ML AGGARWAL CLASS 8 CHAPTER 3 SQUARES AND ROOTS Exercise 3.2
Exercise 3.2
Question 1
Sol :
(i) 2
(i) 2
(ii) 13
(iii) 27
(iv) 88
(v) 243
Question 2
Sol :
(i) 1
(ii) 4
(iii) 1
(iv) 9
(v) 6
(vi) 5
(vii) 9
(viii) 4
(ix) 0
(x) 6
Question 3
Sol :
(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
Sol :
(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
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)=2×12=24
hence , there are 24 natural number between 122 and 132
(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 \times 90= 180
Hence, There are 180 natural numbers between 902 and 912
Question 6
Sol :
(i) 1+3+5+7+9+11+13=72=49
(i) 1+3+5+7+9+11+13=72=49
(ii) 1+3+5+7+9+11+13+15+17+...+29=152=225
sum of first 'n' odd numbers = n2
Question 7
Sol :
(i) 64
(i) 64
64−1=63;63−3=60;60−5=55
55−7=48:48−9=39;39−11=28
28−13=15;15−15=0
∴ 64=1+3+5+7+9+11+13+15=82
(ii) 121
121−1=120;120−3=117;117−5=112;112−7=105;
105−9=96;96−11=85;85−13=72;72−15=57;
57−17=40;40−19=21;21−21=0
∴ 121=1+3+5+7+9+11+13+15+17+19+21=112
Question 8
Sol :
(i) 192=361
"we can express the square of any odd number greater
than 1 as the sum of two consecutive natural numbers."
First number =192−12=180
Second number =192+12=181
192=361=180+181
(ii) 332=1089
First number =332−12=544
Second number =332+12=545
332=1089=544+545
(iii) 472=2209
First number =472−12=1104
Second number =472+12=1105
472=2209=1104+1105
Question 9
Sol :
(i) 312=(30+1)2=(30+1)(30+1)
=30(30+1)+1(30+1)
=900+30+30+1
312=961
(ii) 422=(40+2)2=(40+2)(40+2)
=40(40+2)+2(40+2)
=1600+80+80+4
422=1764
(iii) 862=(80+6)2=(80+6)(80+6)
=80(80+6)+6(80+6)
=6400+480+480+36
862=7396
(iv) 942=(90+4)2=(90+4)(90+4)
=90(90+4)+4(90+4)
=8100+360+360+16
942=8836
Question 10
Sol :
(i) 45
Comparing with a5 where a = 4
4−1(a5)2=a(a+1) hundreds +25
45r=4(4+1) hundreds +25
=20 hundreds +25
452=2025
(ii) 305
Comparing with a5 where a =30
(a5)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
(a5)2=a(a+1) hundreds +25
(525)2=52(52+1) hundreds +25
⇒ 2756 hundreds +25
⇒(525)2=275625
Question 11
Sol :
(i) 8
Given number = 8
⇒let us assume m2−1=8
⇒m2=9
⇒m = 3
Remaining two numbers of Pythagorean triplet are
m2+1,2m
32+1,2×3
10 , 6
The required triplet (6,8,10) with one number
(ii) 15
Given number =15
Let us assume m2−1=15
m2=16
m = 4
Remaining two numbers of Pythagorean triplet are
m2+1,2m
16+1,2×4
17,8
∴ The required triplet (8,15,17) with one number as 15
(iii) 63
Given number 63
Let us assume m2−1=63
m2=64
m=8
Remaining two numbers of Pythagorean triplet are
m2+1,2m
82+12×8
65-16
∴ We required triplet (16,63,65) with one number '63'
(iv) 80
given number 80
let us assume m2−1=80⇒m2=81
m=9
Remaining two numbers of Pythagorean triplet are
m2+1,2m
q2+1,2×9
82,18
∴ The required triplet (18,80,82) wits one number '80'
Question 12
Sol :
212= 441
2012= 40401
20012= 4004001
200012= 40004001
2000012= 4000400001
Question 14
Sol :
72= 49
672= 4489
6672= 444889
66672= 44448889
666672= 4444488889
6666672= 444444888889
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