8th Maths 6.2

Chapter 6

Squares and Square Roots

NCERT Class 8th solution of Exercise 6.1

NCERT Class 8th solution of Exercise 6.3

NCERT Class 8th solution of Exercise 6.4

Exercise 6.2

Q1. Find the square of the following numbers.

i) `32`

ii) `35`

iii) `86`

iv) `93`

v) `71`

vi) `46`

`text{Sol.}`

i) `32^2`

`= (30 + 2)^2`

`= 30(30 + 2) + 2(30 + 2)`

`= 30^2 + 30 times 2 + 2times 30 + 2^2`

`= 900 + 60 + 60 + 4`

`= 1024`

`text{Answer:}`

`1024`.

ii) `35^2`

`= (30 + 5)^2`

`= 30(30 + 5) + 5(30 + 5)`

`= 30^2 + 30times5 + 5times30 + 5^2`

`= 900 + 150 + 150 + 25`

`= 1225`

`text{Answer:}`

`1225`.

iii) `86^2`

`= (80 + 6)^2`

`= 80(80 + 6) + 6(80 + 6)`

`= 80^2 + 80times6 + 6times80 + 6^2`

`= 6400 + 480 + 480 + 36`

`= 7396`

`text{Answer:}`

`7396`.

iv) `93^2`

`= (90 + 3)^2`

`= 90(90 + 3) + 3(90 + 3)`

`= 90^2 + 90times3 + 3times90 + 3^2`

`= 8100 + 270 + 270 + 9`

`= 8649`

`text{Answer:}`

`8649`.

v) `71^2`

`= (70 + 1)^2`

`= 70(70 + 1) + 1(70 + 1)`

`= 70^2 + 70times1 + 1times70 + 1^2`

`= 4900 + 70 + 70 + 1`

`= 5041`

`text{Answer:}`

`5041`.

vi) `46^2`

`= (40 + 2)^2`

`= 40(40 + 2) + 2(40 + 2)`

`= 40^2 + 40times2 + 2times40 + 2^2`

`= 1600 + 80 + 80 + 4`

`= 2116`

`text{Answer:}`

`2116`.

Q2. Write a Pythagorean triplet whose one member is.

i) `6`

ii) `14`

iii) `16`

iv) `18`

`text{Sol.}`

i) 

`text{We can get Pythagorean triplets by}``text{using general form 2m,} text{ m}^2 - 1, text{ m}^2 + 1`

`text{Let  m}^2 - 1 = 6`

`text{m}^2 = 6 + 1` 

`text{m}^2 = 7`

`text{Then the value of m will not be an integer.}`

`text{m}^2 + 1 = 6`

`text{m}^2 = 6 - 1`

`text{m}^2 = 5`

`text{Again the value of  m will not be an integer.}`

`text{2m = 6}`

`text{m = 3 which is an integer}`

`text{thus}`

`text{m}^2  - 1,`

`= 3^2 - 1`

`= 9 - 1 = 8`

`text{m}^2 + 1`

`= 3^2 + 1`

`= 9 + 1 = 10`

`text{Answer:}`

`text{The required tirplet are 6, 8, 10}`

ii) 

`text{We can get Pythagorean triplets by}``text{using general form 2m,} text{ m}^2 - 1, text{ m}^2 + 1`

`text{Let  m}^2 - 1 = 14`

`text{m}^2 = 14 + 1` 

`text{m}^2 = 15`

`text{Then the value of m will not be an integer.}`

`text{m}^2 + 1 = 14`

`text{m}^2 = 14 - 1`

`text{m}^2 = 13`

`text{Again the value of  m will not be an integer.}`

`text{2m = 14}`

`text{m = 7 which is an integer}`

`text{thus}`

`text{m}^2  - 1,`

`= 7^2 - 1`

`= 49 - 1 = 48`

`text{m}^2 + 1`

`= 7^2 + 1`

`= 49 + 1 = 50`

`text{Answer:}`

`text{The required tirplet are 14, 49, 50}`

iii)

`text{We can get Pythagorean triplets by}``text{using general form 2m,} text{ m}^2 - 1, text{ m}^2 + 1`

`text{Let  m}^2 - 1 = 16`

`text{m}^2 = 16 + 1` 

`text{m}^2 = 17`

`text{Then the value of m will not be an integer.}`

`text{m}^2 + 1 = 16`

`text{m}^2 = 16 - 1`

`text{m}^2 = 15`

`text{Again the value of  m will not be an integer.}`

`text{2m = 16}`

`text{m = 8}`

`text{thus}`

`text{m}^2  - 1,`

`= 8^2 - 1`

`= 64 - 1 = 63`

`text{m}^2 + 1`

`= 8^2 + 1`

`= 64 + 1 = 65`

`text{Answer:}`

`text{The required tirplet are 16, 63, 65}`

iv) 

`text{We can get Pythagorean triplets by}``text{using general form 2m,} text{ m}^2 - 1, text{ m}^2 + 1`

`text{Let  m}^2 - 1 = 18`

`text{m}^2 = 18 + 1` 

`text{m}^2 = 19`

`text{Then the value of m will not be an integer.}`

`text{m}^2 + 1 = 18`

`text{m}^2 = 18 - 1`

`text{m}^2 = 17`

`text{Again the value of  m will not be an integer.}`

`text{2m = 18}`

`text{m = 9}`

`text{thus}`

`text{m}^2  - 1,`

`= 9^2 - 1`

`= 81 - 1 = 80`

`text{m}^2 + 1`

`= 8^2 + 1`

`= 81 + 1 = 82`

`text{Answer:}`

`text{The required tirplet are 16, 80, 82}`