In this page we have ** NCERT book Solutions for Class 8th Maths:Square roots ** for
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**Question 1**

Find the square of the following numbers.

(i) 32

(ii) 35

(iii) 86

(iv) 93

(v) 71

(vi) 46

**Answer**

1) 32^{2}

We can find the square using direct multiplication

= 32 x 32 = 1024

But above method can be cumbersome to calculate. We can calculate such values in the another better way f

Since, 32 can be written as (30+2)

So, 32^{2} = (30+2)^{2} = (30+2)(30+2)

Now we know the identity

(a+b)^{2}= a^{2} + b^{2} +2ab

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

= 900 + 120 + 4 = 1024

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

Solving on similar lines as above

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

= 900 + 300 + 25 = 1225

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

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

= 6400 + 960 + 36 = 7396

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

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

= 8100 + 540 + 9 = 8649

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

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

= 4900 + 140 + 1 = 5040 + 1 = 5041

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

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

= 1600 + 480 + 36 = 2080 + 36 = 2116

**Question: 2**

Write a Pythagorean triplet whose one member is:

(i) 6

(ii) 14

(iii) 16

(iv) 18

**Answer**

As we know 2n, n ^{2} + 1 and n^{2} - 1 form a Pythagorean triplet for any number, n > 1.

1) If we take n ^{2} + 1 or n^{2} - 1 to be 6 then then the value of n will not integer(n^{2 } will be 5 or 7)

So we can 2n = 6

Therefore, n = 3

And, n^{2} + 1 = 3^{ 2 }+ 1= 9 + 1 = 10

And,n^{ 2 }- 1 = 3^{ 2 }- 1 = 9 - 1 = 8

Test: 6^{ 2 }+ 8^{ 2 }= 36 + 64 = 100 = 10^{2}

Hence, the triplet is 6, 8, and 10 Answer

2) If we take n ^{2} + 1 or n^{2} - 1 to be 14 then then the value of n will not integer(n^{2} will be 15 or 13)

So we can take 2n= 14, therefore, n = 7

Now, n ^{2 }+ 1 = 7 ^{2 }+ 1 = 49 + 1 = 50

And, n^{2 }- 1 = 7 ^{2 }- 1 = 49 - 1 = 48

Test: 14 ^{2 }+ 48 ^{2 }= 196 + 1304 = 2500 = 50 ^{2 }

Hence, the triplet is 14, 48, and 50 Answer

3) If we take n ^{2} + 1 or n^{2} - 1 to be 16 then then the value of n will not integer(n^{2} will be 17 or 15)

Let us assume 2n = 16, then n = 8

Now, n^{2 }+ 1 = 8^{ 2 }+ 1 = 64 + 1 = 65

And, n ^{2 }- 1 = 8^{ 2 }- 1 = 64 - 1 = 63

Test: 16^{2 }+ 63^{ 2 }= 256 + 3969 = 4225 = 65 ^{2 }

Hence, the triplet is 16, 63, and 65 Answer

4) If we take n ^{2} + 1 or n^{2} - 1 to be 18 then then the value of n will not integer(n^{2} will be 19 or 17)

Let us assume 2n = 18, therefore, n = 9

Now, n ^{2 }+ 1 = 9 ^{2 }+ 1 = 81 + 1 = 82

And, n ^{2 }- 1 = 9 ^{2 }- 1 = 81 - 1 = 80

Test: 18 ^{2 }+ 80 ^{2 }= 324 + 6400 = 6724 = 82 ^{2 }

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