**Download NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.2 – Square and Square roots**

**Q. 1 ****Find the square of the following numbers.**

**(i) 32
(ii) 35
(iii) 86
(iv) 93
(v) 71
(vi) 46**

**Solution:**

**(i) Square of 32**

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

⇒30^{2} + 30 × 2 + 2 × 30 + 2^{2}⇒900 + 60 + 60 + 4

⇒1024

Therefore, square of 32 is 1024

**(ii) Square of 35 **[ 5 in unit’s place ]

⇒35^{2} = (3) (3 + 1) hundreds + 25

⇒(3 × 4) hundreds + 25

⇒1200 + 25

⇒1225

Therefore, square of 35 is 1225

**(iii) Square of 86**

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

⇒80^{2} + 80 × 6 + 6 × 80 + 6^{2}⇒6400 + 480 + 480 + 36

⇒7396

Therefore, square of 86 is 7396

**(iv) Square of 93**

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

⇒90^{2} + 90 × 3 + 3 × 90 + 3^{2}⇒8100 + 270 + 270 + 9

⇒8649

Therefore, square of 93 is 8649

**(v) Square of 71**

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

⇒70^{2} + 70 × 1 + 1 × 70 + 1^{2}⇒4900 + 70 + 70 + 1

⇒5041

Therefore, square of 71 is 5041

**(vi) Square of 46**

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

⇒40^{2} + 40 × 6 + 6 × 40 + 6^{2}⇒1600 + 240 + 240 + 36

⇒2116

Therefore, square of 46 is 2116

**Q.2 ****Write a Pythagorean triplet whose one member is **

**(i) 6
(ii) 14
(iii) 16
(iv) 18**

**Solution:**

We know that, for any natural number m > 1, we have (2m)^{2} + (m^{2} – 1)^{2} = (m^{2} + 1)^{2}.

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

**(i) Pythagorean triplet whose one member is 6**

Let us first take

m^{2} − 1 = 6, then m^{2} = 7

⇒The value of m will not be an integer.

If we try taking

m^{2} + 1 = 6, then m^{2} = 5

⇒Again, the value of m will not be an integer.

So, let’s try

2m = 6, then m = 3

Therefore, the Pythagorean triplets are 2 × 3, 3^{2} − 1, 3^{2} + 1 ie., 6, 8, and 10.

**(ii) Pythagorean triplet whose one member is 14**

Let us first take

m^{2} – 1 = 14, then m^{2} = 15

⇒The value of m will not be an integer

if we try taking

m^{2} + 1 = 14, then m^{2} = 13

⇒Again, the value of m will not be an integer

So, let’s try

2m = 14, then m = 7

Therefore, the Pythagorean triplets are 2 × 7, 7^{2} − 1, 7^{2} + 1 ie., 14, 48, and 50.

**(iii) Pythagorean triplet whose one member is 16**

Let us first take

m^{2} – 1 = 16, then m^{2} = 17

⇒The value of m will not be an integer

if we try taking

⇒m^{2} + 1 = 17,

So, 2m = 16, then m = 8

Therefore, the Pythagorean triplets are 2 × 8, 8^{2} − 1, 8^{2} + 1 ie., 16, 63, and 65.

**(iv) Pythagorean triplet whose one member is 18**

Let us first take

m^{2} – 1 = 18, then m^{2} = 19

⇒The value of m will not be an integer

if we try taking

m^{2} + 1 = 18, then m^{2} = 17

⇒Again, the value of m will not be an integer

So, let’s try

2m = 18, then m = 9

Therefore, the Pythagorean triplets are 2 × 9, 9^{2} − 1, 9^{2} + 1 ie., 18, 80, and 82.

**Download NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.2 – Square and Square roots**

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