ML AGGARWAL CLASS 8 CHAPTER 3 SQUARES AND ROOTS Exercise 3.4

 

EXERCISE 3.4

Question 1

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

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

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

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

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

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 

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 

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 

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 

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 

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

⇒ 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

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

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