NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.2 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.2.

- Squares and Square Roots Class 8 Ex 6.1
- Squares and Square Roots Class 8 Ex 6.3
- Squares and Square Roots Class 8 Ex 6.4

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 6 |

Chapter Name |
Squares and Square Roots |

Exercise |
Ex 6.2 |

Number of Questions Solved |
2 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 8 Maths Chapter 6 Squares 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 (30 + 2) + 2 (30 + 2)

= 30^{2} +30 x 2 + 2 x 30 + 2^{2}

= 900 + 60 + 60 + 4 = 1024

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

= 30 ^{2} + 30 x 5 + 5 x 30 + 5^{2}

= 900 +150 +150 + 25 = 1225

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

= 80^{2} +80 x 6 + 6 x 80 + 62

= 6400 + 480 + 480 + 36 = 7396

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

= 90^{2} + 90 x 3 + 3 x 90 + 92

= 8100 + 270 + 270 + 81 = 8649

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

= 70^{2} +70 x 1 + 1 x 70 + 1^{2}

= 4900 + 70 + 70 +1 = 5041

**
(vi)** = (40 + 6)

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

= 40

^{2}+ 40 x 6 + 6 x 40 +6 x 6

= 1600 +240+240 +36 =2116

**Question 2.**

Write a Pythagorean triplet whose one member is

**(i)** 6

**(ii)** 14

**(iii)** 16

**(iv)** 18

**Solution:**

**(i)** Put m = 3 in 2m, m^{2} -1, m^{2} +1

∴ 2m=6, m^{2} – l =3^{2} – 1= 9 – 1 = 8

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

Thus, 6, 8 and 10 are Pythagorean triplets.

**(ii)** Put m = 7 in 2m, m^{2} -1, m^{2} +1

∴ 2m = 14, m^{2} – 1 = 7^{2} – 1 = 49 – 1 = 48

and m^{2} + 1 = 72 +1 = 49+1 = 50

Thus, 14, 48 and 50 are Pythagorean triplets.

**(iii)** Put m = 8 in 2m, m^{2} – 1, m^{2} +1

∴ 2m =16, m^{2} – 1 = 8^{2} – 1 = 64 – 1 =63

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

Thus, 16, 63 and 65 are Pythagorean triplets.

**(iv)** Put m = 9 in 2m, m^{2} – 1, m^{2} +1

∴ 2m =18, m^{2} – 1 = 9^{2} – 1=81 – 1 = 80,

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

Thus, 18, 80 and 82 are Pythagorean triplets.

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