You can Download KSEEB Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.2 Questions and Answers to help you to revise the complete syllabus.

## KSEEB Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.2

Question 1.

Find the square of the following numbers:

(i) 32

(ii) 35

(iii) 86

(iv) 93

(v) 71

(vi) 46

Solution:

(i) 32^{2} = (30 + 2)^{2}

= (30 + 2)(30 + 2)

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

= 900 + 60 + 60 + 4

= 1024

(ii) 35^{2} = (3 × 4) hundreds + 25

= 1225

(iii) 86^{2} = (80 + 6)^{2}

= (80 + 6)(80 + 6)

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

= 6400 + 480 + 480 + 36

= 6400 + 960 + 36

= 6400 + 946

= 7396

(iv) 93^{2} = (90 + 3)^{2}

= (90 + 3)(90 + 3)

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

= 8100 + 270 + 270 + 9

= 8100 + 540 + 9

= 8649

(v) 71^{2} = (70 + 1)^{2}

= (70 + 1) (70 + 1)

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

= 4900 + 70 + 70 + 1

= 5041

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

= (40 + 6)(40 + 6)

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

= 1600 + 240 + 240 + 36

= 1600 + 480 + 36

= 2116

Question 2.

Write a Pythagorean triplet whose one member is.

(i) 6

(ii) 14

(iii) 16

(iv) 18

Solution:

For any natural number m > 1

We have

(2m)^{2} + (m^{2} – 1)^{2} = (m^{2} – 1)^{2}

So, 2m, m^{2} – 1, m^{2} + 1 forms a Pythagorean triplet.

(i) We take, m^{2} – 1 = 6

m^{2} = 1

m = √7

Then, the value of m is not an integer.

Now, take m^{2} + 1 = 6

m^{2} = 5

m = √5

Again value of m is not an integer

2m = 6

m = 3

Thus, m^{2} – 1 = 3^{2} – 1 = 9 – 1 = 8

m^{2} + 1 = 3^{2} + 1 = 9 + 1 = 10

2m = 2 × 3 = 6

∴ Required triplet = (6, 8, 10)

(ii) Take m^{2} – 1 = 14

m^{2} = 15

m = √15

The value of m is not an integer

Again for m^{2} + 1 = 14

m^{2} = 13

m = √13

Hence, the value of m is not an integer

Now take 2m = 14

m = 7

m^{2} – 1 = 7^{2} – 1 = 49 – 1 = 48

m^{2} + 1 = 7^{2} + 1 = 49 + 1 = 50

2m = 2 × 7 = 14

∴ Required triplet is (14, 48, 50)

(iii) Take m^{2} – 1 = 16

m^{2} = 17

m = √17

The value of m is not an integer.

Again, for m^{2} + 1 = 16

m^{2} = 15

m = √15

Hence, the value of m is not an integer

Now, take 2m = 16

m = 8

m^{2} – 1 = 8^{2} – 1 = 64 – 1 = 63

m^{2} + 1 = 8^{2} + 1 = 64 + 1 = 65

2m = 2 × 8 = 16

∴ Required triplet is (16, 63, 65)

(iv) Take m^{2} – 1 = 18

m^{2} = 19

m = √19

The value of m is not an integer.

For m^{2} + 1 = 18

m^{2} = 17

m = √17

The value of m is not an integer

For 2m = 18

m = 9

∴ Required triplet

m^{2} – 1 = 9^{2} – 1 = 81 – 1 = 80

m^{2} + 1 = 9^{2} + 1 = 81 + 1 = 82

2m = 2 × 9 = 18

∴ Required triplet is (18, 80, 82)