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) 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. Therefore,

35^{2} = (3) (3 + 1) hundreds + 25

= (3 × 4) hundreds + 25

= 1200 + 25 = 1225

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

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

= 70^{2} + 70 × 1 + 1 × 70 + 1^{2}

= 4900 + 70 + 70 + 1

= 5041

(vi) 46^{2 }= (40 + 6)^{2}

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

= 40^{2} + 40 × 6 + 6 × 40 + 6^{2}

= 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, 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

Again 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

The value of *m* will not be an integer.

If we take *m*^{2} − 1 = 14, then *m*^{2} = 15

Again 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 required triplet is 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.

Let 2*m* = 16

*m *= 8

Thus, *m*^{2} − 1 = 64 − 1 = 63 and *m*^{2} + 1 = 64 + 1 = 65

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

(iv) If we take *m*^{2} + 1 = 18,

*m*^{2} = 17

The value of *m* will not be an integer.

If we take *m*^{2} − 1 = 18, then *m*^{2} = 19

Again the value of *m* is not an integer.

Let 2*m *=18

*m *= 9

Thus, *m*^{2} − 1 = 81 − 1 = 80 and *m*^{2} + 1 = 81 + 1 = 82

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