Gujarat BoardĀ GSEB Textbook Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.2

Question 1.

Find the square of the following numbers containing 5 in unit place?

- 32
- 35
- 86
- 93
- 71
- 46

Solution:

1. (32)^{2} = (30 + 2)^{2}

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

= 900 + 120 + 4 = 1024

2. (35)^{2} = (30 + 5)^{2}

= (30)^{2} + 2(30)(5) + (5)^{2}

= 900 + 300 + 25

= 1200 + 25 = 1225

Second method

35^{2} = 3 Ć (3 + 1) Ć 100 + 25

= 3 Ć 4 Ć 100 + 25

= 1200 + 25 = 1225

3. (86)^{2} = (80 + 6)^{2}

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

= 6400 + 960 + 36 = 7396

4. (93)^{2} = (90 + 3)^{2}

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

= 8100 + 540 + 9 = 8649

5. (71)^{2} = (70 + 1)^{2}

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

= 4900 + 140 + 1 = 5041

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

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

= 1600 + 480 + 36

= 2116

Question 2.

Write a Pythagorean tripler whose one member is

- 6
- 14
- 16
- 18

Solution:

1. Let 2n – 6 ā“ n = 3

Now, n^{2} – 1 = 3^{2} – 1 = 8

and n^{2} + 3^{2} + 1 = 10

Thus, the required Pythagorean triplet is 6, 8, 10.

2. Let 2n = 14 ā“ n = 7

Now, n^{2} – 1 = 7^{2} – 1 = 48

and n^{2} + 1 = 7^{2} + 1 = 50

Thus, the required Pythagorean triplet is 14, 48, 50.

3. Let 2n = 16 ā“ n = 8

Now n^{2} – 1

= 64 – 1 = 63

and n^{2} + 1 = 8^{2} + 1

= 64 + 1

= 65

ā“ The required Pythagorean triple (is 16, 63, 65)

4. Let 2n = 18

Now, n^{2} – 1 = 9^{2} – 1

= 81 – 1 = 80

and n^{2} + 1 = 9^{2} + 1

= 81 + 1 = 82

The required Pythagorean triplet is 18, 80, 82.