NCERT Solutions for class 8 Maths chapter 6 Square and square roots

Exercise 6.1  Exercise 6.2  Exercise 6.3  Exercise 6.4

Exercise 6.2

Question 1

Find the square of the following numbers

(i) 32 (ii) 35

(iii) 86 (iv) 93

(v) 71 (vi) 46

Sol :

(i) 322 = (30 + 2)2

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

= 302 + 30 × 2 + 2 × 30 + 22

= 900 + 60 + 60 + 4

= 1024

 

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

352 = (3) (3 + 1) hundreds + 25

= (3 × 4) hundreds + 25

= 1200 + 25 = 1225

(iii) 862 = (80 + 6)2

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

= 802 + 80 × 6 + 6 × 80 + 62

= 6400 + 480 + 480 + 36

= 7396

 

(iv) 932 = (90 + 3)2

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

= 902 + 90 × 3 + 3 × 90 + 32

= 8100 + 270 + 270 + 9

= 8649

 

(v) 712 = (70 + 1)2

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

= 702 + 70 × 1 + 1 × 70 + 12

= 4900 + 70 + 70 + 1

= 5041

 

(vi) 462 = (40 + 6)2

= 40 (40 + 6) + 6 (40 + 6)

= 402 + 40 × 6 + 6 × 40 + 62

= 1600 + 240 + 240 + 36

= 2116

 

Question 2

Write a Pythagorean triplet whose one member is

(i) 6 (ii) 14

(iii) 16 (iv) 18

Sol :

For any natural number m > 1, 2m, m2 − 1, m2 + 1 forms a Pythagorean triplet.

(i) If we take m2 + 1 = 6, then m2 = 5

The value of m will not be an integer.

If we take m2 − 1 = 6, then m2 = 7

Again the value of m is not an integer.

Let 2m = 6

m = 3

Therefore, the Pythagorean triplets are 2 × 3, 32 − 1, 32 + 1 or 6, 8, and 10.

 

(ii) If we take m2 + 1 = 14, then m2 = 13

The value of m will not be an integer.

If we take m2 − 1 = 14, then m2 = 15

Again the value of m is not an integer.

Let 2m = 14

m = 7

Thus, m2 − 1 = 49 − 1 = 48 and m2 + 1 = 49 + 1 = 50

Therefore, the required triplet is 14, 48, and 50.

 

(iii) If we take m2 + 1 = 16, then m2 = 15

The value of m will not be an integer.

If we take m2 − 1= 16, then m2 = 17

Again the value of m is not an integer.

Let 2m = 16

m = 8

Thus, m2 − 1 = 64 − 1 = 63 and m2 + 1 = 64 + 1 = 65

Therefore, the Pythagorean triplet is 16, 63, and 65.

 

(iv) If we take m2 + 1 = 18,

m2 = 17

The value of m will not be an integer.

If we take m2 − 1 = 18, then m2 = 19

Again the value of m is not an integer.

Let 2m =18

m = 9

Thus, m2 − 1 = 81 − 1 = 80 and m2 + 1 = 81 + 1 = 82

Therefore, the Pythagorean triplet is 18, 80, and 82.

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