CHAPTER 5 – EXERCISE 5.2 SOLUTIONS

Q1. Find the square of the following numbers.

(i) \( 32^2 = 32 \times 32 = 1024 \)

(ii) \( 35^2 = 35 \times 35 = 1225 \)

(iii) \( 86^2 = 86 \times 86 = 7396 \)

(iv) \( 93^2 = 93 \times 93 = 8649 \)

(v) \( 71^2 = 71 \times 71 = 5041 \)

(vi) \( 46^2 = 46 \times 46 = 2116 \)

Q2. Write a Pythagorean triplet whose one member is:

Formula: A Pythagorean triplet is of the form \((2m,\; m^2 - 1,\; m^2 + 1)\)

(i) One member = 6
Let \(2m = 6 \Rightarrow m = \frac{6}{2} = 3\).
Now, \(2m = 2 \times 3 = 6\),
\(m^2 - 1 = 3^2 - 1 = 9 - 1 = 8\),
\(m^2 + 1 = 3^2 + 1 = 9 + 1 = 10\).
Therefore, required triplet = (6, 8, 10).
,

(ii) One member = 14
Let \(2m = 14 \Rightarrow m = \frac{14}{2} = 7\).
Now, \(2m = 2 \times 7 = 14\),
\(m^2 - 1 = 7^2 - 1 = 49 - 1 = 48\),
\(m^2 + 1 = 7^2 + 1 = 49 + 1 = 50\).
Required triplet = (14, 48, 50)

(iii) One member = 16
Let \(2m = 16 \Rightarrow m = \frac{16}{2} = 8\).
Now, \(2m = 2 \times 8 = 16\),
\(m^2 - 1 = 8^2 - 1 = 64 - 1 = 63\),
\(m^2 + 1 = 8^2 + 1 = 64 + 1 = 65\).
Required triplet = (16, 63, 65).

(iv) One member = 18
Let \(2m = 18 \Rightarrow m = \frac{18}{2} = 9\).
Now, \(2m = 2 \times 9 = 18\),
\(m^2 - 1 = 9^2 - 1 = 81 - 1 = 80\),
\(m^2 + 1 = 9^2 + 1 = 81 + 1 = 82\).

Required triplet = (18, 80, 82).