Haryana State Board HBSE 8th Class Maths Solutions Chapter 6 Square and Square Roots Ex 6.2 Textbook Exercise Questions and Answers.

## Haryana Board 8th Class Maths Solutions Chapter 6 Square and Square Roots Exercise 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

= 12 × 100 + 25 = 1225

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

= (80+ 6) (80+ 6)

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

= 6400 + 480 + 480 + 36

= 7396

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

= (90 + 3) (90 + 3)

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

= 8100 + 270 + 270 + 9

= 8649

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

= (70)^{2} + 2 × 70 × 1 + (1)^{2}

[Using (a + b)^{2} = a^{2} + 2ab + b^{2}]

= 4900 + 140 + 1 = 5041

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

= (40)^{2} + 2 × 40 × 6 + (6)^{2}

= 1600 + 480 + 36 = 2116.

Question 2.

Write a Pythagorean triplet whose one member is

(i) 6

(ii) 14

(iii) 16

(iv) 18.

Solution:

We have Pythagorean triplet (2m, m^{2} – 1, m^{2} + 1); where m > 1 (natural number).

Let us try one by one. .

Now, we take . .

m^{2} – 1 = 6 ⇒ m^{2} = 6 + 1 = 7

⇒ m = \(\sqrt {7}\) (not an integer)

m^{2} + 1 = 6 ⇒ m^{2} = 5

⇒ m = \(\sqrt {5}\) (not an integer)

2m = 6

⇒ m = \(\frac{6}{2}\) = 3 (is an integer)

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

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

Therefore, the required triplet is (6, 8, 10).

(ii) m^{2} – 1 = 14

or m^{2} = 15

Then the value of m will not be an integer So, we try to take

m^{2} + 1 = 14

or m^{2} = 14 – 1 = 13

or m = \(\sqrt {13}\)

which is also not an integer.

So, take 2m = 14

⇒ m = 7.

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

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

∴ The required triplet is (14, 48, 50).

(iii) Let m^{2} – 1 = 16

⇒ m^{2} = 17

or m = \(\sqrt {17}\)

which is not an integer

m^{2} + 1 = 16

⇒ m^{2} = 15

⇒ m = \(\sqrt {15}\) (not an integer)

and 2m = 16

⇒ m = 8.

Here, the value of m is an integer.

∴ m^{2} – 1 = 8^{2} – 1 = 64 – 1 = 63

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

Therefore, the required triplet is (16, 63, 65)

(iv) Let m^{2} – 1 = 18

⇒ m^{2} = 19

or m = \(\sqrt {19}\)

which is not an integer

Now, take m^{2} + 1 = 18

⇒ m^{2} = 17

or m = \(\sqrt {17}\)

which is also not an integer

So, let us take 2m = 18

⇒ m = 9 which is an integer.

∴ The required triplet is (18, 9^{2} – 1, 9^{2} + 1)

i.e., (18, 80, 81)