ML AGGARWAL CLASS 8 CHAPTER 3 SQUARES AND ROOTS Exercise 3.1

 Exercise 3.1

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Q1 | Ex-3.1 | Class 8 | ML AGGARWAL | SQUARES AND ROOTS | myhelper

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Question 1

Which of the following natural numbers are perfect squares? Give reasons in support of your answer.

(i) 729

(ii) 5488

(iii) 1024

(iv) 243

Sol :

(i) 729

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)

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Question 2 

Show that each of the following numbers is a perfect square. Also, find the number whose square is the given number.

(i) 1296

(ii) 1764

(iii) 3025

(iv) 3969

Sol :

(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=2×2×2×2×3×3×3×3

since 1296 is expressed as the product of pairs of equal prime factors 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 factors.

$3969=3^{4} \times 3^{2} \times 7^{2}$

$3969=(3 \times 3 \times 7)^{2}$

$3969=63^{2}$

Hence, 63 is the number whose square is 3969

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Question 3 

Find the smallest natural number by which 1008 should be multiplied to make it a perfect square.

Sol :

1008

$\begin{array}{r|l}2& 1008\\ \hline 2& 504 \\ \hline 2&252\\ \hline 2& 126\\ \hline 7& 63 \\ \hline 3&9\\  \hline 3&3\\ \hline&1\end{array}$

$1008=2 \times 2 \times 2 \times 2 \times 7 \times 3 \times 3$

Since' 7 ' is left unpaired, so to make 1008 a

Perfect square it should be multiplied by 7

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Question 4 

Find the smallest natural number by which 5808 should be divided to make it a perfect square. Also, find the number whose square is the resulting number.

Sol:

5808

⇒ $\begin{array}{r|l}2& 5808\\ \hline 2& 2904 \\ \hline 2&1452\\ \hline 2& 726\\ \hline 3& 363 \\ \hline 11&121\\  \hline 11&11\\ \hline&1\end{array}$

$5808=2 \times 2 \times 2 \times 2 \times 3 \times 11 \times 11$

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

ML AGGARWAL CLASS 8 CHAPTER 3 SQUARES AND ROOTS Exercise 3.4

 

EXERCISE 3.4

Question 1

Find the square root of each of the following by division method:

(i) 2401

(ii) 4489

(iii) 106929

(iv) 167281

(v) 53824

(vi) 213444

Sol :

(i) 2401

Given number 2401

$\therefore \sqrt{2401}=49$

(ii) 4489

$\therefore \sqrt{4489}=67$

(iii)  106929

$\therefore \sqrt{106929}=327$

(iv) given number 167281

$\therefore \sqrt{167281}=409$

(v) given number 53824

$\therefore \sqrt{213444}=462$

$\therefore \sqrt{53824}=232$

(vi) given number 213444

$\therefore \sqrt{213444}=462$

$\therefore \sqrt{213444}=462$

Question 2

Find the number of digits in the square root of each of the following (without any calculation):

(i) 81

(ii) 169

(iii) 4761

(iv) 27889

(v) 525625

Sol :

(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) Given number $525625=6($ even $)$

∴ The  number of digits in its square root $\frac{6}{2}=3$

Question 3

Find the square root of the following decimal numbers by division method:

(i) 51.84

(ii) 42.25

(iii) 18.4041

(iv) 5.774409

Sol :

(i) Given number 51.84

$\therefore \sqrt{51.84}=7.2$

(ii) 42.25

$\sqrt{42.25}=6.5

(iii)  Given number 18.4041

$\sqrt{18.4041}=4.29$

(iv) Given number 5.774409 

$\therefore \sqrt{5.774409}=2.403$

Question 4

Find the square root of the following numbers correct to two decimal places:

(i) 645.8

(ii) 107.45

(iii) 5.462

(iv) 2

(v) 3

Sol :

(i) 645.8 

$\therefore \sqrt{645.8}=25.412 \approx 25.41$ (correct to 2 decimals )

(ii) 107.45

$\sqrt{107 \cdot 45}=10.365 \approx 10.36$

(iii) 

Given number 5.462

$\therefore \sqrt{5.462}=2.337 \approx 2.34$ (corrected to '2 'decimals)

(iv)

Given number 2 

$\sqrt{2}=1.414 \approx 1.41$

(v)

Given number 3 

$\sqrt{3}=1.732 \approx 1.73$ (Corrected to 2 decimals $)$

Question 5

Find the square root of the following fractions by division method:

(i) $\dfrac{841}{1521}$

(ii) $8\dfrac{257}{529}$

(iii) $16\dfrac{169}{441}$

Sol :

(i) $\frac{841}{1521}$

$=\frac{\sqrt{841}}{\sqrt{1521}}=\frac{29}{39}$

(ii)  $8 \frac{257}{529}=\frac{4489}{529}$

$\sqrt{8 \frac{257}{529}}=\sqrt{\frac{4489}{529}}$

$\sqrt{8 \frac{257}{529}}=\frac{67}{23}$

(iii) $16 \frac{169}{441}=\frac{7225}{441}$

$\sqrt{16 \frac{169}{441}}=\frac{\sqrt{7225}}{\sqrt{441}}=\frac{85}{21}$

Question 6

Find the least number which must be subtracted from each of the following numbers to make them a perfect square. Also find the square root of the perfect square number so obtained:

(i) 2000

(ii) 984

(iii) 8934

(iv) 11021

Sol :

(i) Given number 2000

⇒

⇒ Hence , the least number that must be subtracted from 

2000 so as to make it a perfect square is 64 

∴ Required perfect square numbers =2000 - 64

= $1936=44^{2}$

(ii) Given number 984

⇒

⇒ Hence , the least number that must be subtracted 

  

from 984 so as to make it a perfect square is 23 

∴ Required perfect square numbers = 984 - 23 = 961 = $=31^{2}$

(iii) Given number 8934

⇒

⇒ 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) Given number 11021

⇒

⇒ 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 

Find the least number which must be added to each of the following numbers to make them a perfect square. Also find the square root of the perfect square number so obtained:

(i) 1750

(ii) 6412

(iii) 6598

(iv) 8000

Sol :

(i) Given number 1750 

⇒

⇒ 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) Given number 6412

⇒

⇒$6412>(80)^{2}$

=$81^{2}=6561$

⇒ $\therefore$ 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)  Given number 6598

⇒

⇒ $6598>(81)^{2}$

=$(82)^{2}=6724$

$\therefore$ 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) Given number 8000

⇒

⇒ $8000>89^{2}$

⇒$90^{2}=8100$

⇒ $\therefore$ 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 

Find the smallest four-digit number which is a perfect square.

Sol :

Smallest four digit number =  1000 

⇒

⇒ $1000>31^{2}$

⇒ $32^{2}$ will be next perfect square

⇒ $32^{2}=1024$

⇒ Hence , 1024 is smallest four digit number which perfect square

Question 9 

Find the greatest number of six digits which is a perfect square.

Sol :

Greatest six digit number = 999999 

⇒

⇒ 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 

In a right triangle ABC, ∠B = 90°.

(i) If AB = 14 cm, BC = 48 cm, find AC.

(ii) If AC = 37 cm, BC = 35 cm, find AB.

Sol :

(i) AB = 14 cm 

BC = 48 cm

 

according to Pythagoras theorem 

⇒ $A C^{2}=A B^{2}+B C^{2}$

⇒$14^{2}+48^{2}$

⇒ $A C^{2}=2500$

⇒ $A C=\sqrt{2500}$

⇒ AC=50 cm

(ii) $A C=37 \mathrm{~cm}, B C=35 \mathrm{cm}, A B=?$

 

⇒ According to Pythagoras 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 cm

Question 11 

A gardener has 1400 plants. He wants to plant these in such a way that the number of rows and number of columns remains the same. Find the minimum number of plants he needs more for this.

Sol :

Total plants = 1400 

let no . of rows = x 

no. of columns = x 

⇒ 

$x^{2}=1400$

$1400>(37)^{2}$

$38^{2}=1444$

So To make 1400 a perfect square, we have add

minimum of 44

$\therefore 44$ plants needed more.

Question 12

There are 1000 children in a school. For a P.T. drill they have to stand in such a way that the number of rows is equal to a number of columns. How many children would be left out in this arrangement?

Sol :

⇒ Total no of students = 1000 

⇒ let no of row = no of columns = x

⇒ Total students rows x columns = 1000

⇒

⇒ $x \times x=1000$

$x^{2}=1000$

$x=\sqrt{1000}$

So Remainder $=39$

⇒ hence 39 children will be left out 

Question 13

Amit walks 16 m south from his house and turns east to walk 63 ra to reach his friend’s house. While returning, he walks diagonally from his friend’s house to reach back to his house. What distance did he walk while returning?

Sol :

⇒ 

⇒ Distance that amit walk while returning 

⇒ AC

 In $\triangle A B C$

⇒ According to Pythagoras theorem

⇒ $A C=A B^{2}+B C^{2}$

⇒$A C^{2}=16^{2}+63^{2}$

⇒$A C^{2}=4225$

⇒AC=65 m

ஃ Hence amit walks 65 m while returing to his house

Question 14

A ladder 6 m long leaned against a wall. The ladder reaches the wall to a height of 4.8 m. Find the distance between the wall and the foot of the ladder.

Sol: 

⇒ Length of  ladder = 6m 

height of wall = 4.8m

In $\triangle A B C$

According Pythagoras theorem

⇒ $A C^{2}=A B^{n}+B C^{2}$

$B^{2}=4 \cdot 8^{2}+B C^{2}$

$B C^{N}=12.96$

$B C=\sqrt{12.96}$

BC=3.6 m

⇒ Hence, Distance between wall and foot of ladder

is 3.6 m

ML AGGARWAL CLASS 8 CHAPTER 3 SQUARES AND ROOTS Exercise 3.2

 Exercise 3.2

Question 1

Write five numbers which you can decide by looking at their one’s digit that they are not square numbers.

Sol :

We know that a number which ends with the digits 2,3,7 or 8 at its unit places , is not a perfect square.
(i) 2

(ii) 13

(iii) 27

(iv) 88

(v) 243

Question 2 

What will be the unit digit of the squares of the following numbers?

(i) 951

(ii) 502

(iii) 329

(iv) 643

(v) 5124

(vi) 7625

(vii) 68327

(viii) 95628

(ix) 99880

(x) 12796

Sol :

The unit digit of the square of the following numbers will be 

(i) 951 : Its square will have unit digit 1

(ii) 502 : Its square will have unit digit 4

(iii) 329 : Its square will have unit digit 1

(iv) 643 : Its square will have unit digit 9

(v) 5124 : Its square will have unit digit 6

(vi) 7625 : Its square will have unit digit 5

(vii) 68327 : Its square will have unit digit 9 

(viii) 95628 : Its square will have unit digit 4

(ix) 99880 : Its square will have unit digit 0

(x) 12796 : Its square will have unit digit 6

Question 3 

The following numbers are obviously not perfect. Give reason.

(i) 567

(ii) 2453

(iii) 5298

(iv) 46292

(v) 74000

Sol :

We know that if the square of a number does not have

2, 3, 7, 8 or 0 (in an odd number) as its unit digit.

(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 perfect square.

(ii) 2453

2453 has '3' 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 2453 is not a perfect 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 perfect 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 perfect 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 perfect square.

Question 4 

The square of which of the following numbers would be an odd number or an even number? Why?

(i) 573

(ii) 4096

(iii) 8267

(iv) 37916

Sol :

We know that the square of an odd number is odd and

a square of an even number is even. Therefore:

(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 number has ' 6 ' in the units place, Then its square ends in ' 6 '

(iii) 8267

Square of 8267 is a odd number because, If a number has 7 in the Units place, Then its square ends in ' 9

(iv) 37916

square of 37916 is a even number because if a number has ' 6 ' in the Units place, then it square ends in ' 6 '

Question 5

How many natural numbers lie between the square of the following numbers?

(i) 12 and 13

(ii) 90 and 91

Sol :

(i) 12 and 13

There are 2 non-square numbers between the squares of

two consecutive numbers n and n+1

∴ natural numbers between 12 and $(12+1)=2 \times 12=24$

i

hence , there are 24 natural number between $12^{2}$ and $13^{2}$ 

(ii) 90 and 91

There are 2 non-square numbers between the Squares of two consecutive numbers n and n+1

∴ Natural numbers between 90 and 91=2 × 90= 180

Hence, There are 180 natural numbers between $90^{2}$ and $91^{2}$

Question 6

Without adding, find the sum.

(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29

Sol :
(i) $1+3+5+7+9+11+13=7^{2}=49$

(ii) $1+3+5+7+9+11+13+15+17+...+29=15^{2}=225$

sum of first 'n' odd numbers = $n^{2}$

Question 7

(i) Express 64 as the sum of 8 odd numbers.

(ii) 121 as the sum of 11 odd numbers.

Sol :
(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

Express the following as the sum of two consecutive integers:
(i) 192
(ii) 332
(iii) 472

Sol :

(i) $19^{2}=361$

"we can express the square of any odd number greater

than 1 as the sum of two consecutive 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

Find the squares of the following numbers without actual multiplication:

(i) 31

(ii) 42

(iii) 86

(iv) 94

Sol :

(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

Find the squares of the following numbers containing 5 in unit’s place:

(i) 45

(ii) 305

(iii) 525

Sol :

(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

$(525)^{2}=52(52+1)$ hundreds +25

⇒ 2756 hundreds +25 

⇒$(525)^{2}=275625$

Question 11 

Write a Pythagorean triplet whose one number is

(i) 8

(ii) 15

(iii) 63

(iv) 80

Sol :

(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

Given 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 

Given 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 number '63'

(iv) 80 

given number 80 

let us assume $m^{2}-1=80 \Rightarrow m^{2}=81$

m=9

Remaining 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 

Observe the following pattern and find the missing digits:
212 = 441
2012 = 40401
20012 = 4004001
200012 = 4 – – – 4 – – – 1
2000012 = ————–

Sol :

$21^{2}=$ 441

$201^{2}=$ 40401

$2001^{2}=$ 4004001

$20001^{2}=$ 40004001

$200001^{2}=$ 4000400001

Question 13

Observe the following pattern and find the missing digits:
92 = 81
992 = 9801
9992 = 998001
99992 = 99980001
999992 = 9——–8———01
9999992 = 9——–0———1

Sol :

92 = 81
992 = 9801
9992 = 998001
99992 = 99980001
999992 = 9999800001
9999992 = 9999998000001

Question 14

Observe the following pattern and find the missing digits:
72 = 49
672 = 4489
6672 = 444889
66672 = 44448889
666672 = 4 ———–8 ————– 9
6666672 = 4———–8————8 –

Sol :

$7^{2}=$ 49

$67^{2}=$ 4489

$667^{2}=$ 444889

$6667^{2}=$ 44448889

$66667^{2}=$ 4444488889

$666667^{2}=$ 444444888889