##
**Class 8, CBSE (NCERT) Mathematics **

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**Chapter 6, SQUARES AND
SQUARE ROOTS - Mathematics Exercise 6.2
**

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*(CBSE Guide - NCERT Solutions) *

*(CBSE Guide - NCERT Solutions)*

(Page
98)

__Question.1__: Find the square of the following numbers

(i) 32
(ii) 35

(iii)
86 (iv) 93

(v) 71
(vi) 46

__Solutions__:
(i) 32 = (30 + 2)

=> 32

^{2}= (30 + 2)^{2 }
=> 30 (30 + 2) + 2
(30 + 2)

=> 30

^{2}+ 30 × 2 + 2 × 30 + 2^{2}
=> 900 + 60 + 60 +
4

=> 1024

(ii) The number 35 has
5 in its unit’s place.

Hence, 35

^{2}= (3) (3 + 1) hundreds + 25
=> (3 × 4)
hundreds + 25

=> 1200 + 25

=> 1225

(iii) 86 = 80 + 6

=> 86

^{2 }= (80 + 6)^{2}
=> 80 (80 + 6) + 6
(80 + 6)

=> 80

^{2}+ 80 × 6 + 6 × 80 + 6^{2}
=> 6400 + 480 +
480 + 36

=> 7396

(iv) 93 = (90 + 3)

=> 93

^{2 }= (90 + 3)^{2}
=> 90 (90 + 3) + 3
(90 + 3)

=> 90

^{2}+ 90 × 3 + 3 × 90 + 3^{2}
=> 8100 + 270 +
270 + 9

=> 8649

(v) 71 = (70 + 1)

=> 71

^{2 }= (70 + 1)^{2}
=> 70 (70 + 1) + 1
(70 + 1)

=> 70

^{2}+ 70 × 1 + 1 × 70 + 1^{2}
=> 4900 + 70 + 70
+ 1

=> 5041

(vi) Taking hint from
above solutions, try to solve it yourself.

__Question.2__: Write a Pythagorean triplet whose one member is

(i) 6
(ii) 14

(iii)
16 (iv) 18

**We know that for any natural number**

__Solutions__:*m*> 1, 2

*m*,

*m*

^{2}− 1,

*m*

^{2}+ 1 forms a Pythagorean triplet.

(i) If we take

*m*^{2}+ 1 = 6, then*m*^{2}= 5
The value of

*m*will not be an integer.
If we take

*m*^{2}− 1 = 6, then*m*^{2}= 7
Also, the value of

*m*is not an integer.
Let 2

*m*= 6*=> m*= 3

Therefore, the
Pythagorean triplets are 2 × 3, 3

^{2}− 1, 3^{2}+ 1 or 6, 8, and 10.
(ii) If we take

*m*^{2}+ 1 = 14, then*m*^{2}= 13
Clearly,

*m*will not be an integer.
If we take

*m*^{2}− 1 = 14, then*m*^{2}= 15
Also, the value of

*m*is not an integer.
Let 2

*m*= 14*=> m*= 7

Thus,

*m*^{2}− 1 = 49 − 1 = 48 and*m*^{2}+ 1 = 49 + 1 = 50
Therefore, the Pythagorean
triplets are 14, 48, and 50.

(iii) If we take

*m*^{2}+ 1 = 16, then*m*^{2}= 15
The value of

*m*will not be an integer.
If we take

*m*^{2}− 1= 16, then*m*^{2}= 17
Again the value of

*m*is not an integer.
Say, 2

*m*= 16*=> m*= 8

Thus,

*m*^{2}− 1 = 64 − 1 = 63 and*m*^{2}+ 1 = 64 + 1 = 65
Therefore, the
Pythagorean triplets are 16, 63, and 65.

(iv) Taking hint from
above solutions try to solve it yourself.

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