∴ ${96}^{2}=9216$

#### Page No 49:

${67}^{2}=4489$

#### Page No 49:

${86}^{2}=7396$

#### Page No 49:

${137}^{2}=18769$

#### Page No 49:

${256}^{2}=65536$

#### Page No 50:

By prime factorisation method:

$225=3×3×5×5\phantom{\rule{0ex}{0ex}}\sqrt{225}=3×5=15$

#### Page No 50:

By prime factorisation:

#### Page No 50:

Resolving into prime factors:

$729=3×3×3×3×3×3$

$\sqrt{729}=3×3×3=27$

#### Page No 50:

Resolving into prime factors:

$1296=2×2×2×2×3×3×3×3$

$\sqrt{1296}=2×2×3×3=36$

#### Page No 50:

Resolving into prime factors:

$2025=3×3×3×3×5×5$

$\sqrt{2025}=3×3×5=45$

#### Page No 50:

Resolving into prime factors:
$4096=2×2×2×2×2×2×2×2×2×2×2×2$

$\sqrt{4096}=2×2×2×2×2×2=64$

#### Page No 50:

Resolving into prime factors:

$7056=2×2×2×2×3×3×7×7$

$\sqrt{7056}=2×2×3×7=84$

#### Page No 50:

Resolving into prime factors:

$8100=2×2×3×3×3×3×5×5$

$\sqrt{8100}=2×3×3×5=90$

#### Page No 50:

Resolving into prime factors:

$9216=2×2×2×2×2×2×2×2×2×2×3×3$

$\sqrt{9216}=2×2×2×2×2×3=96$

#### Page No 50:

Resolving into prime factors:

$11025=3×3×5×5×7×7$

$\sqrt{11025}=3×5×7=105$

#### Page No 50:

Resolving into prime factors:

$15876=2×2×3×3×3×3×7×7$

$\sqrt{15876}=2×3×3×7=126$

#### Page No 50:

Resolving into prime factors:

$17424=2×2×2×2×3×3×11×11$

$\sqrt{17424}=2×2×3×11=132$

#### Page No 50:

Resolving into prime factors:

$252=2×2×3×3×7$

Thus, the given number must be multiplied by 7 to get a perfect square.

New number = $252×7=1764$

$\sqrt{1764}=2×3×7=42$

#### Page No 50:

Resolving into prime factors:

$2925=3×3×5×5×13$

13 is the smallest number by which the given number must be divided to make it a perfect square.

New number = $2925÷13=225$

$\sqrt{225}=3×5=15$

#### Page No 50:

Let the number of rows be x.
Therefore, the number of plants in each row is also x.
Total number of plants
${x}^{2}=1225=5×5×7×7\phantom{\rule{0ex}{0ex}}x=\sqrt{1225}=5×7=35$

Thus, the number of rows is 35 and the number of plants in each row is 35.

#### Page No 50:

Let the number of students be $x$.
Hence, the amount contributed by each student is Rs x.

Total amount contributed $=x×x={x}^{2}=1156$

$1156=2×2×17×17\phantom{\rule{0ex}{0ex}}x=\sqrt{1156}=2×17=34$

Thus, the strength of the class is 34.

#### Page No 50:

The smallest number divisible by each of these numbers is their L.C.M.
L.C.M. of 6, 9, 15, 20 = 180

Resolving into prime factors:
$180=2×2×3×3×5$
To make it a perfect square, we multiply it with 5.

Required number = $180×5=900$

#### Page No 51:

The smallest number divisible by each of these numbers is their L.C.M.
L.C.M. of 8, 12, 15, 20 = 120

Resolving into prime factors:
$120=2×2×2×3×5$

To make this into a perfect square, we need to multiply the number with $2×3×5=30$.

Required number = $120×30=3600$

#### Page No 54:

Using the long division method:

∴

#### Page No 54:

Using the long division method:

#### Page No 54:

Using the long division method:

#### Page No 54:

Using the long division method:

∴

#### Page No 54:

Using the long division method:

#### Page No 54:

Using the long division method:

#### Page No 54:

Using the long division method:

∴ $\sqrt{11449}=107$

#### Page No 54:

Using the long division method:

∴ $\sqrt{14161}=119$

#### Page No 54:

Using the long division method:

∴ $\sqrt{10404}=102$

#### Page No 54:

Using the long division method:

#### Page No 54:

Using the long division method:

∴ $\sqrt{19600}=140$

#### Page No 54:

Using the long division method:

∴ $\sqrt{92416}=304$

#### Page No 54:

Using the long division method:

Therefore, the number that should be subtracted from the given number to make it a perfect square is 9.

#### Page No 54:

Using the long division method:

Therefore, the number that should be subtracted from the given number to make it a perfect square is 12.
Perfect square = 7581-12
= 7569

Its square root is 87.

#### Page No 54:

Using the long division method:

Thus, to get a perfect square greater than the given number, we take the square of the next natural number of the quotient, i.e. 78.
${79}^{2}=6241$
Number that should be added to the given number to make it a perfect square $=6241-6203\phantom{\rule{0ex}{0ex}}=38$

The perfect square thus obtained is 6241 and its square root is 79.

#### Page No 54:

Using the long division method:

The next natural number that is a perfect square can be obtained by squaring the next natural number of the obtained quotient, i.e. 91.
Therefore square of (91+1) = ${92}^{2}=8464$
Number that should be added to the given number to make it a perfect square $=8464-8400=64$
The perfect square thus obtained is 8464 and its square root is 92.

#### Page No 54:

Smallest number of four digits $=1000$

Using the long division method:

1000 is not a perfect square.
By the long division method, the obtained square root is between 31 and 32.
Squaring the next integer (32) will give us the next perfect square.

${32}^{2}=1024$

Thus, 1024 is the smallest four digit perfect square.

Also, $\sqrt{1024}=32$

#### Page No 54:

Greatest number of five digits $=99999$

Using the long division method:

99999 is not a perfect square.
According to the long division method, the obtained square root is between 316 and 317.
Squaring the smaller number, i.e. 316, will give us the perfect square that would be less than 99999.

${316}^{2}=99856$

99856 is the required number.
Its square root is 316.

#### Page No 54:

Area of the square field
Length of each side of the square field $=\sqrt{60025}=245m$
Perimeter of the field

The man is cycling at a speed of 18 km/h.

#### Page No 56:

Using long division method:

$\sqrt{1.69}=1.3$

#### Page No 56:

Using long division method:

∴ $\sqrt{33.64}=5.8$

#### Page No 56:

Using long division method:

∴ $\sqrt{156.25}=12.5$

#### Page No 56:

Using long division method:

∴ $\sqrt{75.69}=8.7$

#### Page No 56:

Using long division method:

​∴ $\sqrt{9.8596}=3.14$

#### Page No 56:

Using long division method:

∴  $\sqrt{10.0489}=3.17$

#### Page No 56:

Using long division method:

$\sqrt{1.0816}=1.04$

#### Page No 56:

Using long division method:

$\sqrt{0.2916}=0.54$

#### Page No 56:

Using long division method:

#### Page No 56:

Using long division method:

#### Page No 56:

Using long division method:

#### Page No 56:

Area of the rectangle  sq m
​Thus, area of the square is 46.24 sq m.

Length of each side of the square m

Using long division method:

$\sqrt{46.24}=6.8$

Thus, the length of a side of the square is 6.8 metres.

#### Page No 58:

$\sqrt{\frac{16}{81}}=\frac{\sqrt{16}}{\sqrt{81}}$

∴  $\sqrt{\frac{16}{81}}=\frac{\sqrt{16}}{\sqrt{81}}=\frac{4}{9}$

#### Page No 58:

$\sqrt{\frac{64}{225}}=\frac{\sqrt{64}}{\sqrt{225}}$

Using long division method:
$\sqrt{64}=8$

$\sqrt{225}=15$

#### Page No 58:

$\sqrt{\frac{121}{256}}=\frac{\sqrt{121}}{\sqrt{256}}$

Using division method:

$\sqrt{121}=11$

$\sqrt{\frac{121}{256}}=\frac{\sqrt{121}}{\sqrt{256}}=\frac{11}{16}$

#### Page No 58:

$\sqrt{\frac{625}{729}}=\frac{\sqrt{625}}{\sqrt{729}}$

Using long division method:

$\sqrt{625}=25$

$\sqrt{729}=27$

$\sqrt{\frac{625}{729}}=\frac{\sqrt{625}}{\sqrt{729}}=\frac{25}{27}$

#### Page No 58:

$\sqrt{4\frac{73}{324}}=\sqrt{\frac{1369}{324}}=\frac{\sqrt{1369}}{\sqrt{324}}$

Using long division method:
$\sqrt{1369}=37$

$\sqrt{4\frac{73}{324}}=\frac{37}{18}=2\frac{1}{18}$

#### Page No 58:

$\sqrt{3\frac{33}{289}}=\sqrt{\frac{900}{289}}=\frac{\sqrt{900}}{\sqrt{289}}$

Using long division method:

$\sqrt{289}=17$

And

$\sqrt{3\frac{33}{289}}=\frac{30}{17}=1\frac{13}{17}$

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We have:

$\frac{\sqrt{80}}{\sqrt{405}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{80}{405}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{16}{81}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{16}}{\sqrt{81}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{4}{9}$

We have:

We have:

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(c) 5478

According to the properties of squares, a number ending in 2, 3, 7 or 8 is not a perfect square.

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(d) 2222

According to the property of squares, a number ending in 2, 3, 7 or 8 is not a perfect square.

#### Page No 58:

(a) 1843
According to the property of squares, a number ending in 2, 3, 7 and 8 is not a perfect square.

#### Page No 58:

(b) 4787

By the property of squares, a number ending in 2, 3,7 or 8 is not a perfect square.

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(c) 81000

According to the property of squares, a number ending in odd number of zeroes is not a perfect square.

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(d) 8

According to the property of squares, a perfect square cannot have 2, 3, 7 or 8 as the unit digit.

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(b) smaller than the fraction

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(c) ${n}^{2}$

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(d) (8,15,17)

This can be understood from the property of Pythagorean triplets. According to this property, for a natural number m,  is a Pythagorean triplet.
Here,  m = 4
2m = 8
m2 - 1=15
and    m2 + 1 = 17

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(c) 7

$\left(176-7\right)=169\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\sqrt{169}=13$

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(a) 3

$526+3=529\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}529={23}^{2}$

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(b) 6

$15370+6=15376\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\sqrt{15376}=124$

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(d) 0.94

$\sqrt{0.9}=0.94$

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(c) 0.316

Using long division method:

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(b) 1.2

$\sqrt{0.9}×\sqrt{1.6}=\sqrt{1.44}=1.2$

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(c) $\frac{3}{2}$

$\frac{\sqrt{288}}{\sqrt{128}}=\sqrt{\frac{288}{128}}=\sqrt{\frac{2×2×2×2×2×3×3}{2×2×2×2×2×2×2}}=\sqrt{\frac{3×3}{2×2}}=\frac{\sqrt{3×3}}{\sqrt{2×2}}=\frac{3}{2}$

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(b) $1\frac{1}{2}$

$\sqrt{2\frac{1}{4}}=\sqrt{\frac{9}{4}}=\frac{\sqrt{9}}{\sqrt{4}}=\frac{3}{2}=1\frac{1}{2}$

#### Page No 59:

(a) 196

Square of an even number is always an even number.

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(c) 1369

Square of an odd number is always an odd number.

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Using long division method:

∴  $\sqrt{11236}=106$

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The greatest 5 digit number is 99999.

${316}^{2}=99856$
Thus, this is the greatest 5 digit number.

$\sqrt{99856}=316$

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The least number of 4 digits is 1000.

1024 is the least four digit perfect square and its square root is 32.

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$\sqrt{0.2809}=0.53$

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(d) 1222

A number ending in 2, 3, 7 or 8 is not a perfect square.

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(c) $1\frac{1}{2}$

$\sqrt{2\frac{1}{4}}=\sqrt{\frac{9}{4}}=\frac{\sqrt{9}}{\sqrt{4}}=\frac{\sqrt{3×3}}{\sqrt{2×2}}=\frac{3}{2}=1\frac{1}{2}$

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(c) 1764

The square of an even number is always even.

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(d) 8

$521+8=529\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\sqrt{529}=23$

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(c) 9

$178-9=169\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\sqrt{169}=13$

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(b) 84

$\sqrt{72}×\sqrt{98}=\sqrt{2×2×2×3×3}×\sqrt{2×7×7}=\sqrt{2×2×2×3×3×2×7×7}=2×2×3×7=84$

#### Page No 62:

(i) $1+3+5+7+9+11+13=\left(7{\right)}^{2}$

(ii)

$\sqrt{1681}=41$

(iii) The smallest square number exactly divisible by 2, 4 and 6 is 36.

(iv) A given number is a perfect square having n digits, where n is odd. then, its square root will have $\left(\frac{n+1}{2}\right)$ digits.

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