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

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 9 Chapter 9 Logarithms MCQs

 MCQs correct Solution from the given four options (1 to 7): Question 1 If log√3 27 = x, then the value of x is (a) 3 (b) 4 (c) 6 (d) 9 Sol : ⇒$\log _{\sqrt{3}} 27=x$ ⇒$(\sqrt{3})^{x}=27$ ⇒$(3)^{\frac{1}{2} x x}=3^{3}$ ⇒$3^{\frac{x}{2}}=3^{3}$ ⇒$\frac{x}{2}=3$ ⇒x=6...(c) Question 2 If log 5 (0.04) = x, then the value of x is (a) 2 (b) 4 (c) -4 (d) -2 Sol : ⇒$\log _{5}(0.04)=x$ ⇒$5^{x}=0.04=\frac{4}{100}=\frac{1}{25}=5^{-2}$ ∴x=-2...(d) Question 3 If log 0.5  64 = x, then the value of x is (a) -4 (b) -6 (c) 4 (d) 6 Sol : ⇒$\log _{0.5} 64=x $ ⇒$0.5^{x}=64$ ⇒$=\left(\frac{1}{2}\right)^{x}=2^{6}$ ⇒$2^{-x}=2^{6}$ ∴-x=6⇒x=-6...(b) Question 4 If $\log _{10} \sqrt[3]{5} x=-3$, then the value of x is (a) $\frac{1}{5}$ (b) $-\frac{1}{5}$ (c) -1 (d) 5 Sol : $\log _{\sqrt[3]{5}} x=-3,(\sqrt[3]{5})^{-3}=x$ $x=\left(5^{\frac{1}{3}}\right)^{-3}=5^{\frac{1}{3}(-3)}=5^{-1}$ $x=\frac{1}{5}$ (b) Question 5 If log (3x + 1) = 2, then the value of x is (a) $\frac{1}{3}$ (b) 99 (c) 33 (d) $\frac{19}{3}$ Sol
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference between the actual upper limit and the actual lower li
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S
Read more