NCERT Solutions for Square roots Class 8 Maths Chapter 6 CBSE Part 2

In this page we have NCERT Solutions for Square roots Class 8 Maths Chapter 6 for EXERCISE 2 . Hope you like them and do not forget to like , social share and comment at the end of the page.
Question 1
Find the square of the following numbers.
(i) 32
(ii) 35
(iii) 86
(iv) 93
(v) 71
(vi) 46
1) 322
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, 322 = (30+2)2 = (30+2)(30+2)
Now we know the identity
(a+b)2= a2 + b2 +2ab
= 302 + 2 x 30 x 2 + 22
= 900 + 120  + 4 = 1024
2) (35)2 = (30+5)2
Solving on similar lines as above
= 302 + 2 x 30 x 5  + 52
= 900 + 300 + 25 = 1225
3) 862 = (80 + 6)2
= 802 + 2 x  80 x 6  + 62
= 6400 + 960 + 36 = 7396
4) 932 = (90+3)2
= 90 2 +2 x 90 x 3  + 3 2
= 8100 + 540 + 9 = 8649
5) 71 2 = (70 + 1) 2
= 702 +2 x 70 x 1  + 1 x 1
= 4900 + 140 + 1 = 5040 + 1 = 5041
6) 462 = (40+6)2
= 40 2 + 2 x 40 x 6  + 62
= 1600 + 480 + 36 = 2080 + 36 = 2116

Question: 2
Write a Pythagorean triplet whose one member is:
(i) 6
(ii) 14
(iii) 16
(iv) 18
As we know 2n, n 2 + 1 and n2 - 1 form a Pythagorean triplet for any number, n > 1.
1) If we take n 2 + 1 or  n2 - 1 to be 6 then then the value of n will  not integer(n2  will be 5 or 7)
So  we can  2n = 6
Therefore, n = 3
And, n2 + 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 = 102
Hence, the triplet is 6, 8, and 10 Answer
2) If we take n 2 + 1 or  n2 - 1 to be 14 then then the value of n will  not integer(n2  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, n2 - 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  n2 - 1 to be 16 then then the value of n will  not integer(n2  will be 17 or 15)
Let us assume 2n = 16, then n = 8
Now, n2 + 1 = 8 2 + 1 = 64 + 1 = 65
And, n 2 - 1 = 8 2 - 1 = 64 - 1 = 63
Test: 162 + 63 2 = 256 + 3969 = 4225 = 65 2
Hence, the triplet is 16, 63, and 65 Answer
4) If we take n 2 + 1 or  n2 - 1 to be 18 then then the value of n will  not integer(n2  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