RBSE Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.2

Rajasthan Board RBSE Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.2 Textbook Exercise Questions and Answers.

RBSE Class 8 Maths Solutions Chapter 6 Square and Square Roots Ex 6.2

Question 1.
Find the squares of the following numbers
(i) 32
Answer:
32² = (30 + 2)²
= 30(30 + 2) + 2(30 + 2)
= 30² + 30 × 2 + 2 × 30 + 2²
= 900 + 60 + 60 + 4 = 1024

(ii) 35
Answer:
35² = (30 + 5)²
= 30(30 + 5) + 5(30 + 5)
= 30² + 30 × 5 + 5 × 30 + 5²
= 900 + 150 + 150 + 25 = 1225

(iii) 86
Answer:
86² = (80 + 6)²
= 80(80 + 6) + 6(80 + 6)
= 80² + 80 × 6 + 6 × 80 + 6²
= 6400 + 480 + 480 + 36
= 7396

(iv) 93
Answer:
93² = (90 + 3)²
= 90(90 + 3) + 3(90 + 3)
= 90² + 90 × 3 + 3 × 90 + 3²
= 8100 + 270 + 270 + 9 = 8649

(v) 71
Answer:
71² = (70 + 1)²
= 70(70 + 1) + 1(70 + 1)
= 70² + 70 × 1 + 1 × 70 + 1²
= 4900 + 70 + 70 + 1
= 5041

(vi) 46
Answer:
46² = (40 + 6)²
= 40(40 + 6) + 6(40 + 6)
= 40² + 40 × 6 + 6 × 40 + 6²
= 1600 + 240 + 240 + 36
= 2116

Question 2.
Write a Pythagorean triplet whose one member is.
(i) 6
Answer:
We have Pythagorean triplet (2m, m² - 1, m² + 1);
where m > 1 (natural number)
when m = 3, 2m = 6, m² - 1 = 3² - 1 = 8,
and m² + 1 = 3² + 1 = 10
Thus 6, 8 and 10 are Pythagorean triplets.

(ii) 14
Answer:
when m = 7, 2m = 2 × 7 = 14
m² - 1 = 7² - 1 = 49 - 1 = 48
and m² + 1 = 1² + 1 = 49 + 1 = 50
Thus 14, 48 and 50 are Pythagorean triplets.

(iii) 16
Answer:
when m = 8, 2m = 2 × 8 = 16
m² - 1 = 8² - 1 = 64 - 1 = 63
and m² + 1 = 8² + 1 = 64 + 1 = 65
Thus 16, 63 and 65 are Pythagorean triplets.

(iv) 18
Answer:
when m = 9, 2m = 2 × 9 = 18
m² - 1 = 9² - 1 = 81 - 1 = 80
and m² + 1 = 9² + 1 = 81 + 1 = 82
Thus 18, 80 and 82 are Pythagorean triplets.