Chapter 6 Squares and Square Roots Ex 6.2 Solutions
Question - 1 : - Find the square of the following numbers.
(i) 32
(ii) 35
(iii) 86
(iv) 93
(v) 71
(vi) 46
Answer - 1 : -
(i)32 = 30 + 2
(32)2┬а= (30 + 2)2
= 30(30 + 2) + 2(30 + 2)
= 302┬а+ 30 ├Ч 2 + 2 ├Ч 30+ 22
= 900 + 60 + 60 + 4
= 1024
Thus (32)2┬а= 1024
(ii)35 = (30 + 5)
(35)2┬а= (30 + 5)2
= 30(30 + 5) + 5(30 + 5)
= (30)2┬а+ 30 ├Ч 5 + 5 ├Ч 30+ (5)2
= 900 + 150 + 150 + 25
= 1225
Thus (35)2┬а= 1225
(iii)86 = (80 + 6)
862┬а= (80 + 6)2
= 80(80 + 6) + 6(80 + 6)
= (80)2┬а+ 80 ├Ч 6 + 6 ├Ч 80+ (6)2
= 6400 + 480 + 480 + 36
= 7396
Thus (86)2┬а= 7396
(iv)93 = (90+ 3)
932┬а= (90 + 3)2
= 90 (90 + 3) + 3(90 + 3)
= (90)2┬а+ 90 ├Ч 3 + 3 ├Ч 90+ (3)2
= 8100 + 270 + 270 + 9
= 8649
Thus (93)2┬а= 8649
(v)71 = (70 + 1)
712┬а= (70 + 1)2
= 70 (70 + 1) + 1(70 + 1)
= (70)2┬а+ 70 ├Ч 1 + 1 ├Ч 70+ (1)2
= 4900 + 70 + 70 + 1
= 5041
Thus (71)2┬а= 5041
(vi)46 = (40+ 6)
462┬а= (40 + 6)2
= 40 (40 + 6) + 6(40 + 6)
= (40)2┬а+ 40 ├Ч 6 + 6 ├Ч 40+ (6)2
= 1600 + 240 + 240 + 36
= 2116
Thus (46)2┬а= 2116
Question - 2 : - Write a Pythagorean triplet whose one member is
(i) 6
(ii) 14
(iii) 16
(iv) 18
Answer - 2 : -
(i)Let m2┬атАУ 1 = 6
[Triplets are in the form 2m, m2┬атАУ 1, m2┬а+ 1]
m2┬а= 6 + 1 = 7
So, the value of m will not be an integer.
Now, let us try for m2┬а+ 1 = 6
тЗТ m2┬а= 6 тАУ 1 = 5
Also, the value of m will not be an integer.
Now we let 2m = 6 тЗТ m = 3 which is aninteger.
Other members are:
m2┬атАУ 1 = 32┬атАУ 1 = 8 and m2┬а+ 1 = 32┬а+ 1 = 10
Hence, the required triplets are 6, 8 and 10
(ii)Let m2┬атАУ 1 = 14 тЗТ m2┬а= 1 + 14 = 15
The value of m will not be an integer.
Now take 2m = 14 тЗТ m = 7 which is aninteger.
The member of triplets are 2m = 2 ├Ч 7 = 14
m2┬атАУ 1 = (7)2┬атАУ 1 = 49 тАУ 1 = 48
and m2┬а+ 1 = (7)2┬а+ 1 = 49 + 1 = 50
i.e., (14, 48, 50)
(iii)Let 2m = 16 m = 8
The required triplets are 2m = 2 ├Ч 8 = 16
m2┬атАУ 1 = (8)2┬атАУ 1 = 64 тАУ 1 = 63
m2┬а+ 1 = (8)2┬а+ 1 = 64 + 1 = 65
i.e., (16, 63, 65)
(iv)Let 2m = 18 тЗТ m = 9
Required triplets are:
2m = 2 ├Ч 9 = 18
m2┬атАУ 1 = (9)2┬атАУ 1 = 81 тАУ 1 = 80
and m2┬а+ 1 = (9)2┬а+ 1 = 81 + 1 = 82
i.e., (18, 80, 82)